php 使用 Jquery Ajax 从 Mysql 中检索数据
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Using Jquery Ajax to retrieve data from Mysql
提问by Ravi
list.php: A simple ajax code that I want to display only records of the Mysql table:
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 is the file to fetch records from Mysql.
In the Database are only two fields: 'Name', 'Address'.
Records.php 是从 Mysql 中获取记录的文件。
在数据库中只有两个字段：“姓名”、“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>";
}
?>
This code is not working.
此代码不起作用。
回答by Neha Gandhi
For retrieving data using Ajax + jQuery, you should write the following code:
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>
For mysqli connection, write this:
对于mysqli连接，这样写：
<?php
$con=mysqli_connect("localhost","root","");
For displaying the data from database, you should write this :
为了显示数据库中的数据，你应该这样写：
<?php
include("connection.php");
mysqli_select_db("samples",$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>";
?>
回答by Mariselvam Panneerselvam
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
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。
或者像这样改变你的代码。
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;
}
回答by Taw
Please make sure your $row[1] , $row[2]contains correct value, we do assume here that 1 = Name , and 2 here is your Address field?
请确保您$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？
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
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);
}
});
});
}
In your HTML
在你的 HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
一点说明：
Try learing PDO http://php.net/manual/en/class.pdo.phpsince mysql_* functionsare no longer encouraged..
尝试学习 PDO http://php.net/manual/en/class.pdo.php因为mysql_* functions不再鼓励..
回答by gitaarik
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
需要是：
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
![SQLSTATE[HY000]：PHP 和 PDO 的一般错误](/res/img/loading.gif)