Solved

Need to export all jpeg files (blob entries) to filesystem

Posted on 2008-06-16
5
2,311 Views
Last Modified: 2013-12-13
We have an old Linux box setup years ago as a PHP/MySQL based webserver.  We're upgrading it to a new server, and need to get all the images (jpeg type) that are stored in the sql database out to the filesystem.  I found a PHP script to do just that, but I'm having problems.  I think my problems are that this server is running PHP 4.3.2.  We have no expertise in Linux, or PHP, so we're trying to extract this info without putting effort into upgrading the system first.  I know this has got to be a simple thing to do.  Any ideas?  Oh, the way it's to name the file, is to take and combine the info in 2 of the fields in the table, and append .jpg on the end of it.

I have attached my code.  What's happing when I run it, is that it appears to be bypassing the PHP engine because it simply shows me the PHP code in my browser.  I've tested PHP with the phpinfo() function and that works fine.

Thanks for your help.
< ?php
$location="/var/www/website.com/html/imagearchive/";
 
$db = mysql_pconnect("localhost");
mysql_select_db("database",$db);
 
$sql="select person.name, media.id, media.data from media LEFT JOIN person ON (person.id=media.l_table_name_id)";
 
$rs = mysql_query($sql,$db);
 
echo "<h2>EXPORTING INTO $location</h2>";
$counter=0;
while ($row=mysql_fetch_object($rs)) {
$filename=$location.$row->person.name"."media.id..jpg;
$file=fopen($filename,w');
if (fwrite($file,$row->image)) {
echo $filename.<br />;
$counter++;
}
}
 
echo <h2>$counter images exported</h2>;
? >

Open in new window

0
Comment
Question by:ResortCompanies
5 Comments
 

Expert Comment

by:qaelan
ID: 21798939
You might try removing the spaces between

< ?php

and

? > 

At the start and end of the file. They should be <?php and ?>, respectively
0
 
LVL 18

Expert Comment

by:Matthew Kelly
ID: 21799017
Try this code. The php being output to the screen is caused by the space between the ? and the > sign at the start and end.

There are also a few php syntax errors, which I corrected below.

Let me know if I did not code it right.


<?php
$location="/var/www/website.com/html/imagearchive/";
 
$db = mysql_pconnect("localhost");
mysql_select_db("database",$db);
 
$sql="select person.name, media.id, media.data from media LEFT JOIN person ON (person.id=media.l_table_name_id)";
$rs = mysql_query($sql,$db);
 
echo "<h2>EXPORTING INTO ".$location."</h2>";
 
if ( $rs && mysql_num_rows($rs) > 0 )
{
	$counter=0;
	for ( $i=0; $i < mysql_num_rows($rs); $i++ )
	{
		$name = mysql_result($result,$i,0);
		$id = mysql_result($result,$i,1);
		$data = mysql_result($result,$i,2);
		$filename = $location.$name.$id.".jpg";
		$file=fopen($filename,'w');
		if (fwrite($file,$data)) 
		{
			echo $filename."<br />";
			$counter++;
		}
	}
}
else
{
echo "No files found";
}
 
echo "<h2>".$counter." images exported</h2>";
?>

Open in new window

0
 
LVL 29

Accepted Solution

by:
fibo earned 500 total points
ID: 21800367
Some typos and errors with ' and ".
Code is easier to grasp and debug when indented.

If doing lots of development, you shoud consider using commercial PHP IDEs like Zend or NuCoder: they would help you spotting at once these typos.
They offer trial versions that you can use for free for 2 or 4 weeks: this will help you to determine if the time you save is worth the money. They both have a "personal" version at $100-120.

<?php
$location="/var/www/website.com/html/imagearchive/";
 
$db = mysql_pconnect("localhost");
mysql_select_db("database",$db);
 
$sql="select person.name as my_name, media.id as my_id, media.data as my_data 
from media LEFT JOIN person ON (person.id=media.l_table_name_id)";
 
$rs = mysql_query($sql,$db);
 
echo "<h2>EXPORTING INTO $location</h2>";
$counter=0;
while ($row=mysql_fetch_object($rs)) {
  $filename=$location.$row->my_name . '.' . $row->my_id . '.jpg';
  echo "$filename :";
  $file=fopen($filename,'w');
  if (fwrite($file,$row->my_data)) {
    echo "OK<br />";
    $counter++;
  } else echo 'fail **<br>';
}
 
echo "<h2>$counter images exported</h2>";
?>

Open in new window

0
 

Author Closing Comment

by:ResortCompanies
ID: 31467823
Thanks, worked like a charm.
0
 
LVL 29

Expert Comment

by:fibo
ID: 21805535
Glad it worked, thx for the points
0

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Build an array called $myWeek which will hold the array elements Today, Yesterday and then builds up the rest of the week by the name of the day going back 1 week.   (CODE) (CODE) Then you just need to pass your date to the function. If i…
This article discusses how to create an extensible mechanism for linked drop downs.
The viewer will learn how to dynamically set the form action using jQuery.
This tutorial will teach you the core code needed to finalize the addition of a watermark to your image. The viewer will use a small PHP class to learn and create a watermark.

685 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question