CODEWITHSUNDEEP
jqueryhtmlcssimage
DevB1
DevB1

Reputation: 1575

Setting background image of div using a variable URL (jQuery)

I'm creating a user signup page and I want the user to be able to enter an image URL and then preview the image in a reserved div (which has a default image in it already, applied by a given class). My HTML for the user image display is:

<div id="user-image-preview" class="col-md-4 col-sm-12 user-image-preview">  </div>
<button id="preview-image">Preview image</button>

The CSS for id'user-image-preview' and class 'user-image-preview' are:

#user-image-preview {
height: 300px;
width: 300px;
border-radius: 50%;
position: absolute;
left: 200px;
top: 150px;
background-size: cover;
}

.user-image-preview {
background-image: url("https://cdn1.iconfinder.com/data/icons/unique-round-blue/93/user-512.png");
background-size: cover;
}

Here is the input part of the form where user inputs image URL:

<div class="form-group">
    <i class="fa fa-picture-o" aria-hidden="true"></i>
    <label for="img">Image</label>
        <input id="user-image-set" type = "text" class ="form-control" placeholder = "Enter image URL" name = "image">
 </div>
 <button id="submit-login" type ="submit" class="btn btn-primary btn-lg  disabled">Signup</button>

And then I wrote the following jQuery to remove the default image (class="user-image-preview") and then add whatever is in the input field:

$("#preview-image").on("click", function() {
var newUserImage = $('#user-image-set').val();
    $("#user-image-preview").removeClass("user-image-preview");
    $("#user-image-preview").css("background-image", "url('newUserImage')");
    console.log(newUserImage)
})

(I added the console.log just to check if I am capturing the variable correctly which I am).

I presume the is a problem with my use of quotes within the URL part of the function, but I can't quite work out what the correct way to cite this variable is.

Upvotes: 1

Views: 2220

Answers (2)

Soukaina WARACH
Soukaina WARACH

Reputation: 81

This solution worked for me:

$("#user-image-preview").css("background-image", "url(' " + newUserImage + "')");

you need to add ''

Upvotes: 0

Web pundit
Web pundit

Reputation: 498

As newUserImage is a variable, you don't have to put it in quotes '' so change this line

$("#user-image-preview").css("background-image", "url('newUserImage')");

to 

$("#user-image-preview").css("background-image", "url(" + newUserImage + ")");

Upvotes: 2

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