39

list.php：一个简单的ajax代码，我只想显示Mysql表的记录： 

<html>

<head>
    <script src="jquery-1.9.1.min.js">
    </script>
    <script>
    $(document).ready(function() {
        var response = '';
        $.ajax({
            type: "GET",
            url: "Records.php",
            async: false,
            success: function(text) {
                response = text;
            }
        });

        alert(response);
    });
    </script>
</head>

<body>
    <div id="div1">
        <h2>Let jQuery AJAX Change This Text</h2>
    </div>
    <button>Get Records</button>
</body>

</html>

Records.php 是从 Mysql 获取记录的文件。
数据库中只有两个字段：“名称”、“地址”。

<?php
    //database name = "simple_ajax"
    //table name = "users"
    $con = mysql_connect("localhost","root","");
    $dbs = mysql_select_db("simple_ajax",$con);
    $result= mysql_query("select * from users");
    $array = mysql_fetch_row($result);
?>
<tr>
    <td>Name: </td>
    <td>Address: </td>
</tr>
<?php
    while ($row = mysql_fetch_array($result))
    {
        echo "<tr>";
        echo "<td>$row[1]</td>";
        echo "<td>$row[2]</td>";
        echo "</tr>";
    }   
?>

此代码不起作用。

4

5 回答 5

66

要使用 Ajax + jQuery 检索数据，您应该编写以下代码： 

 <html>
 <script type="text/javascript" src="jquery-1.3.2.js"> </script>

 <script type="text/javascript">

 $(document).ready(function() {

    $("#display").click(function() {                

      $.ajax({    //create an ajax request to display.php
        type: "GET",
        url: "display.php",             
        dataType: "html",   //expect html to be returned                
        success: function(response){                    
            $("#responsecontainer").html(response); 
            //alert(response);
        }

    });
});
});

</script>

<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
   <tr>
       <td> <input type="button" id="display" value="Display All Data" /> </td>
   </tr>
</table>
<div id="responsecontainer" align="center">

</div>
</body>
</html>

对于 mysqli 连接，这样写：

<?php 
$con=mysqli_connect("localhost","root","");

为了显示来自数据库的数据，你应该这样写：

<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);

echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";

while($data = mysqli_fetch_row($result))
{   
    echo "<tr>";
    echo "<td align=center>$data[0]</td>";
    echo "<td align=center>$data[1]</td>";
    echo "<td align=center>$data[2]</td>";
    echo "<td align=center>$data[3]</td>";
    echo "<td align=center>$data[4]</td>";
    echo "</tr>";
}
echo "</table>";
?>
于 2013-06-03T10:24:26.283 回答
4

您不能返回 ajax 返回值。您存储的全局变量在返回后存储您的返回值。
或者像这样更改您的代码。 

AjaxGet = function (url) {
    var result = $.ajax({
        type: "POST",
        url: url,
       param: '{}',
        contentType: "application/json; charset=utf-8",
        dataType: "json",
       async: false,
        success: function (data) {
            // nothing needed here
      }
    }) .responseText ;
    return  result;
}
于 2013-05-23T07:25:35.903 回答
3

请确保您$row[1] , $row[2]包含正确的值，我们在这里假设1 = Name , and 2 here is your Address field？

假设您已从Records.php正确获取记录，您可以执行以下操作： 

$(document).ready(function()
{
    $('#getRecords').click(function()
    {
        var response = '';
        $.ajax({ type: 'POST',   
                 url: "Records.php",   
                 async: false,
                 success : function(text){
                               $('#table1').html(text);
                           }
           });
    });

}

在您的 HTML 中

<table id="table1"> 
    //Let jQuery AJAX Change This Text  
</table>
<button id='getRecords'>Get Records</button>

一点注意事项：

尝试学习 PDO http://php.net/manual/en/class.pdo.php，因为mysql_* functions不再鼓励..

于 2013-05-23T08:26:59.583 回答
1 
$(document).ready(function(){
    var response = '';
    $.ajax({ type: "GET",   
         url: "Records.php",   
         async: false,
         success : function(text)
         {
             response = text;
         }
    });

    alert(response);
});

需要是：

$(document).ready(function(){

     $.ajax({ type: "GET",   
         url: "Records.php",   
         async: false,
         success : function(text)
         {
             alert(text);
         }
    });

});
于 2013-05-23T08:15:52.070 回答
0 
This answer was for @
Neha Gandhi but I modified it for  people who use pdo and mysqli sing mysql functions are not supported. Here is the new answer 

    <html>
<!--Save this as index.php-->
      <script src="//code.jquery.com/jquery-1.9.1.js"></script>
        <script src="//ajax.aspnetcdn.com/ajax/jquery.validate/1.9/jquery.validate.min.js"></script>
    
     <script type="text/javascript">
    
     $(document).ready(function() {
    
        $("#display").click(function() {                
    
          $.ajax({    //create an ajax request to display.php
            type: "GET",
            url: "display.php",             
            dataType: "html",   //expect html to be returned                
            success: function(response){                    
                $("#responsecontainer").html(response); 
                //alert(response);
            }
    
        });
    });
    });
    
    </script>
    
    <body>
    <h3 align="center">Manage Student Details</h3>
    <table border="1" align="center">
       <tr>
           <td> <input type="button" id="display" value="Display All Data" /> </td>
       </tr>
    </table>
    <div id="responsecontainer" align="center">
    
    </div>
    </body>
    </html>

<?php
// save this as display.php


    // show errors 
error_reporting(E_ALL);
ini_set('display_errors', 1);
    //errors ends here 
// call the page for connecting to the db
require_once('dbconnector.php');
?>
<?php
$get_member =" SELECT 
empid, lastName, firstName, email, usercode, companyid, userid, jobTitle, cell, employeetype, address ,initials   FROM employees";
$user_coder1 = $con->prepare($get_member);
$user_coder1 ->execute();

echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";

while($row =$user_coder1->fetch(PDO::FETCH_ASSOC)){
$firstName = $row['firstName'];
$empid = $row['empid'];
$lastName =    $row['lastName'];
$cell =    $row['cell'];

    echo "<tr>";
    echo "<td align=center>$firstName</td>";
    echo "<td align=center>$empid</td>";
    echo "<td align=center>$lastName </td>";
    echo "<td align=center>$cell</td>";
    echo "<td align=center>$cell</td>";
    echo "</tr>";
}
echo "</table>";
?>

<?php
// save this as dbconnector.php
function connected_Db(){

    $dsn  = 'mysql:host=localhost;dbname=mydb;charset=utf8';
    $opt  = array(
        PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
        PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
    );
    #echo "Yes we are connected";
    return new PDO($dsn,'username','password', $opt);
    
}
$con = connected_Db();
if($con){
//echo "me  is connected ";
}
else {
//echo "Connection faid ";
exit();
}
?>
于 2020-12-24T21:16:45.387 回答