Messages

1

Beginners - Learning PHP / Re: Somebody please help

« on: January 11, 2012, 05:11:34 PM »

I'm all sorted now. What I needed to learn is how to pass variables through a url

2

Beginners - Learning PHP / Re: Somebody please help

« on: January 11, 2012, 09:09:59 AM »

Thanks you so much for you reply. Your advice will help me a lot but I don't think I explained what I want to do properly. See how for the url you have used Google. I want to use a php page that will show results based on the row clicked. I need to pass the id onto the next page, I tried using

PHP Code: [Select]

$result = mysql_query("SELECT * FROM property") or die(mysql_error());

while ($row = mysql_fetch_array($result)){

echo $row['short_description'].''."</br>";

echo '<img src="'.$folder.''.$row['picture1'].'">'.''."</br>";

echo '<img src="'.$row['picture2'].'">';

echo '<img src="'.$row['picture1'].''."</br>";

$_SESSION['id'] = $row['id'];

}

But the problem is that the session is always equals to the last row in the array. I'm sorry for my poor explaining but please help me. I can't move on with my project at all and the worst part is I don't even know what to search for to get help.

3

Beginners - Learning PHP / Somebody please help

« on: January 11, 2012, 01:20:22 AM »

I'm creating a real estate website for my final year project. I have a page that displays search results using an array. This is how it looks.

PHP Code: [Select]

$result = mysql_query("SELECT * FROM property") or die(mysql_error());

while ($row = mysql_fetch_array($result)){

echo $row['short_description'].''."</br>";

echo '<img src="'.$folder.''.$row['picture1'].'">'.''."</br>";

echo '<img src="'.$row['picture2'].'">';

echo '<img src="'.$row['picture1'].''."</br>";

}

The thing I want to do is allow the user to click on the short description and be taken to a page that shows the full description based on the link clicked. How can I do this. Please help me guys, I know it's something simple but I'm really struggling with it.

Thanks in advance

4

General PHP Help / Re: Problems displaying image Blob

« on: June 14, 2011, 04:59:36 AM »

Sorry, I forgot to say what the problem was. When I run the php page, I get a broken link

5

General PHP Help / Problems displaying image Blob

« on: June 14, 2011, 04:19:47 AM »

Hi guys

I'm new to PHP. I'm having problems displaying a picture from a Blob. Please please, someone help. This is driving me crazy lol. Here's what I have so far.

PHP Code: [Select]

<?php

// connect to mysql server
$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Not connected : ' . mysql_error());
}
// connect to database server
$db_selected = mysql_select_db('images', $link);
if (!$db_selected) {
    die ('Database error : ' . mysql_error());
}

        //$sql = "SELECT image * FROM gallery";

$sql = "SELECT image FROM gallery";
 
        // the result of the query
        $result = mysql_query($sql) or die("Invalid query: " . mysql_error());
 
        // Header for the image
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0,'image');
 
     
?>