Using PHP to populate javascript arrays

help with a tutorial

[davidhorn01]

Hi All -

I'm trying to follow this tutorial:

[devshed.com...]

and am getting an error message:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in FILENAME on line 25

I've posted the code at the bottom of this email - line 25 is the line which begins:

while ($myrow = mysql_fetch_row($result))

If anyone has any thoughts, I'd really appreciate it. The connection to the database appears fine. When I 'print-error' it's telling me that it's a PARSE error - but I'm not sure if that's the real problem or whether I've implemented the 'print-error' bit of the code wrong!

Anyway, as you can tell, I'm a bit of a novice at this, so any help would be really appreciated. Many thanks,

David

<?php
$db = mysql_pconnect("localhost", "USERNAME", "PASSWORD");
// This establishes a link to MySQL
mysql_select_db("extranet",$db);
// The database is specified

$sql = "SELECT
p.person_id,
s.person_id,
CONCAT(last_name,', ',first_name) AS name,
skill_id ";

$sql .= "FROM
personnel p,
person_skill s
WHERE
p.person_id = s.person_id
ORDER BY
skill_id, name";

$result = mysql_query($sql);

$type = "";
$number2 = "0";
while ($myrow = mysql_fetch_row($result))

{
if ($myrow[3]!= $type) {
if ($number2!= NULL) {
$newnumber2 = ($number2 + "1");
print ("ar[$number2] = new Array();\n");
$number2 = $newnumber2;
$type = $myrow[3];
$number = "0";
}
}
print "ar[" . ($number2 - "1") . "]";
if ($number!= NULL) {
$newnumber = ($number + "1");
print ("[$number]");
$number = $newnumber;
}
print (" = new makeOption(\"$myrow[2]\", \"$myrow[1]$myrow[3]\");\n");
}
?>

[ergophobe]

That usually means your query failed and you are feeding an invalid result to you fetch.

Quick test: Have your script echo the query, then open your mysql client and feed it the same query. I suspect you'll find that it fails.

[ergophobe]

ps, if you won't be able to ensure that the query will be valid (e.g based on user input), but you don't want the script to die, you can enclose everything in an if

if ($result = mysql_query($query)) {
do stuff;
}else{
don't do stuff;
}

In your case, though, your JS arrays would be empty which, I'm guessing, would be bad.

[davidhorn01]

thanks for your help with this - It was a problem with the query - so thanks.

I'm running into some other problems so I'm going to post again shortly!