Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result source

[chemlight]

I'm having a problem. I'm sure I'm going to kick myself over the answer...
 
I have a table that stores vendors and their languages. This table starts out blank. I am querying the table to see if a vendor has been added to the table yet. The problem is, if they haven't been added, I can't seem to get the script to realize that. here is what I am trying to do.
 
$testvend = SELECT language FROM vendor_details WHERE id = $vendorid
 
if(!$testvend){
//script to add vendor to list, with supplied language
}else{
while($row = mysql_fetch_array($testvend)){
//script to get vendor language and display vendor langauge
}
}

The result I get is:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result source in (script name) on line (line with mysql_fetch_array)

*NOTE: I did try searching bytes.com and the web. The search on bytes.com returns a blank page. The web showed my method above, as well as the option to add "or die" after the query. adding or die didn't do anything either. I've also tried manually adding information to the table, so that the script is not working off of a blank table, but I still get the same result if the vendor does not exist. Also, I apologize that the title isn't very clear. I realized after I posted it I should have put a problem description in the title...

[Markus]

You need to suply the mysql_* functions with a result resource. Simply passing it the query string will not do.

// SQL.
$sql = "SELECT * FROM `tbl1`"
 
// Create result resource
$query = mysql_query( $sql );
 
// foreach traverse here.
 

Also, please wrap code with [code] [/code] tags and give your threads a meaningful title.

Moderator.

[TheServant]

I know this has been already said, but took me a second to work out the problem. Your code should be:

$testvend = mysql_query(" SELECT language FROM vendor_details WHERE id = '$vendorid' " );

Or:

$testquery = SELECT language FROM vendor_details WHERE id = '$vendorid';
$testvend = mysql_query( $testquery );

Also, I think that php variables need to be surrounded by single quotes ( ' ), as well as the whole statement being a string - surrounded by double quotes ( " ). Also I see no semicolons " ; "? Was that a mistype or are you decalring variables without that? If so, I don't know why it's not throwing an error up for that but it should.

[Markus]

@TheServant
Deja vu :)

@TheServant
If you surround a variable that your database expects to be an int type, you will get an error. So only variables that are strings need quotes.

@TheServant
I think he just threw it together quickly to show us.

[TheServant]

@Markus
Interesting. I will have to test because I '$quote' all my variables and have not had any trouble. Thanks for the info.

I think he just threw it together quickly to show us.

That's what I thought, but just better make sure ;)

[chemlight]

Yeah, I did just throw it together quickly, thats why they weren't in there.

My actual mistake was that I was leaving out the mysql_query part, so I was assigning the words in the query to the variable, instead of the actual results.

I'm having another issue though.

Here is my code

$testvend = mysql_query("SELECT language FROM vendor_details WHERE id = '" . $vendor . " ' ");
if(!$testvend){
//script if vendor does not exist
}else{
while($row = mysql_fetch_array($testvend)){
//script to do if vendor exists
}
}
 

The problem is, even if the vendor doesn't exist, it executes the code as if it did. I've tried using

if(count(mysql_fetch_array($testvend)) > 0){
//script to do if vendor exists
}
 

But it seems to return a value greater than 0 anyway. I'm working on a test page that will do a better job of showing my issues, and I'll post the code here, and the results, when I get it done.

Thanks again for the responses!

[chemlight]

Here is the code I am testing with:
(copied and pasted)

<?php
 
    require("scripts/db_connect.php");
 
    $vendor = 4486;
 
    $result = mysql_query("SELECT * FROM vendor_details WHERE id = '" . $vendor . "'");
 
    if(!$result){
        echo "no vendor " . $vendor . "<br />";
    }else{
        echo "vendor " . $vendor . " exists<br />";
    }
 
    print_r(mysql_fetch_array($result));
    echo "<br />";
 
    $vendor = 6510;
 
    $result = mysql_query("SELECT * FROM vendor_details WHERE id = '" . $vendor . "'");
 
    if(!$result){
        echo "no vendor " . $vendor . "<br />";
    }else{
        echo "vendor " . $vendor . " exists<br />";
    }
 
    print_r(mysql_fetch_array($result));
?>
 

Here are the results:

vendor 4486 exists
Array ( [0] => 4486 [id] => 4486 [1] => [phone] => [2] => [email] => [3] => [contact] => [4] => German [language] => German [5] => 0 [rating] => 0 )
vendor 6510 exists

And this is my table (copied from phpmyadmin, formatted with | for easier reading - there is only one item it right now)

id | phone | email | contact | language | rating
4486 | | | | German | 0

So, I need to find a way to tell the script that there is no vendor, if there isn't one...

[chemlight]

I also decided to test it by providing a variable to the query string istead. I got the same results

<?php
    require("scripts/db_connect.php");
 
    $vendor = 4486;    
    $result = mysql_query("SELECT * FROM vendor_details WHERE id = " . $vendor . "");
    if(!$result){
        echo "no vendor " . $vendor . "<br />";
    }else{
        echo "vendor " . $vendor . " exists<br />";
    }
    while($row = mysql_fetch_array($result)){
        echo "vendor language " . $row['language'];
    }
    print_r(mysql_fetch_array($result));
    echo "<br />";
 
    $query = "SELECT * FROM vendor_details WHERE id = " . $vendor;
    $result = mysql_query($query);
    if(!$result){
        echo "no vendor " . $vendor . "<br />";
    }else{
        echo "vendor " . $vendor . " exists<br />";
    }
    while($row = mysql_fetch_array($result)){
        echo "vendor language " . $row['language'];
    }
    print_r(mysql_fetch_array($result));
    echo "<br />";
 
    $vendor = 6510;
    $result = mysql_query("SELECT * FROM vendor_details WHERE id = " . $vendor . "");
    if(!$result){
        echo "no vendor " . $vendor . "<br />";
    }else{
        echo "vendor " . $vendor . " exists<br />";
    }
    while($row = mysql_fetch_array($result)){
        echo "vendor language" . $row['language'];
    }
    print_r(mysql_fetch_array($result));
    echo "<br />";
 
    $query = "SELECT * FROM vendor_details WHERE id = " . $vendor;
    $result = mysql_query($query);
    if(!$result){
        echo "no vendor " . $vendor . "<br />";
    }else{
        echo "vendor " . $vendor . " exists<br />";
    }
    while($row = mysql_fetch_array($result)){
        echo "vendor language " . $row['language'];
    }
    print_r(mysql_fetch_array($result));
    echo "<br />";
?>
 

results:
vendor 4486 exists
vendor language German
vendor 4486 exists
vendor language German
vendor 6510 exists

vendor 6510 exists

I took out the print_r because I figured it would be easier to read. I still get the same issue - it keeps telling me the vendor exists when it doesn't.

[Markus]

mysql_num_rows($result_resource) will tell you the amount of rows affected by the query. So, just run an if conditional, checking that mysql_num_rows($result_resource) turns more than 0 rows.

[chemlight]

That worked. I was also just about to post that testing if the resource is null also works.

One of the things I had tried was:

if(count(mysql_fetch_array($result)) > 0)
 

But that returns 1 in both cases. The mysql_num_rows returns the proper count.

Thanks again for your help and patience!

[TheServant]

lol, didn't realize it worked like this but it makes sense. mysql_fetch_array() will fetch an array as the name suggests. If the array is empty and you count it, it counts 1 empty array, returning 1. So mysql_num_rows() is the way to go.

[Markus]

Glad we could be of help. It's always smarter to stick to methods that are there for a specific purpose (using mysql_* functions for mysql related stuff, etc.)

- Mark.