Question

我想知道如何通过ajax调用php函数，如果person匹配参数显示形式。

这是我要调用的php代码，microsub.php：

<?php

$ip = $_SERVER['REMOTE_ADDR'];

$public_ip_ranges = array();

$range = (object) array();
$range->name = 'xxxx xxxxx';
$range->lower = '10.1.0.0';
$range->upper = '10.1.255.255';
$public_ip_ranges[] = $range;

$range = (object) array();
$range->name = 'xxxxxxx';
$range->lower = '141.228.0.0';
$range->upper = '141.228.255.255';
$public_ip_ranges[] = $range;


$range = (object) array();
$range->name = 'xxxxxxxxx';
$range->lower = '146.127.0.0';
$range->upper = '146.127.255.255';
$public_ip_ranges[] = $range;


if (($lngIP=ip2long($ip)) < 0) $lngIP += 4294967296;  

foreach ($public_ip_ranges as $ip_range) {
   if (($lngLow=ip2long($ip_range->lower)) < 0) $lngLow += 4294967296;
   if (($lngHigh=ip2long($ip_range->upper)) < 0) $lngHigh += 4294967296;
   if($lngIP >= $lngLow and $lngIP <= $lngHigh) {
        ?>

以下是我想要进行ajax调用的js：

var utility = require('./utility.js');
var extend = require('extend');
var $ = require('jquery-browserify');

module.exports = {


    init: function() {
        utility.consoleLog('Micro Trial Form init!');
        //utility.getCookie(name);
        //Check if cookie set
        //**AJAX CALL HERE**
        //Display IP Form

        $('#rdm-below-header').append( "<p>Micro Trials Form</p>" );


        //On Submit form via AJAX


        //On Success Hide form, show success message ask to login
        //Set cookie to hide the form
        //utility.setCookie(name);


    }

};

一旦人提交表格（如果显示）我想将数据发送到另一个php文件processer.php来处理数据。

任何帮助都非常感谢。

通过Ajax调用Php函数

0 个答案: