Error

Warning: mysql_fetch_arry(): supplied argument is not a valid MySQL

[CIAimages]

<?php
if ($page) {
$page_sql = "SELECT * FROM SPI_html WHERE page_name=$page";
$page_result = mysql_query($page_sql);
$page_myrow = mysql_fetch_array($page_result);
$body = $page_myrow["body"];
echo $body;
}
?>

I am getting an error and I cannot figure out why?

[johnerazo]

$page_result = mysql_query($page_sql);

Check this out for the right syntax of mysql_query:
[php.net ]

[Timotheos]

Usually if there's something wrong with your sql syntax it will tell you. Another possibility is that you haven't connected with your database.

[CIAimages]

The error I get is:

Warning: mysql_fetch_field(): supplied argument is not a valid MySQL result resource in /usr/local/www/test.php on line 58

I tried

$page_myrow = mysql_fetch_array($page_result);
$page_myrow = mysql_fetch_row($page_result);
$page_myrow = mysql_fetch_field($page_result);

54 <?php
55 if ($page) {
56 $page_sql = "SELECT * FROM SPI_html WHERE page_name=$page";
57 $page_result = mysql_query($page_sql);
58 $page_myrow = mysql_fetch_array($page_result);
59 $body = $page_myrow["body"];
60 echo $body;
61 }
62 ?>

I get the same error all the time.

[edited by: CIAimages at 4:43 pm (utc) on June 14, 2004]

[Philosopher]

and what code do you have on line 58?

Maybe give use the line before and the line after as well.

[jatar_k]

try using this for your query

$page_result = mysql_query($page_sql) or die(mysql_error());

what error message does that give you?

[CIAimages]

Now I'm getting this error:

Unknown column 'home' in 'where clause'

[jatar_k]

looks like your query is not quite right then, checking for a column that doesn't exist

try this now

$page_sql = "SELECT * FROM SPI_html WHERE page_name=$page";
echo $page_sql;

then take a look at the constructed query and see if there are any errors in it. I often take the output query and connect to sql, in my case with putty, and then test the query there.

I usually get the same error but then i can look at the table itself and check all the columns and the data to make sure I am doing what I thought I was doing and it makes it easier to fix the query.

Some people obviously do this with phpmyadmin but I am a command line mysql'er. ;)

[CIAimages]

SELECT * FROM SPI_html WHERE page_name=home

I see no error in the sql statment
I looked at my db via phpMyAdmin and I see no error in my table eather. But I am new at php and mysql so here is how my table look like

Field Type Null
page_name text No
title text No
description text No
keywords text No
body text No

Could the error come from the fact that I use the same connection for 2 different table processing?

// Connect to the mySQl server
// ***************************
$db = mysql_connect($host, $username, $password) or die("Could not connect to host.");
mysql_select_db($database, $db) or die("Could not find database.");

// Get all the Info from the mySQl DB
// **********************************
$sql = "SELECT * FROM SPI_html";
$result = mysql_query($sql);

as well for this script

$page_sql = "SELECT * FROM SPI_html WHERE page_name=$page";

[Timotheos]

Try it with single quotes in the WHERE statement...
SELECT * FROM SPI_html WHERE page_name='home'

If you do a select in phpMyAdmin it will display the SQL statement. Often that gives me clues of what I'm doing wrong.

[m_shroom]

1. I think it is beter to use mysql_pconect.
2. A db conection must be done on each page of your script that it is used.

3. create a function to sinplify it (functions a held in memory)

function condb()
{$db = mysql_pconnect($host, $username, $password) or die("Could not connect to host.");
mysql_select_db($database, $db) or die("Could not find database.");}

then every page you need to connect to the db simply type

condb();