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  #1  
Old February 2nd, 2003, 02:37 PM
lyn lyn is offline
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mysql, databases and me

okay here's the deal

i am setting up my webpage and i am wanting to keep my links out in a database. i made sure the connection to the database was all working and stuff and now that i have set it up so that you have to use the form to select a category it doesn't work.

here's the page: http://www.lunadivina.com/w-d/links.php

code is here: http://www.lunadivina.com/w-d/links.txt

line 44 is
php Code:
Original - php Code
  1. $result =
blah blah blah

any help is much appreciated

oh yeah - the only category that has any links in it is "Actors"

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  #2  
Old February 2nd, 2003, 03:31 PM
filefrog filefrog is offline
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RE: mysql, databases and me

place the database name (as passed to mysql_db_query()) in quotes.

i.e.:
php Code:
Original - php Code
  1. $result = mysql_db_query("lunadivi_links",$query, $connect);

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  #3  
Old February 2nd, 2003, 04:08 PM
lyn lyn is offline
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RE: mysql, databases and me

check did that now i get the following error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/lunadivi/public_html/w-d/links.php on line 46

line 46 is:

php Code:
Original - php Code
  1.  
  2. while ($row = mysql_fetch_array($result)){


i get that error after choosing actors and hitting go

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  #4  
Old February 2nd, 2003, 04:25 PM
lyn lyn is offline
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RE: mysql, databases and me

okay i have come to the conclusion that the problem is in the query string.

this is what i have currently
php Code:
Original - php Code
  1.  
  2. $query = 'SELECT * FROM `links` WHERE `site_cat` ='. $category;


if i hard code Actors into the query string it works properly but i want to be able to take the category from the form so what am i doing wrong??

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  #5  
Old February 2nd, 2003, 04:37 PM
filefrog filefrog is offline
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RE: mysql, databases and me

i imagine the category field is a varchar, right?

if it is, you need to encapsulate the query data in quotes as well. so your PHP/ SQL becomes:
php Code:
Original - php Code
  1. $query = "SELECT * FROM `links` WHERE `site_cat`='$category'"

or possibly:
php Code:
Original - php Code
  1. $query = "SELECT * FROM `links` WHERE `site_cat`='" . $category . "'"

(note the single quotes inside the double quotes, around the $category variable)

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  #6  
Old February 2nd, 2003, 04:42 PM
lyn lyn is offline
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RE: mysql, databases and me

you are fabulous! *HUG* lol it works and i don't know why i didn't think of that! LOL

thank you so much for your help! mucho-appreciated

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  #7  
Old February 2nd, 2003, 11:20 PM
filefrog filefrog is offline
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RE: mysql, databases and me

hey no prob. at least you posted code and weren;t like "okay, im a n00bie, i need a script that does this. thank you for your time" lol

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