CODEWITHSUNDEEP
phpmysqlimage
Simon Carlson
Simon Carlson

Reputation: 1989

Scaling pictures

I'm displaying pictures within a div that is 350px x 350px big. People can upload pictures to my fileserver and a destination link to that picture is created in a database as well as an id. The php document getting the link is in the root folder and all pictures are saved in a subfolder called pictures. I show the picture using something like this:

$piclink = "SELECT `link` FROM `pictures` WHERE `id=`.$id.`";

The $id variable is stored when clicking a link from the previous page.

My problem: Not all pictures are the same format. I want to display them so that the max dimension is 350px, the other dimension according to proportion. How do I do this? Simply doing this:

echo "<img href=" . $piclink . " height='350px' width='350px'/>"

will stretch the picture to 350x350 thus breaking proportions. Any help appreciated.

Upvotes: 1

Views: 125

Answers (1)

ron tornambe
ron tornambe

Reputation: 10780

To retain the aspect ratio, you can retrieve the image dimensions and apply the ratio using the 350px maximums.

Here's an example:

HTML

<div id="img-holder"></div>

Code

var img = new Image();
img.onload = function () {
    alert('Orignal width:'+img.width+', height:'+img.height);
    var width, height;
    if (img.width > img.height) {
        width = (img.width > 350 ? 350 : img.width);
        height = img.height * (350 / img.width);
    } else {
        height = (img.height > 350 ? 350 : img.height);
        width = img.width * (350 / img.height);
    }
    img.width=width;
    img.height=height;
    $("#img-holder").append(img);
}
img.src = "http://mps.thebeautyquotient.com/images/val4a.jpeg"

Here's the jsfiddle: http://jsfiddle.net/aN6Ec/

Upvotes: 1

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