我正在尝试使用jquery的css()方法。但问题是当我使用css()方法显示无它工作正常,但当我将其替换为背景图像属性,然后css()根本不工作。我在做什么错了?请注意:背景图片图片将从托管链接加载,如下面的代码。
工作代码:
$("#register-page").css({"background-image" : "url("https://academy.zikanalytics.com/wp-content/themes/boss-child/buddypress/members/gears.svg")" });
不起作用:
swagger.json
答案 0 :(得分:0)
由于错误使用了'
和$("#register-page").css({"background-image" : "url('https://academy.zikanalytics.com/wp-content/themes/boss-child/buddypress/members/gears.svg')" });
as <- c(1,2) # limit for K
bs <- c(3,4)
cs <- c(2,3) # distribution of food sources ('random'/'clustered')
p_space <- list ()
for (a in as){
for (b in bs){
for (c in cs){
p_space[[length(p_space)+1]] <- c(a,b,c)
}
}
}
made_up <- function(a, b, c){
return(a * b * c)
}
答案 1 :(得分:0)
当您通过纯CSS或JQuery设置背景图像时,显示元素将坚持其先前的值,因为它从未真正更改过。
$("#register-page").css({
"background-image": "url('https://academy.zikanalytics.com/wp-content/themes/boss-child/buddypress/members/gears.svg')",
"display": "block"
});