我使用类似的方法按条件单击页面上的元素:
var findElem = function(elems, text) {
for (var i = 0; i < elems.length; i++) {
if (elems[i].textContent == text) {
return elems[i];
} else {
var result = findElem(elems[i].children, text);
if (result != undefined) {
return result;
}
}
}
return;
}
switch (document.getElementById('my_id').value) {
case "1":
findElem(document.documentElement.children, "blabla1").click();
break;
case "2":
findElem(document.documentElement.children, "blabla2").click();
break;
default:
break;
}
我想在此代码中再添加一个条件。
我正在工作的页面是这样的:
<body id="something">
<div id="something2" class="container">
<div id="header">
<a class="hmm" href="somelink"></a>
<a class="aname" target="_blank" href="anotherlink"></a>
</div>
<div id="anotherthing" class="anotherthing2">
<div class="differentthing " style="left: 2px; ">
<div class="asdafad"></div>
</div>
我必须看看是否 differentthing没有style
简而言之，我想做这样的事情:
switch (document.getElementById('my_id').value) {
case "1":
---if differentthing has no style attrib then---
findElem(document.documentElement.children, "blabla1").click();
break;
所以如果 differentthing页面部分是这样的
<div class="differentthing ">
然后做findElem.click东西。
如果是<div class="differentthing " style="left: 2px; ">然后什么也不做。
我希望我能描述一下。我怎样才能做到这一点？抱歉我的英语不好。
最佳答案
switch (document.getElementById('my_id').value) {
case "1":
// getting an array of elements with "differentthing" class
var elements = document.getElementsByClassName("differentthing");
// if there is just one and it does not have "style" attribute
if (elements.length == 1 && !elements[0].getAttribute("style")){
// perform some action
findElem(document.documentElement.children, "blabla1").click();
}
break;
关于javascript - 如何检查页面中的某个位置？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11038262/