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javascript - jquery非法调用错误

转载 作者:行者123 更新时间:2023-11-28 13:26:28 25 4
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function search()
{
var dateFrom = $("#datepickerFrom").val();
var dateTo = $("#datepickerTo").val();
var sponsers = $("#storeCheck").val();
var organiser = $("#organiser").val();
var sort = $("#sort").val();
$(document).ready(function() {
$.post("searchEvents.php", {
datepickerFrom: datepickerFrom,
datepickerTo: datepickerTo,
storeCheck: sponsers,
organiser: organiser,
sort: sort
},
function(data) {
alert("Data Loaded: " + data);
});
});
}

我试图将变量从 jquery 传递到 php,但它显示未捕获的类型错误:非法调用。请帮忙!

最佳答案

  1. 为什么要将 $(document).ready() 放在 search() 函数中?
  2. 更改变量名称,例如未定义 datepickerFrom

试试这个:

 $(document).ready(function(){
function search() {
var dateFrom = $("#datepickerFrom").val();
var dateTo = $("#datepickerTo").val();
var sponsers = $("#storeCheck").val();
var organiser = $("#organiser").val();
var sort = $("#sort").val();

$.post("searchEvents.php", { dateFrom:dateFrom, dateTo: dateTo, storeCheck: sponsers,organiser:organiser,sort:sort },function(data) {
alert("Data Loaded: " + data);
});


}
});

关于javascript - jquery非法调用错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28770768/

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