在单个ajax请求中将json和js变量数据发送到php脚本

Question

JAVASCRIPT：点击按钮我想要执行以下操作：

group

PHP：此函数将接收值并将执行一些数据库操作。这里没有提到。

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));

      },
});

https://stackoverflow.com/users/4476402/pekka，它不起作用。

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){}

试试这个

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){}

就像这样。