note 117807 added to function.sqlsrv-prepare
| From: | tuxedobob | Date: | Wed, 12 Aug 2015 20:54:05 +0000 |
| Subject: | note 117807 added to function.sqlsrv-prepare | ||
| Groups: | php.notes | ||
| Request: | Send a blank email to php-notes+get-203324@lists.php.net to get a copy of this message | ||
Be careful with your variables once you bind them to a statement with sqlsrv_prepare.
Consider the following:
<?php
$dude = '';
$time = new DateTime();
$sql = "INSERT INTO my_table (person, timein) VALUES (?, ?)";
$stmt = sqlsrv_prepare($conn, $sql, array(&$dude, &$time));
...
// many lines later
foreach ($times as &$time) {
//do stuff
}
// later still...
$time = $times['start'];
if( sqlsrv_execute( $stmt ) === false ) {
die( print_r( sqlsrv_errors(), true));
}
?>
I did something like this. I prepared a statement at the start, used the variable again in the
middle, and then set the value I wanted before running the query.
Trouble is, I used the variable as an iterator instead of a simple scalar. This caused PHP to use a
different location in memory, and the location it was previously bound to was invalid. So SQL simply
inserted a default date/time.
Worse, because SQL just inserted a default, it didn't throw any errors, and in trying to debug
it, I did something like this:
<?php
var_dump($time);
sqlsrv_execute($stmt);
$q = "SELECT * FROM my_table WHERE id=@@IDENTITY";
$r = sqlsrv_query($conn, $q);
$row = sqlsrv_fetch_array($r); $id = $row[0];
var_dump($row['time']);
?>
Having it appear as though you're sending SQL the correct data, and seeing it spitting back
something else entirely is absolutely maddening.
So if SQL seems to be inserting garbage with prepared statements, MAKE SURE YOU'RE NOT USING
THE VARIABLES ANYWHERE ELSE.
----
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Probable Submitter: 96.233.109.34
----
Manual Page -- http://php.net/manual/en/function.sqlsrv-prepare.php
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