Broken images in php?

[Mr Chris]

Hi All,

 

Say I have a database record which holds an image name

 

<img src="http://www.gdgdgdgsks.co.uk/images/<?php echo $row['media_image']; ?>

 

The page will obviously be able to load the image, but say the image can't be found and it is broken ie the above outputs as:

 

<img src="http://www.gdgdgdgsks.co.uk/images/55.jpg

 

Which is a broken image.  Is there any in which I can use something in PHP which says if it's a broken image then output a default image (despite 55.jpg being the value saved in the database)

 

Thanks

 

Chris

 

[phpcodec]

It's not the code that is wrong, it's your image tags

 

Your Code:

<img src="http://www.gdgdgdgsks.co.uk/images/55.jpg">

 

Correct Code:

<img src="http://www.gdgdgdgsks.co.uk/images/55.jpg

[Mr Chris]

No sorry,

 

That's defo not the case, just my poor typing above(!).  Anyone know if I can do this?

 

Thanks

[mescal]

i would strip the .jpg in my database and put that in my php code

like

<img src="http://www.gdgdgdgsks.co.uk/images/<?php echo $row['media_image']; ?>.jpg">

 

does this help?

 

grtZ mescal

[Mr Chris]

Thanks, but I can't do it like that.  I'm running a parser which exports the filenames as filename.jpg and saves it in my database.

[mescal]

try this code

 

<img src="http://www.gdgdgdgsks.co.uk/images/<?php echo $row['media_image']; ?>">

[Wolphie]

Check if the row exists in the database, if it returns true; Check the database against the file system, i.e. check if the file actually exists in the file system. If either return false, specify a default image otherwise output the correct image.

[MadTechie]

try this

<?php
$path = "http://www.gdgdgdgsks.co.uk/images/";
$img = $row['media_image'];
if(!file_exists($path.$img))
{
$img = "broken.jpg";
}

echo "<img src=\"$path.$img\">";
?>

[mescal]

watch for the closing tags of your img src

[Mr Chris]

Hi Guys,

 

Thanks for the reply.  Here's how I have it:

 

  <?php
$path = "http://www.frededeeee.com/images/resize_image.php?image=http://www.frededeeee.com/images/properties";
$img = $prop['media_image_00'];

if(!file_exists($path.$img))
{
$img = "broken.jpg";
}

echo "<img src=\"$path/$img\" class=\"flb\" />";
?>

 

 

So

 

1) The value for 'media_image_00' in the database is outputted

2) The system then checks to see if the file exists

3) However, with the above if there is an image it still shows a broken.jpg image.

 

If I take out the if statement like so:

 

   <?php
$path = "http://www.frededeeee.com/images/resize_image.php?image=http://www.frededeeee.com/images/properties";
$img = $prop['media_image_00'];

if(!file_exists($path.$img))
{
$img = "broken.jpg";
}

echo "<img src=\"$path/$img\" class=\"flb\" />";
?>

 

The code works.  Now I think there is something wrong with the if statement:

 

if(!file_exists($path.$img))

 

 

Needs to be something like:

 

if(!file_exists($path/$img))

 

However if I use that it tries to use divide?

 

Thanks

[MadTechie]

the path is missing the /

so add the / at the end or try this

if(file_exists($path."/".$img))