ID:               29283
 Comment by:       andreas dot perhab at algorithms dot at
 Reported By:      divisor at ad69 dot com
 Status:           Closed
 Bug Type:         MySQL related
 Operating System: FreeBSD 4.10
 PHP Version:      5.0.0
 Assigned To:      georg
 New Comment:

It seems that i have the sam problem with ibase.

my code is:
$datei=fopen($BILD_NEU["tmp_name"], "r");
$ibTrans = interbase_sql_trans();
$BLOB = ibase_blob_import($ibTrans,$datei);
$sql_query = "UPDATE ARTIKEL SET BILD1 = ? WHERE ARTIKELNR = $ANR";
$sth = ibase_prepare($ibTrans,$sql_query);
$ibRid = ibase_execute($sth,$BLOB);
ibase_commit($ibTrans);

and at the end of my file i get the error:
Warning: Unknown: invalid statement handle in Unknown on line 0

the code works only the execute  doesn't return a result identifier it
returns 1

please mail me if you answer my message


Previous Comments:
------------------------------------------------------------------------

[2004-09-02 06:53:37] [EMAIL PROTECTED]

This bug has been fixed in CVS.

Snapshots of the sources are packaged every three hours; this change
will be in the next snapshot. You can grab the snapshot at
http://snaps.php.net/.
 
Thank you for the report, and for helping us make PHP better.



------------------------------------------------------------------------

[2004-09-01 22:18:12] corey_gilmore at dpc dot senate dot gov

I'm experiencing the same issue as well with Apache 1.3.31 and 2.0.50
on XP SP2 and XP SP1 with PHP 5.0, 5.01 and 5.0-DEV from
php5-win32-200408312230.zip.  MySQL 4.1a, 4.1.2, 4.1.3 and now 4.1.4. 
It occurs in a page where I select a large amount of data from one
table, process it, and insert it into multiple tables.  There are 287
rows returned from the first select, and then 400-500 rows inserted.

Line 644 begins with 'if( !$stmt->execute() ) {'

Warning: Statement isn't valid anymore in
...\import\contact.class.inc.php on line 644
Error inserting email (#0): INSERT INTO email (EmailType, Email,
ContactID) VALUES (?, ?, ?);

$sql = "INSERT INTO {$kTABLE_PREFIX}email (EmailType, Email, ContactID)
VALUES (?, ?, ?)";
if( !$stmt = $db->prepare($sql) ) { // prepare it
        printf("Error with %s:<br>%s<br>",$sql, $db->error );
}
if( !$stmt->bind_param("ssi", $addr->EmailType, $addr->Email,
$ContactID ) ) {
        printf("Error (#%d): %s;<br>%s<br>%s : %s : %s<br>", $stmt->errno,
$sql, $stmt->error, $addr->EmailType, $addr->Email, $ContactID );
}
 
if( !$stmt->execute() ) {
        printf("Error inserting email (#%d): %s;<br>%s<br>%s : %s : %i<br>",
$stmt->errno, $sql, $stmt->error, $addr->EmailType, $addr->Email,
$ContactID );
        $this->dump();
        printf("\$stmt = <pre>%s</pre><br>", var_export($stmt, TRUE) );
}

This has been an issue since RC1, and is a bit frustrating.  In my case
reloading the page to re-import works, eventually, but since the entire
process takes about 5 minutes it is painful.

------------------------------------------------------------------------

[2004-08-20 14:47:35] paradox at gmx dot at

i've got the same problem with php 5.0.1 loaded as an apache module -
mysql 4.1.3beta, win2k
it seems that the error occurs more often on much used servers. on my
dedicated test-server it is approximately one error on 100 statements.

------------------------------------------------------------------------

[2004-08-03 08:48:09] alex at avannaproductions dot com

I'll chime in as another voice of reason: this should be 
fixed before 5.0.1. At the moment, prepared statements 
are basically useless, as half the time they give this 
error and fail to return any data. I'm on MacOS X and 
Apache 2, with the same problem.

------------------------------------------------------------------------

[2004-07-25 18:33:46] dev at edwinchu dot info

Hi,

I have got the same problem. I am using PHP5 release with
MySQL4.1.3-beta. 

The code is simple:

$mysqli = new mysqli(); // connected successfully
$stmt = $mysqli->prepare("SOME VALID QUERY");
$stmt->execute(); // No problem here

$mysqli2 = $mysqli;
$mysqli2->query("THE SAME QUERY"); // Still OK
$stmt2 = $mysqli2->prepare("THE SAME QUERY");
$stmt2->execute();

The last line failed and returning "Warning: Statement isn't valid
anymore in xxx".

------------------------------------------------------------------------

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the rest of the comments, please view the bug report online at
    http://bugs.php.net/29283

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