jQuery的图像设置背景渐变不适用于css（）

Question

This is working

 div { width: 100px; height: 100px; background-image: -webkit-linear-gradient(top, rgba(0, 0, 0, 0)0%, rgba(0, 0, 0, 0.65) 100%), url('http://i.imgur.com/Hsban3N.jpg'); } 

 <div id="deco">123</div> 

but why when I try to set it using css() of jquery it doesn't seem working? No error in the console at all :

http://jsfiddle.net/xcso27zk/

Answer 1

From jQuery 1.8.8 on, it prefixes for you, so all you'll need to type is

$('#deco').css({
    'background-image': 'linear-gradient(to bottom, rgba(0,0,0,0) 80%, rgba(0,0,0,.65) 100%), url(' + img + ')'
})

Also. If you type something like this

$('#deco').css({
    'background-image': '-webkit-gradient(linear, left top, left bottom, color-stop(80%, rgba(0,0,0,0)), color-stop(100%, rgba(0,0,0,.65))), url(' + img + ')', 
    'background-image': '-webkit-linear-gradient(top,       rgba(0,0,0,0) 80%, rgba(0,0,0,.65) 100%), url(' + img + ')', 
    'background-image': '   -moz-linear-gradient(top,       rgba(0,0,0,0) 80%, rgba(0,0,0,.65) 100%), url(' + img + ')', 
    'background-image': '     -o-linear-gradient(top,       rgba(0,0,0,0) 80%, rgba(0,0,0,.65) 100%), url(' + img + ')', 
    'background-image': '        linear-gradient(to bottom, rgba(0,0,0,0) 80%, rgba(0,0,0,.65) 100%), url(' + img + ')'
})

The object won't work like you desire in some browsers. You will simply rewrite 'background-image' 4 times, essentially writing useless code.

Answer 2

JQuery's css function doesn't accept a set of objects in a single call.

Instead of

$(...).css({},{},{})

Try

$(...).css({})
  .css({})
  .css({})

 var img = 'http://i.imgur.com/Hsban3N.jpg'; $('#deco').css({ 'background-image': '-moz-linear-gradient(top,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')' }).css({ 'background-image': '-webkit-gradient(linear,left top,left bottom,color-stop(80%,rgba(0,0,0,0)),color-stop(100%,rgba(0,0,0,0.65))), url(' + img + ')' }).css({ 'background-image': '-webkit-linear-gradient(top,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')' }).css({ 'background-image': '-o-linear-gradient(top,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')' }).css({ 'background-image': 'linear-gradient(to bottom,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')' }); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="deco">123</div> 

Answer 3

You are passing multiple javascript objects to the css method, but it only takes one javascript object with multiple name-value pairs. So like this:

$('#deco').css({
    'background-image': '-moz-linear-gradient(top,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')', 
    'background-image': '-webkit-gradient(linear,left top,left bottom,color-stop(80%,rgba(0,0,0,0)),color-stop(100%,rgba(0,0,0,0.65))), url(' + img + ')', 
    'background-image': 'background-image: -webkit-linear-gradient(top,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')', 
    'background-image': '-o-linear-gradient(top,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')', 
    'background-image': 'linear-gradient(to bottom,rgba(0,0,0,0)80%,rgba(0,0,0,0.65)100%), url(' + img + ')'
});

with only one set of brackets