在Post上在jQuery和PHP之间传递变量

我有一个$.ajax函数，它可以发布到php页面并获得一些结果

 function GetLicenseCount(SoftwareNameFK, SoftwareTypeFK,LicenseMethodFK,MinistryFK )
    {

        $.ajax({
            type: "GET",
            url: "modules/SoftwareLicenseAllocations/GetLicenseCount.php",
            data: {SoftwareName : SoftwareNameFK, SoftwareType: SoftwareTypeFK, LicenseMethod: LicenseMethodFK, Ministry: MinistryFK},
            dataType: "json",
            success: function(result) {

               var LicenseCount = document.getElementById("txtLicenseCount");
               var idHidden = document.getElementById("txtId");

               //get id value from string, by using split and getting the first part before the delimeter
               var id = result.split(",")[0];
               //get second part of result using string
               LicenseCount.value = result.split(",")[1];

               idHidden.value = id;

            },
            error: function(reason) {
                console.log(reason);
                alert("Failed to get License Count");
            }
        });


    }

GetLicenseCount.php

<?php
    //Local include path format
    set_include_path('C:\PROJECTS\BRIAN\AMS\Web Application;C:\PROJECTS\BRIAN\AMS\Web Application\inc\pear;C:\PROJECTS\BRIAN\AMS\Web Application\js;');




    //Search Screen
    require_once('vars.php');
    require_once('funcs.php');

    //read the parameter passed through URL
    $SoftwareName = $_GET['SoftwareName'];  
        $SoftwareType = $_GET['SoftwareType'];
        $LicenseMethod = $_GET['LicenseMethod'];
        $Ministry = $_GET['Ministry'];

        //formulate query - get all active departments  
        $queryString = "SELECT DISTINCT id, LicenseCount"
                     . " FROM SoftwareLicenseAllocations WHERE"
                     . " SoftwareTypeFK =" . $SoftwareType 
                     . " AND SoftwareNameFK = " . $SoftwareName 
                     . " AND LicenseMethodFK=" . $LicenseMethod 
                     . " AND MinistryFK=" . $Ministry; 

//        $queryString = "SELECT DISTINCT LicenseCount"
//                     . " FROM SoftwareLicenseAllocations WHERE"
//                     . " SoftwareTypeFK =" . 2 
//                     . " AND SoftwareNameFK = " . 2 
//                     . " AND LicenseMethodFK=" . 2 
//                     . " AND MinistryFK=" . 1 ; 

        $sla = GetOpenItDataSet($queryString);

        //loop on result set to refresh the department options list
        while ($row = mysql_fetch_array($sla))
        {            
            $LicenseCount = $row["LicenseCount"];
            $id = $row['id'];
        }//endWhile

        $_SESSION['id'] = 22;


        //echo $departmentArray;
        echo json_encode($id . "," . $LicenseCount);

        //$_POST["lcID"] = $id;

---

不清楚问题出在哪里？什么不起作用？？？
$\u GET['SoftwareName']
应该起作用。这就是你应该做的。我用问题更新了问题。基本上，我只能在jquery中使用返回的变量，但不能通过php使用