我有这段代码,它适用于所有浏览器,除了Safari:
// Create a connection to the file.
var Connect = new XMLHttpRequest();
// Define which file to open and
// send the request.
Connect.open("GET", "<?php echo $fileName;?>", false);
Connect.setRequestHeader("Content-Type", "text/xml");
Connect.setRequestHeader("Cache-Control", "no-cache");
Connect.send();
// Place the response in an XML document.
var TheDocument = Connect.responseXML;
// Place the root node in an element.
var questions = TheDocument.childNodes[0];
// Retrieve each customer in turn.
for (var i = 0; i < 1; i++) {
console.log(i);
console.log("fake var");
var question = questions.children[0];
console.log(question);
}
变量$filename是指向我正在阅读的XML文件的php链接。我有以下问题,代码运行到For循环,我可以记录var"I",和"假var"。但一旦我定义var问题为问题。Children[0],代码停止。我无法在safari中记录问题变量。
任何想法?(也欢迎反馈,仍在学习javascript/xml)
我的XML示例:
<?xml version='1.0' encoding="UTF-8" ?>
<Beantwoordt_kritiek id="1" admin="Jan Hut" categorie="Omgaan met elkaar" winswf="images/Kado (1).jpg">
<gebruikers_vraag1>
<vraag><![CDATA[Sleep de puzzelstukken naar het juiste vakje.]]></vraag>
<s_vraag><![CDATA[sounds/puzzel.mp3]]></s_vraag>
<thumb1><![CDATA[images/527puzzel1_1.jpg]]></thumb1>
<thumb2><![CDATA[images/527puzzel1_2.jpg]]></thumb2>
<thumb3><![CDATA[images/527puzzel1_3.jpg]]></thumb3>
<thumb4><![CDATA[images/527puzzel1_4.jpg]]></thumb4>
<vraagFoto><![CDATA[images/527puzzel1.jpg]]></vraagFoto>
</gebruikers_vraag1>
</Beantwoordt_kritiek>
您可以使用这个脚本,这样safari会返回相同的结果。
<script>
function getChildren(element) {
if (element.children == undefined) {
var childNodes = element.childNodes
var children = []
for(var i = 1; i < childNodes.length; i += 2) { // take every second element
children.push(childNodes[i]);
}
return children;
} else {
return element.children
}
}
</script>
然后在脚本中调用它
var question = getChildren(questions)[0];