发送JSON和JS可变数据到PHP脚本在一个Ajax请求

Question

JAVASCRIPT：在点击一个按钮，我要做到以下几点：

var paymentPeriodId = $('#paymentPeriodID').val(); 
if(paymentPeriodId < 1){alert('Payment period not set.');e.preventPropagation();return false; } 
if(checkedRows.length < 1){alert('Supervisor not selected.');e.preventPropagation();return false; } 

checkedRows = $.toJSON(checkedRows);  
var formData = new FormData(); 
formData.append("paymentPeriodId", paymentPeriodId); 
formData.append("checkedRows", checkedRows); 

// alert(JSON.stringify(checkedRows)); 
alert(paymentPeriodId); 
alert(checkedRows); 

var qurl = '<?php echo base_url();?>salary_processing/summeriseProduction'; 
$.ajax({ 
url: qurl, 
type: "POST", 
data:formData, 
success: function(data){ 

     alert('successfull'); 
//   var data = $.parseJSON(data); 
//   alert(JSON.stringify(data)); 

     }, 
}); 

PHP：此功能将得到的值，并会做一些数据库操作。这里没有提到。

public function summeriseProduction() { 

// Unescape the string values in the JSON array 
$supervisorIdArray = stripcslashes($this->input->post('checkedRows')); 
$paymentPeriodId = $this->input->post('paymentPeriodId'); 
// Decode the JSON array 
    $supervisorIdArray1 = json_decode($supervisorIdArray, TRUE); 
// $paymentPeriodId1 = json_decode($paymentPeriodId, TRUE); 

echo json_encode($supervisorIdArray1); 
// echo $paymentPeriodId; 

} 

https://stackoverflow.com/users/4476402/pekka，它不工作。

Answer 1

var formData = new FormData(); 
var paymentPeriodId = $('#paymentPeriodID').val(); 
checkedRows = $.toJSON(checkedRows); 
formData.append("paymentPeriodId", paymentPeriodId); 
formData.append("checkedRows", checkedRows); 
var qurl = '<?php echo base_url();?>salary_processing/summeriseProduction'; 
$.ajax({ 
url: qurl, 
type: "POST",  
data: formData,   
success: function(data){} 

尝试这样

Answer 2

var formData = new FormData(); 
var paymentPeriodId = $('#paymentPeriodID').val(); 
checkedRows = $.toJSON(checkedRows); 
formData.append("paymentPeriodId", paymentPeriodId); 
formData.append("checkedRows", checkedRows); 
var qurl = '<?php echo base_url();?>salary_processing/summeriseProduction'; 
$.ajax({ 
url: qurl, 
type: "POST", 
processData: false, 
contentType: false, 
data: formData,   
success: function(data){} 

它炒菜这样。