Hello God(s) Help with an obvious error

[Chezshire]

I've tired playing with this for a bit - and i'm vexed vexed vexed.

 

I keep getting this error (which repeats three times as shown below) which is returned from one of my functions and i'm at a complete lose. Help? Please?

 

 

The Error Message

arning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/z/a/n/zanland/html/xpg/functions.php on line 737

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/z/a/n/zanland/html/xpg/functions.php on line 737

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/z/a/n/zanland/html/xpg/functions.php on line 737

 

 

function listChars () {
global $db;
$theseChars2 = mysql_query("select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC",$db);
$z=0;
WHILE ($tempy = mysql_fetch_assoc($theseChars2)) { //-----THIS IS LINE #737 WHICH IS MAKING MY LIFE ANNOYING------
$codenameList[$z] = $tempy["codename"];
$codenameID[$z] = $tempy["id"];

 

 

 

As always, any help, guidance or offer of soup is appreciated

Thanks guys (I'm just a struggling novice trying to learn)

[Rayhan Muktader]

Run the query: "select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC"  by itself from a command line and see if you get any error.

 

[Chezshire]

I'm not sure how to do that (i'm a real novice).

 

Do i do that by entering this into my code?

 

<? php

$result = mysql_query($select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC());

?>'

 

Thank you for your guidance and help. It's almost like getting soup on a cold new england day

[kenrbnsn]

Replace

<?php
$theseChars2 = mysql_query("select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC",$db);
?>

with

<?php
$q = "select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC";
$theseChars2 = mysql_query($q,$db) or die("Problem with the query: $q<br>" . mysql_error());
?>

and tell us what error is printed, if any.

 

Ken

 

[Chezshire]

Hello and thank you for that.

 

this is the error i received 'Parse error: parse error, unexpected '<' in /home/content/z/a/n/zanland/html/xpg/functions.php on line 735'

 

Nothing but that displayed - it killed my whole site which wave kind of neat. Just that line of black text on a white background. What did it do that? And i'm guessing I have an extra Greater Than (<) sign somewhere in my code on that page?

 

Look forwards to your answers!

-Chex

[Vizor]

Um, the die() function will stop all script execution and in this case output mysql_error().

 

Yes you read the error right you have a < that shouldn't be there or is in the wrong place.

[bluejay002]

i think its not related with the mysql_error() since it didnt showed any mysql_error().

 

i think its a php error... try scan your code... you might have misplaced something foreign.

 

[kenrbnsn]

Please show us the line you change and the ten or so lines before/after it.

 

Also, what is line 735?

 

Ken

[Chezshire]

Thank you - I'm exploring (unsuccessfully mind you but having fun none the less!)

 

ten lines prior to the error line:

// ===============================================
// ==================== LIST CHARACTERS
// ===============================================


function listChars () {
global $db;
$theseChars2 = mysql_query("select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC",$db);
$z=0;

 

 

The line with the error:

WHILE ($tempy = mysql_fetch_assoc($theseChars2)) {

 

Ten lines after as requested:

$codenameList[$z] = $tempy["codename"];
$codenameID[$z] = $tempy["id"];

$appearanceArray=array();

if ($tempy["age"] && $tempy["age"] != $tempy["thisYear"]) { array_push($appearanceArray, $tempy["age"] . " years old"); }
if ($tempy["height"]) { array_push($appearanceArray, $tempy["height"]); }
if ($tempy["weight"]) { array_push($appearanceArray, $tempy["weight"]); }
if ($tempy["hair"]) { array_push($appearanceArray, $tempy["hair"] . " hair"); }
if ($tempy["eyes"]) { array_push($appearanceArray, $tempy["eyes"] . " eyes"); }
if ($tempy["ethnicity"]) { array_push($appearanceArray, $tempy["ethnicity"]); }

 

 

Sorry it took me so long to reply (especially when i'm the one begging for help - I'm dealing with cancer and today was one of my less successful days. But i'm still going!

Thanks for all the help.

 

While getting chemo i did try this: != instead of <>

 

Still had the same error

 

 

[Rayhan Muktader]

Which error are you getting? The unexpected '<' error or the original error? 

Create a separate php file and do the following:

<?php
# FIRST CONNECT TO THE DATABASE
# THEN

$q = "select codename,id,height,weight,eyes,hair,ethnicity,characteristics,(YEAR(NOW()) - YEAR(dateofbirth)) - (MONTH(NOW()) < RIGHT(dateofbirth,5)) AS age, YEAR(NOW()) as thisYear from cerebra WHERE approved <> '' AND approved IS NOT NULL AND codename IS NOT NULL AND codename <> '' ORDER BY LENGTH(codename) DESC";
$theseChars2 = mysql_query($q,$db) or die(mysql_error());

?>

 

It is possible that all error reporting are off in the php settings and thats why you are not getting an error when your query is not executing correctly.  But php does not have a choice about error reporting when fetcharray function gets an invalid parameter.

 

Take care of one problem at a time.  First make sure that this query executes correctly.  And ask your server admin about command line access to mysql.  It will make your life easier.

[Chezshire]

Hi Everyone,

  I figured out what the problem was - and i did it by using PHPfreaks 'unbugging guide!'

  So this is solved but i'm nto sure how to change the status of this to 'solved'.

 

Thank you!

Chez