使用jQuery设置浏览器特定的CSS属性 [英] setting browser-specific CSS properties with jQuery
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问题描述
我正在尝试修改background-image属性中的渐变值,我不能:(
data = 'ff55dd';
$(。el)。css({
'background-image':'-webkit-gradient(linear,left top,left bottom,from(#'+ data + '),to(#aa1133))',
'background-image':'-webkit-linear-gradient(#'+ data +',#aa1133)',
'background-image' :'-moz-linear-gradient(#'+ data +',#aa1133)',
'background-image':'-o-linear-gradient(top,#ff3345,#aa1133)',
'background-image':'-khtml-gradient(linear,left top,left bottom,from(#'+ data +'),to(#aa1133))',
'filter':' progid:DXImageTransform.Microsoft.gradient(startColorstr = \'#'+ data'+'\',endColorstr = \'#aa1133 \',GradientType = 0)',
'background-image' :'linear-gradient(#ff5534,#aa1133)'
});
发生.....
解决方案
看起来像覆盖规则,如果你一次把它们全部。
我逐一添加它们,似乎工作。
$ .e1)
.css('background-image',' - webkit-gradient(linear,left top,left bottom,from(#'+ data +'),to(#aa1133))')
.css('background-image',' - webkit-linear-gradient(#'+ data +',#aa1133)')
.css('background-image',' - moz-linear -gradient(#'+ data +',#aa1133)')
...
I'm trying to modify gradient values in the background-image property and I can't :(
data = 'ff55dd';
$(".el").css({
'background-image' : '-webkit-gradient(linear, left top, left bottom, from(#' + data + '), to(#aa1133))',
'background-image' : '-webkit-linear-gradient(#' + data + ', #aa1133)',
'background-image' : '-moz-linear-gradient(#' + data + ', #aa1133)',
'background-image' : '-o-linear-gradient(top, #ff3345, #aa1133)',
'background-image' : '-khtml-gradient(linear, left top, left bottom, from(#' + data + '), to(#aa1133))',
'filter' : 'progid:DXImageTransform.Microsoft.gradient(startColorstr=\'#' + data' + '\', endColorstr=\'#aa1133\', GradientType=0)',
'background-image' : 'linear-gradient(#ff5534, #aa1133)'
});
Nothing happens.....
解决方案
Looks like its overwriting the rules if you put them in all at once.
I added them one by one and it seems to work.
$(".e1")
.css('background-image','-webkit-gradient(linear, left top, left bottom, from(#' + data + '), to(#aa1133))')
.css('background-image','-webkit-linear-gradient(#' + data + ', #aa1133)')
.css('background-image','-moz-linear-gradient(#' + data + ', #aa1133)')
...
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