在我的一生中，我看不到我错过了什么。我在 drawmenu 函数中硬编码了 selectedIcon 的值。在我将其更改为参数之前，它一直运行良好。我收到以下错误:
Error: this.call is not a function
Source File: http://localhost:9090/tests/jquery-1.3.2.min.js
Line: 19
例程是绘制一个菜单，同时动态调整图像之间的间距。
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<meta name="generator" content="HTML Tidy for Windows (vers 14 February 2006), see www.w3.org" />
<meta http-equiv="Content-Type" content="text/html;charset=utf-8" />
<title>
Test image overlay
</title>
<script type="text/javascript" src="jquery-1.3.2.min.js">
</script>
<script type="text/javascript">
//<![CDATA[
function drawmenu (selectedIcon) {
var offset = 0;
for (x in icons) {
// Add the gap at the top of the image
if (selectedIcon-1 == x) {
offset += icons[x].topBig;
$("#"+icons[x].divid).toggleClass("active");
$("#"+icons[x].imgid).toggleClass("rollit");
} else {
offset += icons[x].topSmall;
}
// Set the location of the image
$("#"+icons[x].divid).css("top",-offset);
// Add the gap at the bottom of the image
if (selectedIcon-1 == x) {
offset += icons[x].bottomBig;
} else {
offset += icons[x].bottomSmall;
}
$("#"+icons[x].imgid).attr("src",icons[x].image);
}
}
var icons = [
{
"divid" : 'icon1', // Home
"imgid" : 'imgicon1',
"image" : 'home.png',
"topSmall" : 5,
"topBig" : 5,
"bottomSmall" : 5,
"bottomBig" : 7
},{
"divid" : 'icon2', // Alert
"imgid" : 'imgicon2',
"image" : 'alert.png',
"topSmall" : 7,
"topBig" : 13,
"bottomSmall" : 0,
"bottomBig" : 0
},{
"divid" : 'icon3', // Question
"imgid" : 'imgicon3',
"image" : 'question.png',
"topSmall" : 4,
"topBig" : 8,
"bottomSmall" : 5,
"bottomBig" : 7
},{
"divid" : 'icon4', // Lightbulb
"imgid" : 'imgicon4',
"image" : 'lightbulb.png',
"topSmall" : 3,
"topBig" : 7,
"bottomSmall" : 5,
"bottomBig" : 7
},{
"divid" : 'icon5', // Blog
"imgid" : 'imgicon5',
"image" : 'blog.png',
"topSmall" : 3,
"topBig" : 6,
"bottomSmall" : 1,
"bottomBig" : 4
},{
"divid" : 'icon6', // Defect
"imgid" : 'imgicon6',
"image" : 'defect.png',
"topSmall" : 7,
"topBig" : 10,
"bottomSmall" : 0,
"bottomBig" : 0
}]
$(document).ready(drawmenu(6));
//]]>
</script>
<style type="text/css">
/*<![CDATA[*/
#iconstack {
position: absolute;
top:30px;left:100px;
}
.icondiv { z-index:1; position: relative;}
.active { height:80px; }
.rollit {
height:50px;
}
.rollit:hover {
background-image:url('highlight.png');
background-color:red;
}
/*]]>*/
</style>
</head>
<body>
<div id="iconstack">
<div id="icon1" class="icondiv">
<img id="imgicon1" class="rollit" src="" alt="" />
</div>
<div id="icon2" class="icondiv">
<img id="imgicon2" class="rollit" src="" alt="" />
</div>
<div id="icon3" class="icondiv">
<img id="imgicon3" class="rollit" src="" alt="" />
</div>
<div id="icon4" class="icondiv">
<img id="imgicon4" class="rollit" src="" alt="" />
</div>
<div id="icon5" class="icondiv">
<img id="imgicon5" class="rollit" src="" alt="" />
</div>
<div id="icon6" class="icondiv">
<img id="imgicon6" class="rollit" src="" alt="" />
</div>
</div>
</body>
</html>
最佳答案
这实际上是一个小问题。你之前做的是这样的:
$(document).ready(drawmenu);
传递了一个函数。你现在正在做的是:
$(document).ready(drawmenu(6));
第二种方式是传递函数的结果而不是函数本身，等价于:
var item = drawmenu(6);
$(document).ready(item);
改为尝试:
$(document).ready(function(){ drawmenu(6); });
关于javascript - this.call 不是函数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/1805121/