Question

如何保存＆＃34; exportURL +数据＆＃34;数据到php变量？ ajax代码在同一个php文件中，我需要将这些数据存储在php变量中。目前它只显示包含数据的警报，但我想将其存储在php变量中。

// URL to Highcharts export server
var exportUrl = 'https://export.highcharts.com/';

// POST parameter for Highcharts export server
var object = {
    options: JSON.stringify(options),
    type: 'image/png',
    async: true
};

// Ajax request
$.ajax({
    type: 'post',
    url: exportUrl,
    data: object,
    success: function (data) {

//Save here in a php variable so i can use the value in the php code that follows in my file

    }
});

Answer 1

如果我了解你想要从提交给外部服务器的ajax中捕获数据并写入你的服务器，那么正确吗？

＆＃13;

    // URL to Highcharts export server
    var exportUrl = 'https://export.highcharts.com/';

    options = {};
    // POST parameter for Highcharts export server
    var object = {
        options: JSON.stringify(options),
        type: 'image/png',
        async: true
    };

    // Ajax request
    $.ajax({
        type: 'post',
        url: exportUrl,
        data: object,
        success: function (data) {
          	//Submit data from your server
             // Ajax request
            $.ajax({
                type: 'post',
                url: 'http://localhost/teste.php',//this your local file
                data: {'data' : exportUrl+data},
                success: function (data2) {
                    //Response from your server
					//if your teste.php print response. echo "" or die("") ;

                }
            });
        }
    });

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

<?php
    //this variable post contains your data sended from ajax
    //print_r($_POST);
    $data = $_POST['data']
?>

如何保存ajax＆＃34;成功＆＃34;变量到php变量？

1 个答案: