ID: 35127
User updated by: sketchdude at gmail dot com
Reported By: sketchdude at gmail dot com
-Status: Open
+Status: Bogus
Bug Type: MySQL related
Operating System: WindowsXP, Linux
PHP Version: 4.4.1
New Comment:
My own bad variable. Sorry for wasting your time.
Previous Comments:
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[2005-11-06 16:12:08] sketchdude at gmail dot com
Description:
------------
The cat_id fields fails to update after an insert and select query.
(see sql example below)
/*
use thisexample;
CREATE TABLE example (
example_id int unsigned not null auto_increment,
cat_id int unsigned not null,
name varchar(35) not null,
total tinyint unsigned not null,
PRIMARY KEY (example_id)
);
INSERT INTO example VALUES (1,1,'One',1);
INSERT INTO example VALUES (2,2,'Two',0);
INSERT INTO example VALUES (3,3,'Three',0);
INSERT INTO example VALUES (4,4,'Four',0);
*/
Reproduce code:
---------------
$name = 'Five';
$query = "INSERT INTO example (example_id, cat_id, name, total) VALUES
('', 0, '$name', 0)";
if (!$result = mysql_query($query)) {
print "Insert query failed ($query) : " . mysql_error();
exit;
}
$query = "SELECT example_id AS new_id FROM example WHERE name
='$name'";
if (!$result = mysql_query($query)) {
print "Select query failed ($query) : " . mysql_error();exit;
}
else {
$new_id = mysql_fetch_row($result);
$query = "UPDATE example SET cat_id = '$new_id[0]' WHERE example_id =
'$name'";
if (!$result = mysql_query($query)) {
print "Update query failed ($query) : " . mysql_error(); exit;
}
else {
print "Script ran successfully! New cat_id = " . $new_id[0];
}
}
Expected result:
----------------
The problem is the third query: cat_id field should be updated to 5, or
else the script should die and print out the query.
Actual result:
--------------
The script executes successfully, but the cat_id field remains at 0.
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Edit this bug report at http://bugs.php.net/?id=35127&edit=1
