Php to return value

Question

EDIT: Check at the end of this for the solution

I am new to php, ajax and all other things :)

My question is: is there a way for a php file to return a value?

I have:

a file "loadImages.php" containing the script to get the paths to images from the database.
an image gallery where I want to load images via ajax.
a database containing the path to the images (/images/image1.jpg, /images/image2.jpg).
4 categories of images.

What i'm trying to do is :

When clicking on a link (example, first category), I want to call via jquery's ajax(), loadImages.php with the category passed via POST (cat0, cat1, cat2, ...)

I want to return the value from this php file for example : <img src="image/image1.jpg" />, so via javascript, I can retrieve this string. Using only return in my php file returns XMLHttpRequest when i'm putting the ajax() function in a variable instead of the string <img>.

Is there a way to do it? Or maybe a better way, as I don't fully understand ajax.

Thank you! Sorry for my bad grammar.

Below is a more precise map of what i'm trying to do.

JavaScript:

var test = $.ajax({  
type: "POST",  
    url: "loadImages12.php",  
    data: "category=0",  
    complete: function() {  
        alert("COMPLETE");  
    },  
    error: function (){  
        alert("NOT LOADED");  
    }  
});

PHP (loadImages.php)

function createThumb() { 
        if(isset($_POST['category'])) {  
            $imageQuery = mysql_query("SELECT * FROM t_pictures WHERE p_category = 0");
            $thumbHtml = '';

            while ($tempImageQueryFetch = mysql_fetch_assoc($imageQuery)){
                $thumbHtml .= '<a href="#" class="thumbnail"><img src="ressources/images/' . $tempImageQueryFetch["p_fileName"] . 'Small.jpg" /></a>';
            }

        return $thumbHtml;
        }
        else {
            $noCategory = "NO CATEGORY TEST";
            return $noCategory ;
        }
    }

    createThumb();

SOLVED

Php File

function createThumb(){
//Mysql request
$someVar = //Result of request
return $someVar
}

echo createThumb()
Javascript

$("#someDiv").load("loadImages.php", {category:0});

I'm new to AJAX, too. I use $.post to send XHRs and in my PHP file, I "echo" the response (I don't use functions). Try using the $.post method. — afaolek
– afaolek, Commented Jun 14, 2011 at 3:39
Thanks, that's what I did! The load function is easier to use finally. — Elggetto
– Elggetto, Commented Jun 14, 2011 at 14:53

deceze · Accepted Answer · 2011-06-14 03:33:00Z

5

An AJAX request to PHP is exactly the same as a normal request for a website. You request data from example.com/foo/bar and in return you receive text. The return statement of PHP has nothing to do with what's returned to the client. Only things you echo or otherwise output will be received by the client. That's the same for normal pages and AJAX requests.

So, to "return" <img src="image/image1.jpg" /> to your Javascript AJAX call, echo that string in PHP.

answered Jun 14, 2011 at 3:33

deceze♦

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1 Comment

Elggetto Over a year ago

Thank you for the answer. I used : $("someDiv").load("loadImages", {category:0}) instead. And I echo(ed) the function in my php file.

dimirc · Accepted Answer · 2011-06-14 05:36:55Z

2

if you want to use the php file that you already have (loadImages.php) just echo the result of the function:

echo createThumb();

answered Jun 14, 2011 at 5:36

dimirc

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Comments

Vamsi Krishna B · Accepted Answer · 2011-06-14 03:44:37Z

0

Just make the php script to return the path to thumbnail.

and then do something like this in js

somediv.html('<a href="#" class="thumbnail"><img src="">');

and just use the path returned by the php file for the img src ..

answered Jun 14, 2011 at 3:44

Vamsi Krishna B

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