[SOLVED] Query help

[ninedoors]

I am stumped.  Maybe I hsve been looking at this too long, who knows.  Anyways I can't get this query to work.  I keep getting this error:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\Ergo\admin\training\training_delete.php on line 209

 

Which means there is something wrong with my query statement, here is the code I am using:

 

<?php
$query = "SELECT training.flienumber, training.date, training.title, trainingupload.name, trainingupload.size ".
	"FROM training t LEFT JOIN trainingupload ta ON t.filenumber = ta.filenumber";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
?>

 

The two tables I am querying are training and trainingupload.  And I want pull out all the training records and only the trainingupload records that have matching filenumbers to the training table. 

 

Not sure what I have wrong.  Any help would be great.

 

 

[friedemann_bach]

"SELECT training.flienumber" should read "SELECT training.filenumber", I think.

 

[ninedoors]

You are correct, I changed that and still no luck.  Any other thoughts.

 

Nick

[zenag]

can u try this one ...

<?php
$query = "SELECT training.flienumber, training.date, training.title, trainingupload.name, trainingupload.size ".
	"FROM training LEFT JOIN   trainingupload  ON training.filenumber = trainingupload.filenumber";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
?>

[aschk]

Your query is failing, run it through MySQL command line, or phpMyAdmin and you should see it doesn't return any results and should point to your incorrect syntax. Also, use "echo mysql_error()" in your PHP to see why the query failed.

[friedemann_bach]

Try a more simple query first (with less specified values):

SELECT t.* FROM training t LEFT JOIN trainingupload tu ON t.filenumber = tu.filenumber

 

If this does not work, reduce your query until it works, then extend it step by step, and as soon as an error occurs, you will know which element is wrong.

 

Alternatively, arrange the same query without JOIN:

SELECT t.* FROM training t, trainingupload tu WHERE t.filenumber=tu.filenumber

 

[GingerRobot]

Indeed - do some error checking!

 

<?php
$query = "SELECT training.filenumber, training.date, training.title, trainingupload.name, trainingupload.size ".
	"FROM training t LEFT JOIN trainingupload ta ON t.filenumber = ta.filenumber";
$result = mysql_query($query) or die(mysql_error());
if(mysql_num_rows($result) > 0){
    $row = mysql_fetch_array($result);
}else{
    echo 'No results!';
}
?>