how to auto generate contents on a page ?

[lukelee]

hi, guys, Here is my codes:

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

on this page, it shows a list of images, when people click on a image, it will link to house.php, on house.php, there will be the details of the image people click. so the contents in house.php should be dynamic, anyone know how to do this?

I wish people understand what I mean. what I am doing here is to make a cms, people can add a new product, each product has a description and image, all descriptions and images are stored in database. and those information will be showed on house.php, so different image people click, different contents display on house.php

 

I wish I got a correct concept of doing this. Can anyone help? thanks

[justinh]

do you have the image details stored in a database?

[rhodesa]

you need to pass the unique id for that image to house.php...i'm going to assume the column is 'id'

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?id=<?php echo $row['id']; ?>"><img src="upload/<?php echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

then in house.php

<?php
require_once('db.php');
$sql = sprintf("SELECT * FROM image WHERE id = '%s'",mysql_real_escape_string($_GET['id']));
$query = mysql_query($sql) or die(mysql_error());
if(!mysql_num_rows($query))
  die("Image not found");
$row = mysql_fetch_assoc($query);

//Now you can echo the data from $row
?>

[waynewex]

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

What you need to learn about is GET values. Obviously, each image in your database has a unique primary key /index key? Well, when you're printing them out, place that id in a get variable so so:

 

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?image_id=<?php echp $row['image_id']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

Then on house.php.

 

<?php

if(isset($_GET['image_id']) && is_numeric($_GET['image_id'])){
$id = mysql_real_escape_string($_GET['image_id']);
$get_details_of_image = mysql_query("SELECT title FROM image WHERE image_id = $id") or die(mysql_error());
}

?>

?>

[justinh]

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {

?>
<ul>
<li><a href="house.php?id=<?php echo $row['id']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

house.php

<?php 

require_once('db.php');

$id = $_GET['id'];
query = mysql_query("SELECT * FROM image WHERE id = $id);
while($row = mysql_fetch_array($query)){
$name = $row['name'];
$size = $row['size']; 
}
//then just put it in a table 

 

 

I think this would work as you are describing

 

 

 

[rhodesa]

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {

?>
<ul>
<li><a href="house.php?id=<?php echo $row['id']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

house.php

<?php 

require_once('db.php');

$id = $_GET['id'];
query = mysql_query("SELECT * FROM image WHERE id = $id);
while($row = mysql_fetch_array($query)){
$name = $row['name'];
$size = $row['size']; 
}
//then just put it in a table 

 

 

I think this would work as you are describing

 

 

 

 

 

don't use this code for the following reasons:

#1) It doesn't provide any protection from SQL injection

#2) The while loop isn't needed and a waste of resources

 

edit: and the string inside mysql_query() isn't closed

[lukelee]

you need to pass the unique id for that image to house.php...i'm going to assume the column is 'id'

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?id=<?php echo $row['id']; ?>"><img src="upload/<?php echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

then in house.php

<?php
require_once('db.php');
$sql = sprintf("SELECT * FROM image WHERE id = '%s'",mysql_real_escape_string($_GET['id']));
$query = mysql_query($sql) or die(mysql_error());
if(!mysql_num_rows($query))
  die("Image not found");
$row = mysql_fetch_assoc($query);

//Now you can echo the data from $row
?>

thanks for help, I see what you mean. here is my database looks like:

pid          title          address          thumb          imagedata

 

1            house        carnegie        1              4352fdase234.jpg

2            farm          carnegie          2              423fdsg43tghgh45.jpg

1            bed            oakleigh          1              hytuy56retgy4.jpg

1            hotel          springvale        1              554thser4fs.jpg

 

because each house may have more than 1 image, so I use address to identify different houses. thumb = 1 mean this image will be used as a thumbnail for people to click, thumb = 2 will be the images on house.php if people click the carnegie house, on the house.php, 'house' and 'farm' will be displayed on house.php

here is the changed codes:

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?id=<?php echo $row['address']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

<?php
require_once('db.php');

$sql = sprintf("SELECT * FROM image WHERE address = '%s'",mysql_real_escape_string($_GET['address']));
$query = mysql_query($sql) or die(mysql_error());
if(!mysql_num_rows($query))
  die("Image not found");
$row = mysql_fetch_assoc($query);

//Now you can echo the data from $row
?>

it keeps give me the message of Image not found, btw, what is '%s'

[justinh]

you should also have an auto_increment id number to help sort different housing

[lukelee]

you should also have an auto_increment id number to help sort different housing

you mean give each house a number? and all images for this house have the same number? but why cant i use address to identify them? I do have auto_increment id, that is pid.

[lukelee]

seems I cant get this work, I dont know why, the image just cant be displayed on house.php. I have tryed all the methods people told me, there must be something wrong with my codes or the database.

[waynewex]

Post your code.

[lukelee]

Post your code.

here is my database looks like:

pid          title          address          thumb          imagedata

 

1            house        carnegie        1              4352fdase234.jpg

2            farm          carnegie          2              423fdsg43tghgh45.jpg

1            bed            oakleigh          1              hytuy56retgy4.jpg

1            hotel          springvale        1              554thser4fs.jpg

 

because each house may have more than 1 image, so I use address to identify different houses. thumb = 1 mean this image will be used as a thumbnail for people to click, thumb = 2 will be the images on house.php if people click the carnegie house, on the house.php, 'house' and 'farm' will be displayed on house.php

here is the changed codes:

 

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?id=<?php echo $row['address']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

house.php

<?php
require_once('db.php');

$sql = sprintf("SELECT * FROM image WHERE address = '%s'",mysql_real_escape_string($_GET['address']));
$query = mysql_query($sql) or die(mysql_error());
if(!mysql_num_rows($query))
  die("Image not found");
$row = mysql_fetch_assoc($query);

//Now you can echo the data from $row
?>

it keeps give me the message of Image not found, btw, what is '%s'

[lukelee]

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

What you need to learn about is GET values. Obviously, each image in your database has a unique primary key /index key? Well, when you're printing them out, place that id in a get variable so so:

 

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?image_id=<?php echp $row['image_id']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

Then on house.php.

 

<?php

if(isset($_GET['image_id']) && is_numeric($_GET['image_id'])){
$id = mysql_real_escape_string($_GET['image_id']);
$get_details_of_image = mysql_query("SELECT title FROM image WHERE image_id = $id") or die(mysql_error());
}

?>

?>

 

<?php
require_once('db.php');
$query = mysql_query("SELECT * FROM image WHERE thumb = '1'");
while($row = mysql_fetch_assoc($query)) {
?>
<ul>
<li><a href="house.php?id=<?php echo $row['pid']; ?>"><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></a></li>
</ul>
<?php } ?>

 

<?php

if(isset($_GET['pid']) && is_numeric($_GET['pid'])){
$id = mysql_real_escape_string($_GET['pid']);
$get_details_of_image = mysql_query("SELECT * FROM image WHERE pid = $id") or die(mysql_error());}
while($row = mysql_fetch_assoc($get_details_of_image)) {
?>
<ul>
<li><img src="upload/<?PHP echo $row['imagedata']; ?>" width="150" height="150" border="1" /></li>
</ul>
<?php } ?>

 

I have tried in this way, still not working...

 

[lukelee]

I tryed this in house.php

<?php  
require_once('db.php');
$address=$_GET['address'];
echo "this is $address!";
?>

the output is

 

this is !

 

seems the address wasnt been passed to house.php

[lukelee]

done it, i made a stupid mistake. thanks guys