loadURL = "detail_data.php? xsafecode = 123456 How to pass parameters?

Quote

var gridOptionDetails={
   id : grid_details_id,
   loadURL = "detail_data.php?xsafecode = 123456" ;
   width: "850",
   height: "260",
   container : 'gridboxDetails',
   replaceContainer : true,
   //editable : true,
   groupable : true,
   pageSizeList : [5,10,15,20],
   dataset : dsOptionDetails ,
   columns : colsOptionDetails ,
   //clickStartEdit : true ,
   defaultRecord : ["","",0,0,0.0,0,0.0,"Customer","UPS",""],
   pageSize:30,
   toolbarContent : 'reload | print | state',
   showGridMenu : true,
   allowCustomSkin   : true ,
   allowFreeze   : true ,
   allowHide   : true ,
   allowGroup   : true
};

Server-side php program

Quote

<?php
header('Content-type:text/javascript;charset=UTF-8');
include ("global.php");
$xnewcommon = new common();

$json=json_decode(stripslashes($_POST["_gt_json"]));
$pageNo = $json->{'pageInfo'}->{'pageNum'};
$pageSize = 10;//10 rows per page

$xsafecode = $_GET["xsafecode"];
$xopdate = $_GET[xopdate];
$xcodeno = $_GET[xcodeno];
$Selectywzhl = $_GET[Selectywzhl];
$xbj_date = $_GET[xbj_date];
$Selectop = $_GET[Selectop];
$file="date_sql.txt";
file_put_contents($file,$xsafecode);

I did not get parameter value, ask for help