我在此代码中收到上述错误:
var inputs = {
$("<input>").attr({"type":"hidden","name":"collegeID"}).val(collegeID),
$("<input>").attr({"type":"hidden","name":"collegeID"}).val(collegeID),
$("<input>").attr({"type":"hidden","name":"collegeID"}).val(collegeID),
$("<input>").attr({"type":"hidden","name":"collegeID"}).val(collegeID),
$("<input>").attr({"type":"hidden","name":"collegeID"}).val(collegeID),
$("<input>").attr({"type":"hidden","name":"collegeID"}).val(collegeID)
}
这对我来说看起来很有效。我想将变量
.append()
放到div上,我不想创建一堆不同的变量来做到这一点。 最佳答案
inputs
必须是Array
。使用[]
代替{}
var collegeID = "";
var inputs = [
$("<input>").attr({
"type": "hidden",
"name": "collegeID"
}).val(collegeID),
$("<input>").attr({
"type": "hidden",
"name": "collegeID"
}).val(collegeID),
$("<input>").attr({
"type": "hidden",
"name": "collegeID"
}).val(collegeID),
$("<input>").attr({
"type": "hidden",
"name": "collegeID"
}).val(collegeID),
$("<input>").attr({
"type": "hidden",
"name": "collegeID"
}).val(collegeID),
$("<input>").attr({
"type": "hidden",
"name": "collegeID"
}).val(collegeID)
]
console.log(JSON.stringify(inputs));
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>