Javascript 错误:无法将对象转换为原始值
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Javascript Error: Cannot Convert Object to Primitive Value
提问by Eric
I'm receiving this error using the following javascript code:
我使用以下 javascript 代码收到此错误:
function tempTest(evt) {
alert(evt.currentTarget.id);
ct = document.getElementById(evt.currentTarget.id);
rslt = document.getElementById('rslt');
var props;
for (var prop in ct) {
if (ct.hasOwnProperty(prop)) {
propVal = ct[prop];
var propDat = prop + ' = ' + propVal;
props += propDat + '<br/>';
}
}
rslt.innerHTML = props;
}
This one has me puzzled. Any ideas?
这个让我很疑惑。有任何想法吗?
采纳答案by Itay Moav -Malimovka
Not all the properties of a HTML element are primitives. for example, parent, childs etc are also HTML elements. You can't just use them as strings or numbers.
You need to add there a condition and use that property accordingly.
并非 HTML 元素的所有属性都是基元。例如,父、子等也是 HTML 元素。您不能仅将它们用作字符串或数字。
您需要在那里添加一个条件并相应地使用该属性。
回答by Eliran Malka
(The OP:)
(OP:)
Just wanted to post the updated snippet for anyone who stumbles onto this post...
只是想为任何偶然发现这篇文章的人发布更新的片段...
function tempTest(evt) {
alert(evt.currentTarget.id);
ct = document.getElementById(evt.currentTarget.id);
rslt = document.getElementById('rslt');
var props;
for (var prop in ct) {
if (ct.hasOwnProperty(prop)) {
var propVal = ct[prop];
props += prop + ' (' + typeof(prop) + ')' + ' = ';
if (typeof(ct[prop]) == 'string') {
propVal += ct[prop];
} else {
if (propVal != null && propVal.toString) {
props += propVal.toString();
} else {}
}
props += '<br/>';
}
}
rslt.innerHTML = props;
}
回答by Glen Pierce
If the object in question is json, you can call JSON.stringify(thingThatIsJson)which will return a String. .toString()does not work on json.
如果有问题的对象是 json,您可以调用JSON.stringify(thingThatIsJson)which 将返回一个字符串。.toString()不适用于 json。
This is a message to those of you dealing with something like req.bodywhich will work in console.log()which is rather confusing since it may not otherwise behave like a String (like when you're trying to add it to another String).
这是给你们中那些处理类似的东西的人的消息,req.body这会console.log()很混乱,因为它可能不像字符串那样表现(比如当您尝试将它添加到另一个字符串时)。
回答by Ankit
The problem lies with the propValpart of your code. Since that may not be converted into a string.
问题在于propVal您的代码部分。因为那可能无法转换为字符串。

