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PHP MSSQL > Updating DB entries

Asked by getinked

at 2024-02-16 14:18:36

Point:500 Replies:32 POST_ID:828885USER_ID:11789

Topic:

PHP Scripting Language;Microsoft IIS Web Server;MS SQL Server

Morning Experts,
I have the following script attempting to update fields in my MSSQL table.

<?php$Shift_Start 	= $_POST['Shift_Start'];$Shift_End 	= $_POST['Shift_End'];$Lunch_Start 	= $_POST['Lunch_Start'];$Lunch_End	= $_POST['Lunch_End'];$Agent_ID	= $_POST['Agent_ID'];$Update		= array($Shift_Start,$Shift_End,&$Agent_ID);$serverName = "---";$connectionInfo = array( "Database"=>"---", "UID"=>"---", "PWD"=>"---");$conn = sqlsrv_connect( $serverName, $connectionInfo);if( $conn === false) {    die( print_r( sqlsrv_errors(), true));}$sql = "UPDATE dbo.Roster        SET 		Shift_Start = ?,			Shift_End = ?,        WHERE    Agent_ID = ?";$stmt = sqlsrv_prepare( $conn, $sql, $Update);if( !$stmt ) {    die( print_r( sqlsrv_errors(), true));}    if( sqlsrv_execute( $stmt ) === false ) {          die( print_r( sqlsrv_errors(), true));    }?>

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This works fine when I give the array information to it directly (rather than using the $_POST) but when I try to use $_POST to deliver the information from the previous page it gives me the following error:

Array ( [0] =>" Array ( [0] =>" IMSSP [SQLSTATE] =>" IMSSP [1] =>" -15 'code' =>" -15 [2] =>" An invalid direction for parameter 1 was specified. SQLSRV_PARAM_IN, SQLSRV_PARAM_OUT, and SQLSRV_PARAM_INOUT are valid values. [message] =>" An invalid direction for parameter 1 was specified. SQLSRV_PARAM_IN, SQLSRV_PARAM_OUT, and SQLSRV_PARAM_INOUT are valid values. ) )

My form is giving time strings as per the following:

<"td>"<"input name='Shift_Start' type='text' id='Shift_Start' value='".($row['Shift_Start']->"format('H:i:s'))."'>"<"/td>"<"td>"<"input name='Shift_End' type='text' id='Shift_End' value='".($row['Shift_End']->"format('H:i:s'))."'>"<"/td>"<"td>"<"input name='Lunch_Start' type='text' id='Lunch_Start' value='".($row['Lunch_Start']->"format('H:i:s'))."'>"<"/td>"<"td>"<"input name='Lunch_End' type='text' id='Lunch_End' value='".($row['Lunch_End']->"format('H:i:s'))."'>"<"/td>"

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And my DB table is as follows:

DB.JPG

20 KB
Snapshot of DB Table

Hope I have provided enough information, but if not please feel free to ask!
Thanks in advance
Luke

Expert: Dave Baldwin replied at 2024-02-17 16:56:04

Yes it is!! Glad to help!

Expert: Chris Stanyon replied at 2024-02-17 16:06:56

Now that's one happy coder :)

Author: getinked replied at 2024-02-17 16:02:07

Thank you to everyone that assisted in this, Especially Dave for your tremendous assistance and Chris for the Solution!

Author: getinked replied at 2024-02-17 15:58:57

YEEEEEHAAAAA

Awesome! Thank you to both Chris and Dave!
It is working!

Accepted Solution

Expert: Chris Stanyon replied at 2024-02-17 15:36:07

400 points EXCELLENT

Your HTML is naming the fields incorrectly. What we need is: details[agentID][fieldName]

What you have is details[fieldName]agentID

Change your code to this:

<td><input name='details[". $row['Agent_ID']. "][Shift_Start]' type='text' value='" . $row['Shift_Start']. "'></td><td><input name='details[". $row['Agent_ID']. "][Shift_End]' type='text' id='Shift_End' value='" . $row['Shift_End']. "'></td><td><input name='details[". $row['Agent_ID']. "][Lunch_Start]' type='text' id='Lunch_Start' value='" . $row['Lunch_Start']. "'></td><td><input name='details[". $row['Agent_ID']. "][Lunch_End]' type='text' id='Lunch_End' value='" . $row['Lunch_End']. "'></td>

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You're also including a form on every row of the table - if you want to handle all rows at the same time, then you only need one form - put the opening form tag before your table and the closing form tag after the table.

