CODEWITHSUNDEEP
phpmysqlhtml
Harry Mason
Harry Mason

Reputation: 39

PHP images from function

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.

Upvotes: 2

Views: 87

Answers (2)

Craig Taub
Craig Taub

Reputation: 4179

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.

Upvotes: 1

Jorge Olivares
Jorge Olivares

Reputation: 1492

The problem is file_exists is only for local files.

For example:

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

Upvotes: 0

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