PHP images from function

Question

I am a php newbie so please go easy.

I created queries for imgFld and imageFldName. I am trying to find why my images from my db are not being displayed.

I have the below code:

image_show(stripslashes($row['imgFld']),stripslashes($row['imageFldName']));

echo ' '.$records_num;

function image_show($name_image, $alt_tag) {

    if (file_exists("mywebsite.co.uk/images/'$name_image'")) {
        $img = getimagesize('mywebsite.co.uk/images/'.$name_image);
        echo '<img src="mywebsite.co.uk/images/'.$name_image.'" alt = '.$alt_tag.' border=0 align="bottom"';
        echo 'width = '. $img[0] .' height = ' .$img[1] . ' />';
    } else {
        echo 'Add an image here';
    }

}

Im getting the image names from a column inside my db and each column has an 'image.jpg', connecting it with the img src script from HTML so that I can display the images from the db.

However no images are being displayed and I cant find the error. Doesnt seem like anything is wrong. When I echo $name_image nothing is produced.

shouldn't you add http:// before the image location? :)

Vlad Preda
– Vlad Preda

2013-01-29 12:28:55 +00:00
Commented Jan 29, 2013 at 12:28 — Vlad Preda
– Vlad Preda, Commented Jan 29, 2013 at 12:28
the image location is a local static folder

Harry Mason
– Harry Mason

2013-01-29 12:47:01 +00:00
Commented Jan 29, 2013 at 12:47 — Harry Mason
– Harry Mason, Commented Jan 29, 2013 at 12:47

Harry Mason · Accepted Answer · 2013-01-29 12:58:54Z

1

In that case it means that there is nothing populating the $name_image variable.

Assuming that website directory is a local one, the cause of this is most likely your arguments when you call the image_show function. They do not match the order you have specified.

The first argument should be the name and the second the alt text, as defined:

function image_show($name_image, $alt_tag)

However you are passing the id to $name_image and the name as $alt_tag.

That should be it.

edited Jan 29, 2013 at 12:58

Harry Mason

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answered Jan 29, 2013 at 12:28

Craig Taub

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Comments

Marcel Korpel · Accepted Answer · 2013-01-29 12:44:36Z

0

The problem is file_exists is only for local files.

For example:

if (file_exists($_SERVER['DOCUMENT_ROOT'] . '/path/to/files/image.jpg')) {
...
}

edited Jan 29, 2013 at 12:44

Marcel Korpel

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answered Jan 29, 2013 at 12:33

Jorge Olivares

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2 Comments

Harry Mason Over a year ago

the directory of the images is local..I had given it a fake name to post on here. The directory is not mywebsite.co.uk

Jorge Olivares Over a year ago

another typo ? hehe... try to use this, including documen_root. Greatings!

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