I'm not overly familiar with the MS SQL library, but normally (at least when using mySQL/PDO), you only need to prepare the SQL statement once. You then execute that prepared statement inside your loop as many times as you need. While your script won't 'break' it is preparing the query over and over again, probably unnecessarily.

Make those changes and give it another whirl

Author: getinked replied at 2024-02-17 15:21:24

The Form

<html><head><style>body {font-family:Verdana;font-size:10px;}table {border-collapse:collapse;}th {font-family:Verdana;font-size:10px;text-align:right;padding: 3px;background-color:#3396FF;color:#ffffff;border:1px solid #3396FF;}td {font-family:Verdana;font-size:10px;text-align:right;border-width:thin;border:1px solid #B3B3B3;padding:2px 8px 2px 2px;}</style></head><body ><!--<div id="loading" style="position:absolute; width:100%; text-align:center; top:50px;"><img src="http://ndssl.vbi.vt.edu/gis/dbv/v2.0/images/loading_circle_fully_working.gif" width"50" border="0"></div>--><?phpset_time_limit(10);//Start Tableecho "<p>&nbsp;</p><table id='update' align='center' width='950'>";$server = "---";$connectionInfo = array( "Database"=>"---", "UID"=>"---", "PWD"=>"---" );$conn = sqlsrv_connect( $server, $connectionInfo );$sql = "SELECT * FROM dbo.Roster";$params = array();$options =  array( "Scrollable" => SQLSRV_CURSOR_KEYSET );$stmt = sqlsrv_query( $conn, $sql , $params, $options );$rowC = sqlsrv_num_rows($stmt);echo $rowC;while( $row = sqlsrv_fetch_array( $stmt) ) {echo"<tr><form name='details' action='roster_update_roster.php' method='post'><td>".$row['Agent_Name']."</td><td><input name='details[Shift_Start]". $row['Agent_ID']. "' type='text' value='" . $row['Shift_Start']. "'></td><td><input name='details[Shift_End]". $row['Agent_ID']. "' type='text' id='Shift_End' value='" . $row['Shift_End']. "'></td><td><input name='details[Lunch_Start]". $row['Agent_ID']. "' type='text' id='Lunch_Start' value='" . $row['Lunch_Start']. "'></td><td><input name='details[Lunch_End]". $row['Agent_ID']. "' type='text' id='Lunch_End' value='" . $row['Lunch_End']. "'></td></tr>";}echo "<tr><td colspan='5'><input type='submit' name='Submit' value='Submit'></td></tr></table>";?>

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The Process

<?phpset_time_limit(15);$serverName = "---";$connectionInfo = array( "Database"=>"---", "UID"=>"---", "PWD"=>"---");$conn = sqlsrv_connect( $serverName, $connectionInfo);if( $conn === false) {    die( print_r( sqlsrv_errors(), true));}//var_dump($_POST['details']);$sql = "UPDATE dbo.Roster        SET			Shift_Start = ( ?),			Shift_End = ( ?),			Lunch_Start = ( ?),			Lunch_End = ( ?)		WHERE Agent_ID = ( ?)";foreach ($_POST['details'] as $agentID => $info):   echo $agentID;   echo $info['Shift_Start'];   echo $info['Shift_End'];   echo $info['Lunch_Start'];   echo $info['Lunch_End'];$stmt = sqlsrv_prepare( $conn, $sql, array(&$info['Shift_Start'],&$info['Shift_End'],&$info['Lunch_Start'],&$info['Lunch_End'],&$agentID));	if( !$stmt ) {    die( print_r( sqlsrv_errors(), true));}( sqlsrv_execute( $stmt ));    if( sqlsrv_execute( $stmt ) === false ) {          die( print_r( sqlsrv_errors(), true));    }endforeach;?>

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Expert: Chris Stanyon replied at 2024-02-17 15:00:59

Generally in programming, something either works or it doesn't. It doesn't work twice and then fail, unless changes were made to your code after the second run :)

I have no idea what lines 22-27 of your code are!! Can you post your roster_update_roster.php script, along with the part of your HTML code that generates the form fields so we can make sure they tie up.

