not a valid MySQL result...have no idea!

wannasub

Hi especially to Dereck

If you visit www.adultinfo4u.co.uk/new/index.php and goto the 'look for an escort' link then when the menu opens out u will see a list of areas I have added this to the link

display.php?location=london

that should then pass the arguments through to the display.php page the code below

<?php
include "common.php";
$link = mysql_connect("$server", "$username", "$password") 
or die("Could not connect"); 
mysql_select_db("$db") or die("Could not select database"); 
if (empty($page)) $page = "";
if (empty($search)) $search = "";
if ($page == search) {
if ($search == 1) {
$result = mysql_query("SELECT * FROM user WHERE location = '$location'");
} 
echo "Bad Search";
}
echo "<table align=center><tr><td><font face=arial size=2>Escort</td><td><font face=arial size=2></td><td><font face=arial size=2>City</td><td><font face=arial size=2>Age</td></tr>";
(line 15) while ($i = mysql_fetch_array($result)) {
echo "<tr><td><font face=arial size=2><a href=view.php?id=".$i[id].">".$i[user]."</a></td><td><font face=arial size=2>".$i[location]."</td><td><font face=arial size=2>".$i[city]."</td><td><font face=arial size=2>".$i[age]."</td></tr>";
}
echo "</table>";

?>

but once clicked I am getting the following error why is this?

Warning: Supplied argument is not a valid MySQL result resource.....at line 15

Thanks

Aaron

mjr

what does mysql_error() say?

wannasub

the error says Warning: Supplied argument is not a valid MySQL result resource in *********/www.adultinfo4u.co.uk/public_html/new/display.php on line 15

dcjones

if I go to the URL in your thread this is the error I get

Warning: Failed opening 'affiliate.php' for inclusion (include_path='.:/usr/local/lib/php') in /home/4523/dreams24/www.adultinfo4u.co.uk/public_html/new/index.php on line 1

which states that it cannot find your inculsion file.

Dereck

wannasub

True thats just another part! I must have confused you.

Goto the escorts link on the menu...this should then dropdown and show some more options...if you then mouse over the 'look for an escort' that then expands to give you the areas covered over the UK ....London....North East etc etc I have added the following to the London link

display.php?location=london

This if i'm correct should then pass those values through to the display.php page if you follow it through you will end up with the error above?

Hope u can follow

dcjones

The code below, is it from : display.php

is display.php the calling page or the called page

<?php
include "common.php";
$link = mysql_connect("$server", "$username", "$password")
or die("Could not connect");
mysql_select_db("$db") or die("Could not select database");
if (empty($page)) $page = "";
if (empty($search)) $search = "";
if ($page == search) {
if ($search == 1) {
$result = mysql_query("SELECT * FROM user WHERE location = '$location'");
}
echo "Bad Search";
}
echo "<table align=center><tr><td><font face=arial size=2>Escort</td><td><font face=arial size=2></td><td><font face=arial size=2>City</td><td><font face=arial size=2>Age</td></tr>";
(line 15) while ($i = mysql_fetch_array($result)) {
echo "<tr><td><font face=arial size=2><a href=view.php?id=".$i[id].">".$i[user]."</a></td><td><font face=arial size=2>".$i[location]."</td><td><font face=arial size=2>".$i[city]."</td><td><font face=arial size=2>".$i[age]."</td></tr>";
}
echo "</table>";

Dereck

wannasub

the code is from display.php which is called once the link is clicked on the main menu.

dcjones

Try

echo $location;

on your called page to see what it contains. It should contain London.

Just had a look at your calling page, when you pass the cursor over London the task bar gives "display.php?location=london

I would expect to see "display.php?$location=London

Have a look

Dereck

wannasub

ok man thanks remember you're talking to a novice plus thats what someone told me to put so Im just taking guys like you on word value.

dcjones

Your calling URL should be ?$location=london

Have you change it, does it now work

Dereck

wannasub

ok its not echoed what u said so is something going wrong?

All I want is to pull all the escorts from each individual location in the database that corresponds with the link. ie all escorts from london only and display them down the page

dcjones

On your calling page you need to change your URL append

there is a "$" missing

Dereck

wannasub

u sure man ive chaged it...on the index.php the calling page hover over the link and it shows in the task bar
'display.php?$location=location'

if thats correct then it doesnt work

dcjones

Right

Go to your server and run the query in the SQL software.

then you will know if it is your script or your data.

Dereck

wannasub

it would be ok if I knew how? Ive got phpmyadmin can i do it through that? But I still dont know what to type...sorry im stupid lol

i use msn messenger if u do it might be easier just use my email aaronmccarton@hotmail.com

dcjones

You will be presented with a GUI which will allow you to run your query.

There is a query box .

Dereck

wannasub

ok have done that it said query completed successfully but it didnt show any data im not sure if it was meant to

dcjones

if you run the query from within side mysqladmin it should show you the results. if there are no results then there is no data matching your query. Which is why your script does not show any.

No data match. You need to look at your data or what your asking your query to return.

Dereck

wannasub

ok man ill have a think about it and email you if I have any further problems

thanks

aaron

Rodders

...this looks like a data issue and not a code issue. Check that you have 'london' in the location field, or is it 'LONDON' or 'London' ?