Question

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，但它显示我未捕获TypeError：非法调用。请帮助！

Answer 1

为什么要将$(document).ready()放在search()函数中？
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);
           });


       }
    });

Answer 2

更改变量dateFrom和dateTo。尝试像这样改变

$(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", { datepickerFrom:dateFrom, datepickerTo:dateTo, storeCheck: sponsers,organiser:organiser,sort:sort }, function(data) {
      alert("Data Loaded: " + data);
   });

 }
 });

非法调用错误jquery

2 个答案: