Using the $_GET variable to view certain records

[yobo]

Hello,

 

First let me explain my problem, I have 2 pages the first page pull a list of Guide titles from a database for example if i had a guide called "Install windows" it would be pulled from the database with a link next to the guide called "read this Guide" now that works fine.

 

the problem that i am having is when a user clicks the "Read this guide" it will goto the next page and pull the content from the database that releates to the guide.  however nothing is being displayed on the next page. My code is shown below for the 2 pages and my database structure.

 

Guides.php (1st page)

<?php


/**
* @author Joe Moore
* @copyright 2010
*/

//Include the database connection details, to allow us to communicate with the database
include 'dbcon.php';


echo '<p> Here are all the articles in the database </p>';

$result = @mysql_query('SELECT guidetitle, id FROM guides');

if (!$result) 
{
    exit('<p>Error performing query: ' . mysql_error() . '</p>');
}

// Display the text of each guide in a paragraph
//with a "Delete this Guide" link next to each
while ($row = mysql_fetch_array($result))
{
   $id = $row['id'];
    $guidetitle = $row['guidetitle'];
    echo "<p>$guidetitle " .
         "<a href=read.php?id=$id>Read this guide</a></p>";
         
}

?>

 

read.php (2nd page)

<?php
//include('auth.php');

/**
* @author Joe Moore
* @copyright 2010
*/
?>
<h1>Welcome <?php // echo $_SESSION['SESS_FIRST_NAME'];?></h1>
<?php
//Include the database connection details, to allow us to communicate with the database
include 'dbcon.php';

$id = $_GET['id'];

$guide = mysql_query("SELECT guidetext, id FROM guides WHERE id='$id'");
    
if (!$guide) 
{
    exit('<p>Error Fetching Guide Details: ' .
        mysql_error() . '</p>');
}

$guide = mysql_fetch_array($guide);
?>

 

my database structure (table)

 

id guidetext                                                                         guidedate guidetitle                                   author

1 This is sample text and will be updated accordinly         2010-01-08 Secure Passwords                         1

2 this guide will talk abut web servers and how to i... 2010-01-08 Install web servers on windows 1

3 this guide will talk about password storage                 2010-01-08 How to secure passwords 2         1

5 HOW TO INSTALL LITESPEED ON LINUX                         2010-01-09 Litespeed webserver                 1

 

Thanks

 

- Joe

[JAY6390]

You fetched the data but haven't done anything with it.

put this after the fetch

echo '<pre>'.print_r($guide, true).'</pre>';

and see if anything gets displayed

[yobo]

You fetched the data but haven't done anything with it.

put this after the fetch

echo '<pre>'.print_r($guide, true).'</pre>';

and see if anything gets displayed

 

Hey Jay6390

 

That works but it outputs an array structure which i dont want to happen, I would like it to just output the content in guidetext for example lets say guidetext had a value in the database table called "Lets begin" I would like that outputted to the screen

[yobo]

I have modifed my code on the read.php page to this.

<?php
//include('auth.php');

/**
* @author Joe Moore
* @copyright 2010
*/
?>
<h1>Welcome <?php // echo $_SESSION['SESS_FIRST_NAME'];?></h1>
<?php
//Include the database connection details, to allow us to communicate with the database
include 'dbcon.php';

$id = $_GET['id'];

$guide = mysql_query("SELECT guidetext FROM guides WHERE id='$id'");
    
if (!$guide) 
{
    exit('<p>Error Fetching Guide Details: ' .
        mysql_error() . '</p>');
        
}

while ($guide = mysql_fetch_array($guide))
{
$read = $guide['guidetext'];
echo "$read";
}
?>

 

it works but I get an error which says this

 

Warning: mysql_fetch_array() expects parameter 1 to be resource, array given in C:\wamp\www\blog\read.php on line 25

[wildteen88]

You're getting that error because your overwriting the $guide variable

while ($guide = mysql_fetch_array($guide))

 

That line should be

while ($row = mysql_fetch_array($guide))

 

In your while loop use $row['guidetext'] instead of $guide['guidetext']

[JAY6390]

Yeah that was just to test that you were getting data from the db succesfully. OK, well all you need to do is use the format

echo $guide['column_name_here'];

in your first code and it will work fine

[yobo]

Thanks all for helping me out I shall remember what you all said for future refernce