Javascript ajax使用jquery发送多个值

我在这个脚本中发送了一个值。但我尝试发送多个值。使用ajax

files.php

   $(".search_button").click(function()
    {
var search_word = $("#search_box").val();
var dataString = 'search_word='+ search_word;
  if(search_word=='')
  {
  }
 else
  {
 $.ajax({
 type: "GET",
 url: "include/search.php",
  data: dataString,
  cache: false,

 if(isset($_POST['search_word'])){  

$serach_word = $_POST['search_word'];
$serach_word1 = $_POST['search_word1'];
    echo "$serach_word<br/>$serach_word1";

 }

search.php

   $(".search_button").click(function()
    {
var search_word = $("#search_box").val();
var dataString = 'search_word='+ search_word;
  if(search_word=='')
  {
  }
 else
  {
 $.ajax({
 type: "GET",
 url: "include/search.php",
  data: dataString,
  cache: false,

 if(isset($_POST['search_word'])){  

$serach_word = $_POST['search_word'];
$serach_word1 = $_POST['search_word1'];
    echo "$serach_word<br/>$serach_word1";

 }

if（isset（$\u POST['search\u word']）{
$serach_word=$_POST['search_word']；
$serach_word1=$_POST['search_word1']；
回声“$serachu word
$serachu word1”；
}

---

我有一些错误，请解决此查询。

您可以使用

data

属性发送多个数据，如

  data: { search_word:search_word ,search_word1:search_word1 }

您还需要将类型更改为

POST

 type: "post",

试试这个：

$.ajax({
 type: "POST",
 url: "include/search.php",
  data: {'dataString': dataString , "value2" : value2,"value3":value3 },
  cache: false,
  success : function()
  {
  }
});

使用

type=“post”

在“数据”中使用

$\u post

访问值时，可以添加一个或多个如下参数：

 type: "post",

 data: {'name': val , "name1" : val1,"name2":val2 }

 $.ajax({
      type: "POST",
      url: "include/search.php",
      data: {'dataString': dataString , "val1" : value1,"val2":value2 },
      cache: false,
      success : function()
      {
       alert("success");
      }
    });

$_REQUEST['dataString'];
$_REQUEST['val1'];
$_REQUEST['val2'];

数据：{'paramA'：valueA，'paramB'：valueB，'paramC'：valueC，…}

)()

--将类型设置为POST，就像您使用的那样

$\u POST

--multilple数据格式如下：

 data: {'name': val , "name1" : val1,"name2":val2 }

 $.ajax({
      type: "POST",
      url: "include/search.php",
      data: {'dataString': dataString , "val1" : value1,"val2":value2 },
      cache: false,
      success : function()
      {
       alert("success");
      }
    });

$_REQUEST['dataString'];
$_REQUEST['val1'];
$_REQUEST['val2'];

您的jquery应该是这样的：

 data: {'name': val , "name1" : val1,"name2":val2 }

 $.ajax({
      type: "POST",
      url: "include/search.php",
      data: {'dataString': dataString , "val1" : value1,"val2":value2 },
      cache: false,
      success : function()
      {
       alert("success");
      }
    });

$_REQUEST['dataString'];
$_REQUEST['val1'];
$_REQUEST['val2'];

在php中，您可以通过以下名称获取所有值：

 data: {'name': val , "name1" : val1,"name2":val2 }

 $.ajax({
      type: "POST",
      url: "include/search.php",
      data: {'dataString': dataString , "val1" : value1,"val2":value2 },
      cache: false,
      success : function()
      {
       alert("success");
      }
    });

$_REQUEST['dataString'];
$_REQUEST['val1'];
$_REQUEST['val2'];

---

以及在search.php文件中编写的脚本。与您编写的代码相同。由于您需要

POST

数据，您可以仅使用

POST

获取它们right@user3329459：

$serach_word=$\u POST['search_word']$serach_word1=$_POST['search_word1']SO上已存在此类问题。。请先努力搜索一下。。。