JavaScript alert dynamically created variable name

Question

I got the follow code:

var image1 = 'aaa.jpg';
var image2 = 'bbb.jpg';
var image3 = 'ccc.jpg';
var image4 = 'ddd.jpg'; // all dynmically created

current = 1; // dynamically changing value, in this case between 1 and 4

What I need now is to alert alert(image1); but on this way: alert(image + current);

Is it even possible this way? I hope I explained clearly what I need.

Thats a nice solution too Jack, not easy to create with PHP but would work :D — donbingo, Jul 09 '15 at 22:32
How do you create the variables using PHP currently? Seems a bit like a [XY problem](http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem). — insertusernamehere, Jul 09 '15 at 22:34

score 1 · Accepted Answer · answered Jul 09 '15 at 22:36

1

Do it this way:

this.image1 = "aaa.jpg";

alert(this["image" + "1"]);

answered Jul 09 '15 at 22:36

Joerg

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That works too so I accept that :) – donbingo Jul 09 '15 at 22:43

Miles · Answer 2 · 2015-07-09T22:32:59.447

0

From what I see, I think you've on the right path. But here are some things to consider...

Put the dynamically created variable in quotes to make them a string:

var image1 = 'aaa.jpg';

When you out put in the alert box put the word 'image' in quotes also. That way it will be a string too:

alert(eval('image'+current));

The alert box won't actually show the image. It will only show the name of the image that is stored in the variable. I think this is what you're going for.

Good Luck!

I see you edited the question to put it the dynamic variable in quotes. Well done.

edited Jul 09 '15 at 22:32

answered Jul 09 '15 at 22:26

Miles

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1: I did oO 2: It alerts "image1" but not the value of image1 3: wth xD – donbingo Jul 09 '15 at 22:29

score 0 · Answer 3 · edited Aug 23 '15 at 13:06

0

I solved it the way one user posted it and deleted his answer (why?):

alert(eval('image'+current));

edited Aug 23 '15 at 13:06

vaultah

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answered Jul 09 '15 at 22:26

donbingo

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insertusernamehere · Answer 4 · 2015-07-09T22:57:04.200

As it was suggested in the comments and based on your PHP code from the comments: Create a JavaScript array to store your data instead of creating lots of variables. You can accomplish it for example like this:

Create string containing all images

<?php
    $images = '';
    while($row = mysqli_fetch_assoc($result)) {
        $images .= ( $images ? ',' : '' ) . '"' . $row['image'] . '"';
    }
?>

Output

var images = [<?php print $images; ?>];
alert (images[0]);

score 0 · Answer 5 · answered Jul 10 '15 at 00:42

You're clearly working with a collection, so you should create an array structure in this case:

<?php 

$images = [];
while ($row = mysqli_fetch_assoc($ergebnis)) {
    $images[] = $row['image'];
}

?>
<script type="text/javascript">
var images = <?php echo json_encode($images) ?>;
</script>

You can then reference each image path via, e.g. images[0].

JavaScript alert dynamically created variable name

5 Answers5