How can I pull a random image from MySQL with PHP? - php

Question

I'm learning PHP/MySQL while working on a web app. Part of this app calls for echoing out a random image from an image directory titled 'htdocs/full'. The database is holding the path to each image in a column titled 'image_path' in addition to columns for 'first_name' 'last_name'.
I'm able to make it call a random row from the database and I can make it display literal data from the 'first_name' and 'last_name' columns. But I'm struggling to make it take the path from the 'image_path' column and converting it into an image rather than a literal.
I've been experimenting with a number of methods over the last 24 hours, but most everything else I've been seeing on Stack and Google is for pulling an image straight out of the directory without using the database row (which is needed due to the corresponding 'first_name' and 'last_name').
Here is what I have so far (although I also have many different variations, this I think is the closest I've come):
<?php
$username="root";$password="*********";$database="offenders";
mysql_connect('localhost',$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$sSQLQuery = "SELECT image_path FROM offenders ORDER BY RAND() LIMIT 1";
$aResult = mysql_query($sSQLQuery);
WHILE($aRow = mysql_fetch_array($aResult)):
{
header("Content-Type:image/jpeg");
echo "<img src='php/imgView.php?imgId=".$aRow['image_path']."' />";
}
ENDWHILE;
?>
It produces a page with a tiny broken image at the top left. I've tried this with echoing $aRow as well as $aRow['image_path'] as seen above and a number of other tweaks that seems to cause error. Any ideas what I'm doing wrong?
Thanks a lot!
Updated code:
<?php
$username="root";$password="*******";$database="offenders";
mysql_connect('localhost',$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$src="full/";
//if you have any problem in the solution check this line,
//means weather you are providing right path for your files or not.
$sSQLQuery = "SELECT image_path FROM offenders ORDER BY RAND() LIMIT 1";
$aResult = mysql_query($sSQLQuery);
$aRow = mysql_fetch_array($aResult);
$file_path=$src.$aRow["image_path"];
//echo "<a href='php/imgView.php?imgId=".$aRow['image_path']."'>"."<img src=".$file_path." />"."</a>".'<br/>';
echo "<img src=".$file_path." />".'<br/>';
//echo "<img src=full/1b2ba2f9a8b8ee9c062b09767535d69bd954e125.jpg>";
/* if(isset($_GET['imgId']))
{
//i am query only first_name but you can add for field as you want
//and i also used $_get['imgId']; in the query which is not safe but you can make it for //secure.
$sSQLQuery = "SELECT first_name FROM offenders where image_path='".$_GET['imgId']."'";
$aResult = mysql_query($sSQLQuery);
WHILE($aRow = mysql_fetch_array($aResult)):
{
echo $aRow['first_name'];
}
ENDWHILE;
}
*/
?>

Answer 1

There are several things wrong here.
You have your database output in a loop. Instead of
WHILE($aRow = mysql_fetch_array($aResult)):
{
...
}
ENDWHILE;
use:
$aRow = mysql_fetch_array($aResult);
...
You are returning what looks like HTML, but you have a content type of JPG. Either return HTML and use an appropriate content type or return a JPG, and return that.
If you want to return the JPG itself (and not the HTML), use CURL to grab the write image and return it.

Answer 2

In my view i dont know why you are running the while loop because in your query you have limit of one so you dont have to run the while loop.
lets say you have your images in the folder name php/image/than your images comes
I am writing my solution on the assumption that your images are in the image folder, if they are not you can simply change the $src value in the solution.
and i am also thinking you want to use these images as a link, so you can get the first_name, and last_name when some one click on the image, reason i am thinking like this, that you try to give id in the img src="",
any way i am writing two solution see which one work for you. first one gonna make them link and second one only gonna pull the images from the database.
First one on the assumption you want to pull the first_name and last_name, from the database, when some one click on the image.
<?php
$username="root";$password="";$database="offenders";
mysql_connect('localhost',$username,$password);
mysql_se
lect_db($database) or die( "Unable to select database");
$src="php/image/";//if you have any problem in the solution check this line,
//means weather you are providing right path for your files or not.
$sSQLQuery = "SELECT image_path FROM offenders ORDER BY RAND() LIMIT 1";
$aResult = mysql_query($sSQLQuery);
$aRow = mysql_fetch_array($aResult);
$file_path=$src.$aRow["image_path"];
echo "<a href='php/imgView.php?imgId=".$aRow['image_path']."'>"."<img src=".$file_path." />"."</a>".'<br/>';
if(isset($_GET['imgId']))
{
//i am query only first_name but you can add for field as you want
//and i also used $_get['imgId']; in the query which is not safe but you can make it for //secure.
$sSQLQuery = "SELECT first_name FROM offenders where image_path='".$_GET['imgId']."'";
$aResult = mysql_query($sSQLQuery);
WHILE($aRow = mysql_fetch_array($aResult)):
{
echo $aRow['first_name'];
}
ENDWHILE;
}
?>
Second solution only pull the images from the database
remove the isset condition and while loop in the isset condition.
and change this line
echo "<a href='php/imgView.php?imgId=".$aRow['image_path']."'>"."<img src=".$file_path." />"."</a>".'<br/>';
To
echo "<img src=".$file_path." />".'<br/>';
//if you notice the only different is this that i
//remove the <a> tag in its path, so not it gonna pull images
// from the database but you cant use them as link.
Thanks
this solution work fine if it does not please post your folder structure, means where you have images and this file imgView. and second note i removed your while loop from the solution,
//