使用php和jquery ajax从mysql数据库中获取数据

Question

I want to get data from mysql database using php and jquery ajax. 我想使用php和jquery ajax从mysql数据库中获取数据。 'process.php' is the php file which connects to database and get mysql data. “ process.php”是连接到数据库并获取mysql数据的php文件。 It works when it is run separately, but when called using ajax it doesn't work. 当它单独运行时它可以工作，但是当使用ajax调用时它不起作用。 Can someone please help to correct error? 有人可以帮忙纠正错误吗？ Here is my html file: 这是我的html文件：

<html>
<head>
<script type="text/javascript" src="js/jquery-1.11.0.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
    function showRoom(){
        $.ajax({
            type:"POST",
            url:"process.php",
            data:{action:showroom},
            success:function(data){
                $("#content").html(data);
            }
        });
    }
    showRoom();
});
</script>
</head>
<body>
<div id="content"></div>
</body>
</html>

Here is my process.php file 这是我的process.php文件

<?php
$link=mysqli_connect("localhost","root","raspberry","homebot");

if (mysqli_connect_errno())
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

$action=$_POST["action"];
if($action=="showroom"){
    $query="SELECT * FROM user";
    $show=mysqli_query($link,$query) or die ("Error");
    while($row=mysqli_fetch_array($show)){
        echo "<li>$row['name']</li>";
    }
}
?>

Answer 1

There are two syntax errors in your ajax call: 您的ajax调用中有两个语法错误：

$(document).ready(function(){
    function showRoom(){
        $.ajax({
            type:"POST",
            url:"process.php",
            data:{action:"showroom"},
            success:function(data){
                $("#content").html(data);
            }
        });
    }
    showRoom();
});

Keep in mind that jQuery's ajax expects an object as parameter. 请记住，jQuery的ajax需要一个对象作为参数。 Inside an object the syntax is 在对象内部的语法是

{ key : value }

You had type="POST" which is correct in declarative syntax, but incorrect when defining an object key. 您输入的类型=“ POST”在声明性语法中正确，但是在定义对象键时不正确。

Second, the data property of the aforementioned object should be an object too. 其次，上述对象的数据属性也应该是一个对象。 So instead of action=showroom it should be 因此，与其说是action = showroom，不如说是

{action:"showroom"}

Answer 2

you did mistake in your code: 您在代码中犯了错误：

 echo "<li>$row['name']</li>";

This should be: 应该是：

 echo "<li>".$row['name']."</li>";

try that... 试试...

使用php和jquery ajax从mysql数据库中获取数据

问题描述

2 个解决方案

解决方案1
4 已采纳 2014-03-25 18:53:28

解决方案2
1 2014-12-06 23:36:29

使用php和jquery ajax从mysql数据库中获取数据

问题描述

2 个解决方案

解决方案1 4 已采纳 2014-03-25 18:53:28

解决方案2 1 2014-12-06 23:36:29

解决方案1
4 已采纳 2014-03-25 18:53:28

解决方案2
1 2014-12-06 23:36:29