Line 29:
$sql2="UPDATE inventory SET inv_status="assigned" where inv_tag='$inv_tag' ";
will give a php error - it should read either
$sql2="UPDATE inventory SET inv_status='assigned' where inv_tag='$inv_tag' ";
or
$sql2="UPDATE inventory SET inv_status=\"assigned\" where inv_tag='$inv_tag' ";
to be syntactically correct in php.
Balaji Ankem schrieb am Samstag, 28. Juli 2001, 10:41:21:
> Hi dearest friend,
> can we update the same row immediately after selection.
> please see the following code at line number 29 it is giving error.
> code
> ------------------------------------
> <?php
> // Connect to MySQL
> $connection = mysql_connect( 'localhost', 'balaji', 'pingpong' )
> or die ( 'Unable to connect to server.' );
> // Select database on MySQL server
> mysql_select_db( 'imac' )
> or die ( 'Unable to select database.' );
> // Formulate the query (inv_tag will come by post method
> $sql1 = "SELECT * FROM inventory WHERE inv_tag = '$inv_tag'";
> // Execute the query and put results in $result
> $result1 = mysql_query( $sql1 )
> or die ( 'Unable to select the data from inventory.' );
> if ( $result1 )
> {
> //change the status of the inventory item in inventory table
> $sql2="UPDATE inventory SET inv_status="assigned" where inv_tag='$inv_tag'
>"; // This is line number 29.
> $result2 = mysql_query( $sql2 )
> or die ( 'Unable to execute query for upadation of inventory
>table.' );
> $sql3 = "insert into status
>values('$inv_name','$inv_type','$inv_tag','$issued_to','$issued_by','$issue_date')";
> $result3 = mysql_query( $sql3 )
> or die ( 'Unable to execute query for insertion in to status
>table.' );
> }
> else
> {
> // header( 'WWW-Authenticate: Basic realm="Private"' );
> // header( 'HTTP/1.0 401 Unauthorized' );
> echo '$inv_tag does not exist';
> exit;
> }
> ?>
> please inform if any errors.
> Thanks in advance.
> Regards
> -Balaji
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