Question

SQL problem

Asked by: DancingFighterG

Hello, I'm getting the following error on the sql below:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Here is the code that I am using:

$converge_info = $this->query("select m.member_id, m.member_group_id, m.email,
          m.members_pass_hash, m.members_pass_salt, field_1 as chapter from ibf_members m
            join ibf_pfields_content pf on pf.member_id = m.member_id  
              where name = \"" .       $this->username ."\"");

$converge_info = $this->query("select m.member_id, m.member_group_id, m.email,
    	m.members_pass_hash, m.members_pass_salt, field_1 as chapter from ibf_members m
		join ibf_pfields_content pf on pf.member_id = m.member_id  
 	 	where name = \"" . 	$this->username ."\"");

                                  
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Asked On
2009-10-30 at 14:42:54ID24859815
Topic

PHP Scripting Language

Participating Experts
3
Points
500
Comments
14

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Answers

 

by: leakim971Posted on 2009-10-30 at 14:56:29ID: 25707079

Hello DancingFighterG,

Do you have "name" and "field_1" fields in the table ?

Try :

$converge_info = $this->query("select m.member_id, m.member_group_id, m.email,
    	m.members_pass_hash, m.members_pass_salt, field_1 as chapter from ibf_members m
		join ibf_pfields_content pf on pf.member_id = m.member_id  
 	 	where name = '$this->username'");

                                              
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by: bportlockPosted on 2009-10-30 at 16:07:26ID: 25707346

If $this->username is a simple text field then the following should work

$converge_info = $this->query("select m.member_id, m.member_group_id, m.email,
        m.members_pass_hash, m.members_pass_salt, field_1 as chapter from ibf_members m
                join ibf_pfields_content pf on pf.member_id = m.member_id  
                where name = '$this->username' ");

Remember to check that $this->username is escaped

http://www.php.net/mysql_real_escape_string

 

by: bportlockPosted on 2009-10-30 at 16:08:46ID: 25707353

Oops! The perils of not maximizing the window - I did not see leakim971's answer. My apologies....

 

by: SammayePosted on 2009-10-30 at 16:10:29ID: 25707360

where does name come from?, you must specify where name comes from ie m.name else it won't work

 

by: SammayePosted on 2009-10-30 at 16:13:04ID: 25707373

I also noticed you got a field_1 (dunno why you named it that but hey) that also needs a table to read from.

The reason for this is because you have a join so the field names no longer associate themselves with one table and unless MySQL is told which table to read from for the fields it will read nout. This always annoys me when I make these errors I wish MySQL had better error reporting

 

by: SammayePosted on 2009-10-30 at 16:15:38ID: 25707381

Third post, just a code snippet, this is leakim's solution modified a little so accomodate the table naming:

$converge_info = $this->query("select m.member_id, m.member_group_id, m.email,
        m.members_pass_hash, m.members_pass_salt, m.field_1 as chapter from ibf_members m
                join ibf_pfields_content pf on pf.member_id = m.member_id  
                where m.name = '$this->username'");

                                              
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by: DancingFighterGPosted on 2009-10-30 at 16:46:18ID: 25707516

It's not the sql because I have tested the sql and it returns the values that I want. There is something going on with the syntax for some reason. Maybei it's the php version?

 

by: SammayePosted on 2009-10-30 at 16:59:16ID: 25707597

a. PHP version? (most likely not though)
b. And tested the SQL where because I know that fields are different in PHP there is a different SQL handler for PHP than say from console on MySQL server.
c. Have you check that username actually contains something?

Can you give us some debug output maybe?

 

by: DancingFighterGPosted on 2009-10-30 at 17:19:15ID: 25707672

The sql is returning data because with all the other validation I have arond the code I verified that it seeing if a login is incorrect via username and password.

I tested the sql on my server where the code is running. From the database the query returns rows just fine.

This is what I have right now and the sql works. The error is coming from line 1 of the sql command:

$converge_info = $this->query("SELECT m.member_id, m.member_group_id, m.email,
     m.members_pass_hash, m.members_pass_salt, field_1 as chapter from ibf_members m
     join ibf_pfields_content pf on pf.member_id = m.member_id  
     where m.name = \"" . 	$this->username ."\"");

                                              
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by: SammayePosted on 2009-10-30 at 17:45:16ID: 25707801

I really puzzled now, That works perfect, just quickly made up the table and it works gives me the data in php fine:

$converge_info = "SELECT m.member_id, m.member_group_id, m.email,
     m.members_pass_hash, m.members_pass_salt, field_1 as chapter from ibf_members m
     join ibf_pfields_content pf on pf.member_id = m.member_id  
     where m.name = \"" .       $username ."\""; 
$result = $db->getRow($converge_info);

                                              
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by: SammayePosted on 2009-10-30 at 17:58:07ID: 25707852

can you by any chance show the rest of the function/page?, I'm making a wild guess but I think it could be:

a. broken function having a knock on effect
b. you look like your using an abstract layer, maybe you are fetching data from the resulting query incorrectly. If your using php::DB then making an incorrect loop to scroll through an query function can result in a SQL error on that query...

If the data is grabbing fine when you run the query in say, phpmyadmin or console then my only guess is that its gotta be the php around the function.

 

by: DancingFighterGPosted on 2009-10-30 at 18:00:36ID: 25707858

Ok, I figure it. There was another column name that I needed to change. What happen was that we just upgraded our invision board and the new version merged some tables make our custom app not to work right. Mis named a couple of columns. Thanks for the help people!!

 

by: SammayePosted on 2009-10-30 at 18:07:19ID: 25707883

of course there is also:

c. You accidentally have changed $this->username to an object instead of say a string. Replace your $this->username with just a string variable of something like I have if that don't work you know you haven't accidentally assigned it to an object. So the last line would now equal (or something close):

where name = 'bob'");

I know it seems something that impossible but sometimes it does happen.

 

by: SammayePosted on 2009-10-30 at 18:07:41ID: 25707884

oh ok, :)

20120131-EE-VQP-002

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