At least we're making progress :)

Author: getinked replied at 2024-02-17 14:48:37

Thanks Chris, It worked... Twice, Then it died...
I am getting this error:

Shift_Start
Warning: Illegal string offset 'Shift_Start' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 22
5
Warning: Illegal string offset 'Shift_End' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 23
5
Warning: Illegal string offset 'Lunch_Start' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 24
5
Warning: Illegal string offset 'Lunch_End' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 25
5
Warning: Illegal string offset 'Shift_Start' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 27

Fatal error: Cannot create references to/from string offsets in C:InetpubwwwrootDevelopment oster_update_roster.php on line 27
PHP Warning: Illegal string offset 'Shift_Start' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 22 PHP Warning: Illegal string offset 'Shift_End' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 23 PHP Warning: Illegal string offset 'Lunch_Start' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 24 PHP Warning: Illegal string offset 'Lunch_End' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 25 PHP Warning: Illegal string offset 'Shift_Start' in C:InetpubwwwrootDevelopment oster_update_roster.php on line 27 PHP Fatal error: Cannot create references to/from string offsets in C:InetpubwwwrootDevelopment oster_update_roster.php on line 27

We are so close!

Expert: Chris Stanyon replied at 2024-02-17 04:27:11

When you're submitting rows of form field like this, you can make your life a whole lot easier by naming your form fields correctly. This means using an array for the names, and the easiest way to loop through your records in a PHP script is to name then something like this:

details[agentID][Shift_Start]
details[agentID][Shift_End]
details[agentID][Lunch_Start]
details[agentID][Lunch_End]

In your form building script, use something like this:

<tr><td><input name='details[" . $row['Agent_ID'] . "][Shift_Start]' type='text' value='" . $row['Shift_Start']->format('H:i:s') . "'></td><td><input name='details[" . $row['Agent_ID'] . "][Shift_End]' type='text' id='Shift_End' value='" . $row['Shift_End']->format('H:i:s') . "'></td><td><input name='details[" . $row['Agent_ID'] . "][Lunch_Start]' type='text' id='Lunch_Start' value='" . $row['Lunch_Start']->format('H:i:s') . "'></td><td><input name='details[" . $row['Agent_ID'] . "][Lunch_End]' type='text' id='Lunch_End' value='" . $row['Lunch_End']->format('H:i:s') . "'></td></tr>

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Then when the whole form is posted, you have a POST key called 'details' that you can loop through:

foreach ($_POST['details'] as $agentID => $info):   echo $agentID;   echo $info['Shift_Start'];   echo $info['Shift_End'];   echo $info['Lunch_Start'];   echo $info['Lunch_End'];   //execute your Update in here :)endforeach;

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Expert: Dave Baldwin replied at 2024-02-16 21:41:35

Your 'sqlsrv_prepare' statement needs to be inside the loop but the $sql doesn't. That's because the 'sqlsrv_prepare' function tries to 'bind' the variable to the SQL statement variables. That not what you want here because the variables for the different rows are not in the 'same place' that the '&' pass-by-reference wants them to be.

Author: getinked replied at 2024-02-16 20:14:04

Hi Guys,
So it looks like I have worked it out almost.

the following is working for the first record in the array:

<?php$Shift_Start 	=($_POST['Shift_Start']);$Shift_End 		=($_POST['Shift_End']);$Lunch_Start 	=($_POST['Lunch_Start']);$Lunch_End		=($_POST['Lunch_End']);$Agent_ID		=($_POST['Agent_ID']);$Update			= array($Shift_Start[0], $Shift_End[0], $Lunch_Start[0], $Lunch_End[0]);$serverName = "---";$connectionInfo = array( "Database"=>"---", "UID"=>"---", "PWD"=>"---");$conn = sqlsrv_connect( $serverName, $connectionInfo);if( $conn === false) {    die( print_r( sqlsrv_errors(), true));}$sql = "UPDATE dbo.Roster        SET			Shift_Start = ( ?),			Shift_End = ( ?),			Lunch_Start = ( ?),			Lunch_End = ( ?)";$stmt = sqlsrv_prepare( $conn, $sql, array(&$Update[0],&$Update[1],&$Update[2],&$Update[3]));	if( !$stmt ) {    die( print_r( sqlsrv_errors(), true));}( sqlsrv_execute( $stmt ));    if( sqlsrv_execute( $stmt ) === false ) {          die( print_r( sqlsrv_errors(), true));    }?>

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What I need to do now is get the update to loop through the available results.
I think i need to get the following to increment by one for every result (or row counted), what are your thoughts?

$Update = array($Shift_Start[0], $Shift_End[0], $Lunch_Start[0], $Lunch_End[0]);

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Author: getinked replied at 2024-02-16 19:50:47

Thanks Dave, I do want all of the rows to appear, not sure this is going to resolve my issue.

Assisted Solution

Expert: Dave Baldwin replied at 2024-02-16 19:37:26

100 points EXCELLENT

You're getting all of the data in part because every form has the same name. And since each item is an array, you're getting an error message. If you used 'foreach' to separate them into 'sets' of variables, you could do a loop that would update each row that way.

If you used a counter ($formcnt) and added it to the name of the form, then you should get only the row with that submit button instead of all of them.

$formcnt++;echo"<tr><form name='update$formcnt' action='roster_update_roster.php' method='post'>

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There are other things you can do to separate the rows.

Author: getinked replied at 2024-02-16 18:24:22

This may make more sense...
When I click on Submit, I expect only the 1 set of results to flow to the next page
ie; if i Select the first row, only the changes for "Luke Chilcott" should come through but as you can see from the var_dump I am getting all fields from all agents.
*note* It would be AWESOME if I could have a single submit to change all fields for all agent, in case that is easier.

Here is the Screen shot I have put together to show the issue in more detail...

issue.JPG

121 KB
The Issue

Author: getinked replied at 2024-02-16 18:02:25

array(5) { [0]=> string(8) "09:00:00" [1]=> string(8) "08:30:00" [2]=> string(8) "08:30:00" [3]=> string(8) "08:30:00" [4]=> string(8) "08:30:00" } array(5) { [0]=> string(8) "17:00:00" [1]=> string(8) "17:00:00" [2]=> string(8) "17:00:00" [3]=> string(8) "17:00:00" [4]=> string(8) "17:00:00" } array(5) { [0]=> string(8) "12:30:00" [1]=> string(8) "12:30:00" [2]=> string(8) "12:30:00" [3]=> string(8) "12:30:00" [4]=> string(8) "12:30:00" } array(5) { [0]=> string(8) "13:29:00" [1]=> string(8) "13:29:00" [2]=> string(8) "13:29:00" [3]=> string(8) "13:29:00" [4]=> string(8) "13:29:00" } array(5) { [0]=> string(1) "1" [1]=> string(1) "2" [2]=> string(1) "3" [3]=> string(1) "4" [4]=> string(1) "5" }

Expert: Dave Baldwin replied at 2024-02-16 17:53:04

Show me the dump. If, for example, $_POST['Shift_Start'] was actually an array itself, then I can see where it would cause a problem.

Author: getinked replied at 2024-02-16 17:45:54

Hi guys, I think I have found the issue (or part there of...)
I added "Var_Dump" to my roster_update_roster.php and found that I am getting mutliple results from the form page...

Any ideas?

Author: getinked replied at 2024-02-16 17:30:22

Do you need something more from me?

Expert: Dave Baldwin replied at 2024-02-16 17:27:03

That tells me you're using SQL Server 2008 (32-bit) and the SQL Native client that came with it. Nothing else yet.

Author: getinked replied at 2024-02-16 16:06:10

Dave:
PHP Version 5.4.9

Windows NT --- 5.2 build 3790
(Windows Server 2003 R2 Standard Edition Service Pack 2) i586

SQLServerVersion: 10.00.1600

Does that help?

Expert: Dave Baldwin replied at 2024-02-16 15:58:13

What Windows version? What PHP version? SQLSRV2 or SQLSRV3? I am using SQL Server Native Client 10.0 also.

Author: getinked replied at 2024-02-16 15:52:10

Dave

Yep here is is, just tried it again to be sure:

$stmt = sqlsrv_prepare( $conn, $sql, array(&$Shift_Start, &$Shift_End, &$Agent_ID));	if( !$stmt ) {    die( print_r( sqlsrv_errors(), true));

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Error:

Warning: Variable parameter 1 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value. For more information, see sqlsrv_prepare or sqlsrv_query in the API Reference section of the product documentation. in C:InetpubwwwrootDevelopment oster_update_roster.php on line 25

Warning: Variable parameter 2 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value. For more information, see sqlsrv_prepare or sqlsrv_query in the API Reference section of the product documentation. in C:InetpubwwwrootDevelopment oster_update_roster.php on line 25
Array ( [0] => Array ( [0] => 07002 [SQLSTATE] => 07002 [1] => 0 "code" => 0 [2] => [Microsoft][SQL Server Native Client 10.0]COUNT field incorrect or syntax error [message] => [Microsoft][SQL Server Native Client 10.0]COUNT field incorrect or syntax error ) ) PHP Warning: Variable parameter 1 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value. For more information, see sqlsrv_prepare or sqlsrv_query in the API Reference section of the product documentation. in C:InetpubwwwrootDevelopment oster_update_roster.php on line 25 PHP Warning: Variable parameter 2 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value. For more information, see sqlsrv_prepare or sqlsrv_query in the API Reference section of the product documentation. in C:InetpubwwwrootDevelopment oster_update_roster.php on line 25

Expert: Dave Baldwin replied at 2024-02-16 15:47:34

Are you sure you copied my post? Because it does not create an error on this machine. Note the '&' in front of the variable names. The '&' is necessary because this function 'binds' the SQL variables to the PHP variables. That means that if the value of the PHP variables changes, those changes are 'automatically' picked up the SQL execute statement (for the duration of this script).

$stmt = sqlsrv_prepare( $conn, $sql, array(&$Shift_Start, &$Shift_End, &$Agent_ID));

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Expert: duncanb7 replied at 2024-02-16 15:43:07

Digesting it.sorry about that At the beginning I thought the form is from html code . now it
is shown on php code that is why it works for value='".($row['Agent_ID'])."' on php code
not html and not javascript code.

Duncan

Author: getinked replied at 2024-02-16 15:37:54

Duncan

This is the Form:

<?phpset_time_limit(10);//Start Tableecho "<p>&nbsp;</p><table id='update' align='center' width='950'>";$server = "---";$connectionInfo = array( "Database"=>"---", "UID"=>"---", "PWD"=>"---" );$conn = sqlsrv_connect( $server, $connectionInfo );$sql = "SELECT * FROM dbo.Roster";$params = array();$options =  array( "Scrollable" => SQLSRV_CURSOR_KEYSET );$stmt = sqlsrv_query( $conn, $sql , $params, $options );$row_count = sqlsrv_num_rows( $stmt );while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {echo"<tr><form name='update' action='roster_update_roster.php' method='post'><td><input name='Agent_ID' type='text' id='Agent_ID' readonly value='".($row['Agent_ID'])."'></td><td>".$row['Agent_Name']."</td><td><input name='Shift_Start' type='text' id='Shift_Start' value='".($row['Shift_Start']->format('H:i:s'))."'></td><td><input name='Shift_End' type='text' id='Shift_End' value='".($row['Shift_End']->format('H:i:s'))."'></td><td><input name='Lunch_Start' type='text' id='Lunch_Start' value='".($row['Lunch_Start']->format('H:i:s'))."'></td><td><input name='Lunch_End' type='text' id='Lunch_End' value='".($row['Lunch_End']->format('H:i:s'))."'></td><td><input type='submit' name='Submit' value='Submit'></td></tr>";}echo "</tr></table>";?>

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This is the "roster_update_roster.php" processing file

<?php$Shift_Start 	= $_POST['Shift_Start'];$Shift_End 	= $_POST['Shift_End'];$Lunch_Start 	= $_POST['Lunch_Start'];$Lunch_End	= $_POST['Lunch_End'];$Agent_ID	= $_POST['Agent_ID'];$Update		= array(&$Shift_Start,&$Shift_End,&$Agent_ID);$serverName = "---";$connectionInfo = array( "Database"=>"---", "UID"=>"---", "PWD"=>"---");$conn = sqlsrv_connect( $serverName, $connectionInfo);if( $conn === false) {    die( print_r( sqlsrv_errors(), true));}$sql = "UPDATE dbo.Roster        SET			Shift_Start = ?,			Shift_End = ?        WHERE Agent_ID = ?";$stmt = sqlsrv_prepare( $conn, $sql, array( $Shift_Start, $Shift_End, &$Agent_ID));	if( !$stmt ) {    die( print_r( sqlsrv_errors(), true));}    if( sqlsrv_execute( $stmt ) === false ) {          die( print_r( sqlsrv_errors(), true));    }?>

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Author: getinked replied at 2024-02-16 15:34:59

Dave
Updated as per your post, Error as follows:
Warning: Variable parameter 1 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value. For more information, see sqlsrv_prepare or sqlsrv_query in the API Reference section of the product documentation. in C:InetpubwwwrootDevelopment oster_update_roster.php on line 23

Warning: Variable parameter 2 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value. For more information, see sqlsrv_prepare or sqlsrv_query in the API Reference section of the product documentation. in C:InetpubwwwrootDevelopment oster_update_roster.php on line 23

Expert: duncanb7 replied at 2024-02-16 15:30:42

Thanks for your reply.

What is Agent_ID's value in your form input tag that doesn't show in your post ?
Could you talk more information of Agent_ID ?

And at input tag
<input name='Shift_Start' type='text' id='Shift_Start' value='".($row['Shift_Start']->format('H:i:s'))."'>

($row['Shift_Start']->format('H:i:s')) from which how it works ? could you post your format()
function code to us ?

please check whether The number OR string type of variable parameter in $sql is not matched to $update or not ?

If send us the full html and php code(not just partial) , that will be easier for us

Duncan

Expert: Dave Baldwin replied at 2024-02-16 15:30:05

Try using it this way and let me know what happens.

$stmt = sqlsrv_prepare( $conn, $sql, array( &$Shift_Start, &$Shift_End, &$Agent_ID));

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Author: getinked replied at 2024-02-16 15:25:09

Thanks Dave, I also fount that very confusing!
I had been over that recourse several times with no joy.

As I mentioned, I know that the array is fine, and the update is correct, For some reason though, When i try to $_POST the data for the update it falls over!

Author: getinked replied at 2024-02-16 15:23:41

Thanks Duncan, User is submitting as 18:06:03 and this works if i don't transmit the output with $_POST.

The reason for the &$Agent_ID is because this is the only way to pass the ID, not sure why, but in the array this is the only way it is accepted, when I remove the & i get the following error:
Variable parameter 3 not passed by reference (prefaced with an &). Variable parameters passed to sqlsrv_prepare or sqlsrv_query should be passed by reference, not by value.

Expert: duncanb7 replied at 2024-02-16 14:58:12

Dear getinked,

What is format that user input Shift_Start, Shift_End,Lunch_Start,Lunch_End, for example,
they will input 18:06:03 at input tag, Right ? Or in time stamp format ?

You better do javascript time format convention before submit the form to yourform.com,

If the input is in time stamp format , you need javascript function of toHHMMSS to convert the input value to H:i:s format. I mean convert it using format() and then submit
the form to http://your form.com, for just example only and not tested as follows

and please read the link for onsubmit at http://www.w3schools.com/jsref/tryit.asp?filename=tryjsref_form_onsubmit

<form method="post" action="http://yourform.com" onsubmit="format();">......</form>

function format(){
$("#Start_Start).val($("#Start_Start).val().toHHMMSS());
$("#Start_End).val($("#Start_End).val().toHHMMSS());
$("#Lunch_Start).val($("#Lunch_Start).val().toHHMMSS());
$("#Lunch_End).val($("#Lunch_End).val().toHHMMSS());

String.prototype.toHHMMSS = function () {
var sec_num = parseInt(this, 10); // don't forget the second param
var hours = Math.floor(sec_num / 3600);
var minutes = Math.floor((sec_num - (hours * 3600)) / 60);
var seconds = sec_num - (hours * 3600) - (minutes * 60);

if (hours < 10) {hours = "0"+hours;}
if (minutes < 10) {minutes = "0"+minutes;}
if (seconds < 10) {seconds = "0"+seconds;}
var time = hours+':'+minutes+':'+seconds;
return time;
}
}

Or if what I said that is not correct above, it might be whether it is related
to sqlsrv_prepare( $conn, $sql, $Update) ?

The number OR string type of variable parameter in $sql is not matched to $update ?
and what is "&" in &Agent_ID in array of $Update=array($Shift_Start,$Shift_End,&$Agent_ID); ?

Hope understand your question completely, if not please pt it out .
Give us more detail that will help on it

Duncan

Expert: Dave Baldwin replied at 2024-02-16 14:49:50

The problem is with passing an array as the data for the sqlsrv_prepare function. This page http://us1.php.net/manual/en/function.sqlsrv-prepare.php shows an array structure but frankly, I find it confusing when compared to their example.

&nbps;

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