使用Jquery Ajax从Mysql检索数据

  • Z时代
  • 2024-01-10
  • 分类：问答

list.php：我只想显示Mysql表记录的简单ajax代码：

<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是用于从Mysql提取记录的文件。

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>";
    }   
?>

该代码不起作用。

回答：

为了使用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，请编写以下代码：

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

为了显示数据库中的数据，您应该这样编写：

<?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>";
?>

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