Question

我现在正在研究ajax几天，它仍然无法工作。所以我决定使用ajax下载一个示例程序，但它仍然无法工作。我的服务器有问题吗？这样做不会返回任何错误。但它没有显示我的jquery。

client.php

<html>
  <head>
    <script language="javascript" type="text/javascript" src="jquery.js"></script>
  </head>
  <body>

  <!---------------------------------------------------------------------------------------------
  1) Create some html content that can be accessed by jquery
  ---------------------------------------------------------------------------------------------->
  <h2> Client example </h2>
  <h3>Output: </h3>
  <div id="output">this element will be accessed by jquery and this text will be replaced</div>

  <script id="source" language="javascript" type="text/javascript">

  $(function () 
  {

    //-------------------------------------------------------------------------------------------
    // 2) Send a http request with AJAX http://api.jquery.com/jQuery.ajax/
    //-------------------------------------------------------------------------------------------
    $.ajax({                                      
      url: 'api.php',                  //the script to call to get data          
      data: "",                        //you can insert url argumnets here to pass to api.php for example "id=5&parent=6"
      dataType: 'json',                //data format      
      success: function(data)          //on recieve of reply
      {
        var id = data[0];              //get id
        var vname = data[1];           //get name
        //--------------------------------------------------------------------------------------
        // 3) Update html content
        //--------------------------------------------------------------------------------------
        $('#output').html("<b>id: </b>"+id+"<b> name: </b>"+vname);     //Set output element html
        //recommend reading up on jquery selectors they are awesome http://api.jquery.com/category/selectors/
      } 
    });

  }); 
  </script>

  </body>
</html>

api.php

<?php 

  //------------------------------------------------------------------------
  //--------------------------------------------------------------------------
  $host = "localhost";
  $user = "root";
  $pass = "root";

  $databaseName = "ajax01";
  $tableName = "variables";

  //--------------------------------------------------------------------------
  // 1) Connect to mysql database
  //--------------------------------------------------------------------------

  $con = mysql_connect($host,$user,$pass);
  $dbs = mysql_select_db($databaseName, $con);

  //--------------------------------------------------------------------------
  // 2) Query database for data
  //--------------------------------------------------------------------------
  $result = mysql_query("SELECT * FROM $tableName");            //query
  $array = mysql_fetch_row($result);                          //fetch result    

  //--------------------------------------------------------------------------
  // 3) echo result as json 
  //--------------------------------------------------------------------------
  echo json_encode($array);

?>

Answer 1

我猜你的链接很好，但是放了console.log(data)并使用浏览器开发者工具检查来自该链接的内容。

哦，在PHP中，在第一行使用标题 header('Content-type: application/json');

Answer 2

如果你有空的返回体，那么你可能在PHP代码中有错误，这是由php.ini

沉默的

尝试插入：

<dom-module id="logo-standard">
 <style>
:host {
  display: block;
}
</style>

<template>

<div class="logo-wrap">

  <div style="width: {{logoWidth}}px;">
    Some awesome logo
  </div>

</template>

<script>

(function() {

Polymer({
  is: 'logo-standard',

  properties: {

    logoWidth: {
      type: String,
      value: '400'
    }
  }

});
})();
</script>
</dom-module>

位于PHP文件的顶部，看看您的浏览器网络面板中发生了什么（打开开发人员工具 - F12）

如何在wamp服务器中运行ajax

2 个答案: