require urgent help to retrieve image from database

[rime]

I have written the following code to retrieve image from mysql database:

<?php
include('dbinfo.inc.php');
include('../functions.php');
mysql_connect("localhost","username","password")
or die("Failure to communicate");
mysql_select_db("db")
or die("Could not connect to Database");

$query = "SELECT content,size,type FROM upload WHERE id=1;";
$result=mysql_query($query);
mysql_close();

$content=mysql_result($result,0,"content");
$size=mysql_result($result,0,"size");
$type=mysql_result($result,0,"type");

header('Content-Type: '.$type);
print $content;
?>

It is working.. Now I would like to retrieve multiple images at a time..
what shall I do?
please help.. its urgent

[ithelp]

You can retrieve only one image at a time.

[ryan_vietnow]

Try to make this code as a function.
I'll give you an example:

<?php
include('dbinfo.inc.php');
include('../functions.php');
mysql_connect("localhost","username","password")
or die("Failure to communicate");

mysql_select_db("db")
or die("Could not connect to Database");

function displayimage($id)
{
$query = "SELECT content,size,type FROM upload WHERE id='$id';";//$id is the id of your pic or whatever...
$result=mysql_query($query);
mysql_close();

$content=mysql_result($result,0,"content");
$size=mysql_result($result,0,"size");
$type=mysql_result($result,0,"type");

header('Content-Type: '.$type);
imagejpeg($content); //I have changed "print $content;" to this one...
}
$one=1;
$two=2;
$three=3;

//just enter in the parameter of your function whatever id you want..
$one1=displayimage($one);
$two2=displayimage($two);
$three3=displayimage($three);

echo '<img src="'.$one1.'">'; 
echo '<img src="'.$two2.'">';
echo '<img src="'.$three3.'">';

?>

I think that's it.
3,2,1...ZERO.

use fetch_array command with while circle.

[rime]

Thanks ryan_vietnow for replying..

Basically I have a database where each row has three images..

And each row has an id, i.e. all three images have only one id..

so I am confused about the part in your code

$one=1;$two=2;$three=3;

what shall I do to make the code run for me.

Please Help

i will send you the code that i write soon. it support up to hundreds of images on one row...

<?php
$Get = $_GET;
$GetID = "select * from dbtable where ID='" . $Get . "'";
$Result = mysql_query($GetID);
$NumberofPicture = 3 // Define here have many columns stores your pictures data.

while($Write = mysql_fetch_array($Result))
{
$ID = $Write;

for($i = 1; $i < $NumberofPictures; $i++)
{
$Picture[] = $Write; // Name of column will appear like Picture1, Picture2
}
}
?>

<?php
// to print the data out from array
echo $Picture[1];
?>

i could not find the exact code i wanted to give you but i hope this would help you.

[rime]

Thanks fatihpiristine for replying.

can you please explain me the line

$Get = $_GET;

what is NewsID meant for?

ahaa :) i forgot to change. it is belong to mine

NewsID stands for the row id, requests and assing the data to $Get which is posted

[rime]

Try to make this code as a function.
I'll give you an example:

<?php
include('dbinfo.inc.php');
include('../functions.php');
mysql_connect("localhost","username","password")
or die("Failure to communicate");

mysql_select_db("db")
or die("Could not connect to Database");

function displayimage($id)
{
$query = "SELECT content,size,type FROM upload WHERE id='$id';";//$id is the id of your pic or whatever...
$result=mysql_query($query);
mysql_close();

$content=mysql_result($result,0,"content");
$size=mysql_result($result,0,"size");
$type=mysql_result($result,0,"type");

header('Content-Type: '.$type);
imagejpeg($content); //I have changed "print $content;" to this one...
}
$one=1;
$two=2;
$three=3;

//just enter in the parameter of your function whatever id you want..
$one1=displayimage($one);
$two2=displayimage($two);
$three3=displayimage($three);

echo '<img src="'.$one1.'">'; 
echo '<img src="'.$two2.'">';
echo '<img src="'.$three3.'">';

?>

I think that's it.
3,2,1...ZERO.

thanks ryan_vietnow.
I have tried your code but it is not working.
Is there any other way to solve the problem?

[rime]

ahaa :) i forgot to change. it is belong to mine

NewsID stands for the row id, requests and assing the data to $Get which is posted

Thank you BUT
Still its not working..

[ryan_vietnow]

Is that so?I have tested it and it is working on mine.
Anyway,try to seperate the function to another page and include it,but do not use it as a function,make it just as an include page.Wish I could help you....