Query works in phpMyAdmin but doesn't when using php code

[five]

Hi all,

I've been trying to get this sql query to work from php, but i cannot seem to get it to. I am hoping that someone else has experienced this and knows a solution or can point out my mistakes.

I have a table in my sql db titled Parent_Child, and as you may imagine is just two IDs associating the two together. The two fields are Parent and Child.

I want to run a query like this:

PHP Code: [Select]

SELECT * FROM Parent_Child WHERE Child ='8C8FDBCB-0984-49D8-BE14-1942E26600C4'

It works when I perform it in phpMyAdmin, but does not when I attempt to run the query using mysql_query.

PHP Code: [Select]

//$g is the ID of the current child.
$sql= "SELECT * FROM Parent_Child WHERE Child ='".$g."'";
$result= mysql_query($sql);
while ($row = mysql_fetch_array($result))
 

{
 

echo $row['Parent'];
 

}


It's weird because when i echo $sql, it outputs the query, and when I paste that directly into phpMyAdmin query, it works.. If I passed $g as 8C8FDBCB-0984-49D8-BE14-1942E26600C4, it would give me my desired output in the variable $sql as: SELECT * FROM Parent_Child WHERE Child ='8C8FDBCB-0984-49D8-BE14-1942E26600C4'

note: I have the mysql_connect stuff also but that isn't the focus of the problem. I know it connects to the DB it just doesn't execute the query properly.

Any thoughts?

[Smokey PHP]

Hi there,

Is it completing the query at all? Above the script put:

PHP Code: [Select]

error_reporting(E_ALL);

And change this bit:

PHP Code: [Select]

$result= mysql_query($sql);
while ($row = mysql_fetch_array($result))

to:

PHP Code: [Select]

$result= mysql_query($sql);
if($result === false) echo mysql_error();
while ($row = mysql_fetch_array($result))

[five]

Hiya,

Just tried that and it doesn't return anything. It just outputs a blank line. What would be stopping it from completing the query? Here's the entire function in case its somewhere else that I'm going wrong with what you asked me to do.

PHP Code: [Select]

function getParent($g)
 

{
 
 

error_reporting(E_ALL);
 

 

 

$con=mysql_connect("localhost", "someuser", "somepass");
 

if(!$con)
 

{
 

die('Error: ' . mysql_error());
 

}

 

@mysql_select_db("someproject", $con) or die("Unable to connect to database");
 

 

$sql= "SELECT * FROM Parent_Child WHERE Child ='".$g."'";
 

 

$result=mysql_query($sql);
 

if($result === false) echo mysql_error();
 

while ($row = mysql_fetch_array($result))
 

{
 

echo $row['Parent'];
 

}
}


The output is:

PHP Code: [Select]

8C8FDBCB-0984-49D8-BE14-1942E26600C4
(and a blank line here)


[Smokey PHP]

If that's the output of the code you have posted, then surely the "8C8FDBCB-0984-49D8-BE14-1942E26600C4" is being done by "echo $row['Parent'];"

To make sure that the echo is being called try changing that line to say:

PHP Code: [Select]

echo '['.$row['Parent'].']';

Oh and I'm assuming of course that the die() functions didn't get called and that no errors were displayed after adding error_reporting?

[five]

Oh oops, i accidentally deleted a line from the previous post.. so that output is from the echo which is just $g. I printed it previously to help myself debug and see if it was even getting the right variable.

PHP Code: [Select]

echo $g

[five]

To make sure that the echo is being called try changing that line to say:

PHP Code: [Select]

echo '['.$row['Parent'].']';

Oh and I'm assuming of course that the die() functions didn't get called and that no errors were displayed after adding error_reporting?

Changed the line. still no luck. And yeah the die() functions were not called and no errors are displayed. 

[Smokey PHP]

By still no luck I will assume that you mean that there is no output.

Firstly, try changing the sql to just "SELECT * FROM `Parent_Child`" (and make sure that this table name is EXACTLY the same as it appears in phpmyadmin - same capitals etc)

Secondly, have you ever managed to get an SQL query running on the server you are using?

[five]

Right, still no output.

I did change the SQL statement to "Select * FROM `Parent_Child`", and it did return all of the Parent/Child ID's from the database. I'm sure that I checked on the case sensitivity of them.

And yeah, I'm able to run other SQL queries which makes me believe that something is wrong with my syntax in $sql.

[Smokey PHP]

In which case it might have been the use of ` quotes or maybe the fact that the WHERE clause was left out. As a test try using the same code you started with but use the ` quotes:

PHP Code: [Select]

$sql = "SELECT * FROM `Parent_Child` WHERE `Child` = '".$g."'";

[five]

In which case it might have been the use of ` quotes or maybe the fact that the WHERE clause was left out. As a test try using the same code you started with but use the ` quotes:

PHP Code: [Select]

$sql = "SELECT * FROM `Parent_Child` WHERE `Child` = '".$g."'";

Still no output 

I also tried to remove the usage of ` around Child and from Parent_Child and one with `` and the other without. Unfortunately, still no output.

[Smokey PHP]

And there is DEFINITELY a row in the table `Parent_Child` where the field `Child` is EXACTLY equal to '8C8FDBCB-0984-49D8-BE14-1942E26600C4' ?

This is very strange that this is not working...

Once you have checked that the row I mentioned does exist, and it does, try changing the query to say:

PHP Code: [Select]

$sql = "SELECT * FROM `Parent_Child` WHERE `Child` LIKE '%8C8FDBCB%'";

And in the loop do:

PHP Code: [Select]

echo $row['Parent'];
echo '<br />';
echo $row['Child'];

[five]

Once you have checked that the row I mentioned does exist, and it does, try changing the query to say:

PHP Code: [Select]

$sql = "SELECT * FROM `Parent_Child` WHERE `Child` LIKE '%8C8FDBCB%'";

And in the loop do:

PHP Code: [Select]

echo $row['Parent'];
echo '<br />';
echo $row['Child'];

Changed the query to what you suggested and it gave me the Parent and Child.

PHP Code: [Select]


33BD0276-50E2-4F6D-A5DD-0B588D1DD33F
8C8FDBCB-0984-49D8-BE14-1942E26600C4


I also tried taking what you did using LIKE and replacing the = with it, and it gave me no output.

PHP Code: [Select]


$sql = "SELECT * FROM `Parent_Child` WHERE Child LIKE '".$g."'";

[Smokey PHP]

Okay, I think I know the solution then... fingers crossed!

PHP Code: [Select]

$sql = "SELECT * FROM `Parent_Child` WHERE `Child` LIKE '%".$g."%'";

Try that and let me know!

[five]

Nope didn't work.

[Smokey PHP]

Damn, okay...

PHP Code: [Select]

$sql = "SELECT * FROM `Parent_Child` WHERE `Child` LIKE '%".trim(strtoupper($g))."%'";