Question

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获取记录的文件在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>";
    }   
?>

此代码无效。

Answer 1

要使用Ajax + jQuery检索数据，您必须编写以下代码：

 <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，请写下：

<?php 
$con=mysqli_connect("localhost","root",""); 
if (!$con)
{
    die('Could not connect: ' . mysqli_error());
}   
?>

要显示数据库中的数据，您必须写下：

<?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>";
?>

Answer 2

您无法返回ajax返回值。您存储的全局变量会在返回后存储您的返回值或者像这样更改你的代码。

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;
}

Answer 3

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

假设您已从 Records.php 中正确提取了记录，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，因为不再鼓励mysql_* functions。

Answer 4

$(document).ready(function(){
    var response = '';
    $.ajax({ type: "GET",   
         url: "Records.php",   
         async: false,
         success : function(text)
         {
             response = text;
         }
    });

    alert(response);
});

需要：

$(document).ready(function(){

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

});

Answer 5

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();
}
?>

使用Jquery Ajax从Mysql中检索数据

5 个答案: