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Views: 820 | Replies: 16 | Solved
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Join Date: May 2008
Posts: 23
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Newbie Poster
i am extemly newbie and don’t know whether I should post my problem here or somewhere else, however…
i have following problem in one of my php-mysql application
header link through browser is,
"http://localhost/test/reply.php?id=852"
and query is as follow,
The application does not pass any value related to $q_id in database table, the table view is as follow, The query does not insert any value to tid column except default value. What should I do…
The table specification is as follow,
I have tried every method, I know but in vain.
Please any body helps me….
Shuja
i have following problem in one of my php-mysql application
header link through browser is,
"http://localhost/test/reply.php?id=852"
and query is as follow,
$name=$_POST['name']; //from the name form $q_id=$_GET['id']; $sql="INSERT INTO `test`.`reply`(`rid`,`tid`,`name`)VALUES(NULL,'$q_id','$name')"; $result=mysql_query($sql);
rid tid name 1 0 karam 3 0 drupal 4 0 Shuja-u-Rehman
The table specification is as follow,
Field Type Null Default Extra rid int(11) No auto_increment tid int(11) No name varchar(25) No
I have tried every method, I know but in vain.
Please any body helps me….
Shuja
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Join Date: Nov 2007
Location: Bangalore, India
Posts: 3,057
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Rep Power: 8
Solved Threads: 229
Hmm.. Does it give any error ? print out the query and see whats being passed as values..
What method have you specified for the form ? POST or GET ?
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$name=$_POST['name']; //from the name form
$q_id=$_GET['id'];
What method have you specified for the form ? POST or GET ?
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
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Join Date: May 2008
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Newbie Poster
it does not give any error, $_GET['id'] is extracting from "http://localhost/test/reply.php?id=852"
and $_POST['name'] is coming from form using POST method, it is working well. But the $q_id=$_GET['id'] is not being inserted in database table.
and $_POST['name'] is coming from form using POST method, it is working well. But the $q_id=$_GET['id'] is not being inserted in database table.
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Join Date: Nov 2007
Location: Bangalore, India
Posts: 3,057
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So, is your form action is like this ?
If this is the case, then it should work fine. If not, then you should post your code..
php Syntax (Toggle Plain Text)
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
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Join Date: Aug 2005
Location: somewhere in time
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Junior Poster in Training
You need to isolate the problem. At this point, you don't know if the GET variable is not picking up the value or if the value is not being inserted into thte table.
add echo $q_id;
just after
$q_id=$_GET['id'];
also add echo $sql;
just before you assign the INSERT statement to $sql.
That will tell you right away where the problem lies.
add echo $q_id;
just after
$q_id=$_GET['id'];
also add echo $sql;
just before you assign the INSERT statement to $sql.
That will tell you right away where the problem lies.
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Join Date: May 2008
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Newbie Poster
i can not tell you, how much i am thankful to you. yes it was the error. thanks for your great hint. application is working like rocket...
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Join Date: Nov 2007
Location: Bangalore, India
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Umm.. so, what was the error ?
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
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Join Date: Dec 2007
Location: Somewhere in Germany
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Junior Poster in Training
His problem is quite simple actually. Here's his original code
All he needs to do is remove the single quotes (' ') from around the $q_id in his SQL.
So his new code should look like this:
He has declared the tid field as an integer and he is trying to pass it as a string. MySQL probably doesn't like it.
ALSO
Just to be on the safe side you really need to run the $_GET['id'] and the $_POST['name'] variables through the mysql_real_escape_string() function to prevent someone from doing nasty things to your database. Example:
Hope that helps you.
$name=$_POST['name']; //from the name form $q_id=$_GET['id']; $sql="INSERT INTO `test`.`reply`(`rid`,`tid`,`name`)VALUES(NULL,'$q_id','$name')"; $result=mysql_query($sql);
All he needs to do is remove the single quotes (' ') from around the $q_id in his SQL.
So his new code should look like this:
$name=$_POST['name']; //from the name form $q_id=$_GET['id']; $sql="INSERT INTO `test`.`reply`(`rid`,`tid`,`name`)VALUES(NULL, $q_id ,'$name')"; $result=mysql_query($sql);
He has declared the tid field as an integer and he is trying to pass it as a string. MySQL probably doesn't like it.
ALSO
Just to be on the safe side you really need to run the $_GET['id'] and the $_POST['name'] variables through the mysql_real_escape_string() function to prevent someone from doing nasty things to your database. Example:
php Syntax (Toggle Plain Text)
Hope that helps you.
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Join Date: Nov 2007
Location: Bangalore, India
Posts: 3,057
Reputation:
Rep Power: 8
Solved Threads: 229
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He has declared the tid field as an integer and he is trying to pass it as a string. MySQL probably doesn't like it.
I still believe its the form action which was causing the problem!
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
*PM asking for help will be ignored*
*PM asking for help will be ignored*
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