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Newbie

 
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  #1
Nov 22nd, 2007
Can someone please help, I have this code below and every time i execute it i get this error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /var/www/dbconnect1.php on line 14

I can work out what i am doin wrong.


<?php

$dbhost = "localhost";
$dbuser = "root";
$dbpassword= "spewguts";
$dbdatabase - "productsdb";

$db = mysql_connect($dbhost, $dbuser, $dbpassword);
mysql_select_db($database, $db);

$sql = "SELECT * FROM products;";
$result = mysql_query($sql);

while ($row = mysql_fetch_assoc($result)) {
echo $row['product'];
}

?>
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Re: Newbie

 
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  #2
Nov 22nd, 2007
this is your error dude:


$sql = "SELECT * FROM products;";


change that one to this one:


$sql = "SELECT * FROM products";
http://www.ryantetek.com/
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Re: Newbie

 
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Nov 22nd, 2007
if i may ask why is using a semi colon at the end of an SQL statement a problem ? this is valid sql markup
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Re: Newbie

 
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  #4
Nov 23rd, 2007
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Originally Posted by Fungus1487 View Post
if i may ask why is using a semi colon at the end of an SQL statement a problem ? this is valid sql markup
The problem is that he has a semicolon inside the quotation marks and at the end of the statement. Check it out.
Last edited by TheZert; Nov 23rd, 2007 at 3:17 am.
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Re: Newbie

 
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Nov 23rd, 2007
i realise there is a semi colon inside the sql statement and after which is correct syntax for sql string then he appends the php line with a semi colon ?

does this affect the way php reads it?
if so then i have learned something new today.
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Re: Newbie

 
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  #6
Nov 24th, 2007
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i realise there is a semi colon inside the sql statement and after which is correct syntax for sql string then he appends the php line with a semi colon ?
does this affect the way php reads it?
Yes, php's mysql_query "automaticaly adds semicolon" after the query string.
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Re: Newbie

 
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  #7
Nov 24th, 2007
i saw someother errors in the code. did you solve the problem?
Last edited by kkeith29; Nov 24th, 2007 at 4:10 am.
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Re: Newbie

 
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  #8
Nov 24th, 2007
yeah,this one:


$dbdatabase - "productsdb";

$db = mysql_connect($dbhost, $dbuser, $dbpassword);
mysql_select_db($database, $db);


should be:


$dbdatabase = "productsdb";

$db = mysql_connect($dbhost, $dbuser, $dbpassword);
mysql_select_db($dbdatabase, $db);
http://www.ryantetek.com/
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Re: Newbie

 
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  #9
Dec 8th, 2007
No i still have the same problem.
I have made some changes but sill the same error.

PHP Syntax (Toggle Plain Text) 
  1. <?php
  2.  
  3.         $dbhost = "localhost";
  4.         $dbuser = "root";
  5.         $dbpassword= "spewguts";
  6.         $dbdatabase = "productsdb";
  7.  
  8.         $db = mysql_connect($dbhost, $dbuser, $dbpassword);
  9.         mysql_select_db($database, $db);
  10.  
  11.                 $sql = "SELECT * FROM products";
  12.                 $result = mysql_query($sql);
  13.  
  14.                 while ($row = mysql_fetch_assoc($result)) {
  15.                         echo $row['product'];
  16.                         }
  17.  
  18. ?>

Any suggestions would be great.
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Re: Newbie

 
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  #10
Dec 8th, 2007
use mysql_error()...

oh, and you named your vars wrong....
see if it works now

php Syntax (Toggle Plain Text) 
  1. <?php
  2.   $dbhost = "localhost";
  3.   $dbuser = "root";
  4.   $dbpassword = "spewguts";
  5.   $dbdatabase = "productsdb";
  6.  
  7.   $db = mysql_connect($dbhost, $dbuser, $dbpassword) or die("Error: ".mysql_error());
  8.         mysql_select_db($dbdatabase, $db);
  9.  
  10.                 $sql = "SELECT * FROM products";
  11.                 $result = mysql_query($sql) or die("Error: ".mysql_error());
  12.  
  13.                 while ($row = mysql_fetch_assoc($result)) {
  14.                         echo $row['product'];
  15.                         }
  16. ?>
Last edited by hacker9801; Dec 8th, 2007 at 8:13 pm.
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