displaying varying amounts of images

[paramedic]

Hi guys

I am having a bit of a problem figuring out how to display random amounts of pictures

I allow people to upload up to 5 photos to my website. I store the file name in a database and the actual photos seperately.

At the moment i am displaying the photos using the following code.


Print "<td class=border width=20% valign=top rowspan=2 ><img src=http://www.propertyadsdirect.com/images/".$info['pic'] ."></a> <br>";
Print "<img src=http://www.propertyadsdirect.com/images/".$info['pic2'] ."></a> <br>";
Print "<img src=http://www.propertyadsdirect.com/images/".$info['pic3'] ."></a> <br>";
Print "<img src=http://www.propertyadsdirect.com/images/".$info['pic4'] ."></a> <br>";
Print "<img src=http://www.propertyadsdirect.com/images/".$info['pic5'] ."></a> <br>";
Print "</td >";

However if the user has not uploaded the full 5 photos a image is displayed in its place to indicate that an image is supposed to be there. How can I check that a photo is present and if not stop the site displaying this other image.

Many thanks for your help.

[World Wide Weird]

Use file_exists to check whether the named file exists.

[paramedic]

Thanks for the reply. I expected something more complicated than that. I will give it a go tomorrow and let you know how it went.

Many thanks

[World Wide Weird]

Complicating your life is one thing PHP doesn't try to do.

A bit of advice: download the CHM (Windows Help) version of the PHP Manual and keep it open while coding. You can usually find a function in seconds that does whatever you want to do.

Good luck!

[paramedic]

Hi I have tried adding file_exists and a loop ( not sure if it is right as it is the first time I have used a loop) and I am reciving the message

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in E:\domains\p\propertyadsdirect.com\user\htdocs\showall.php on line 110

Line 10
refers to Print "<td class=border width=20% valign=top rowspan=2 ><img src=http://www.propertyadsdirect.com/images/".info.$piccy"></a> <br>";


******************
CODE
$piccy = array("pic","pic2","pic3","pic4","pic5");

if (file_exists($piccy))
{
Print "<td class=border width=20% valign=top rowspan=2 ><img src=http://www.propertyadsdirect.com/images/".info.$piccy"></a> <br>";
}
else
{

any ideas what I am doing wrong?


Many thanks

[spikeZ]

Hi
file_exists looks for the file on the server so:

$pic1 = "http://www.propertyadsdirect.com/images/". $info['pic2']; 

if(file_exists($pic1)) {
    echo  '<img src="'. $pic1 .'" />';
} 

Something like that should work.....

*disclaimer - I'm goosed so that might be right!*

[Jake Arkinstall]

$piccies = array('pic', 'pic2', 'pic3', 'pic4', 'pic5');
$base = 'http://www.propertyadsdirect.com/images';
echo '<td>';
foreach($piccies as $piccy){
    if (file_exists($piccy)){
        echo "<img src=\"{$base}/{$piccy}\" /><br />";
    }
}
echo '</td>'; 

*disclaimer: It's right

[spikeZ]

*disclaimer: It's right

[paramedic]

Hi thanks for your help. I have changed my code to the code you suggested, but unfortunately no pictures are displayed.
I used the following code to get the info from the database

$data = mysql_query("SELECT * FROM propertydetails WHERE id = '$houseId'")

or die(mysql_error());


while($info = mysql_fetch_array( $data ))

Should I use $info in the code you gave me?


Many thanks