Javascript 如何将数据返回到格式化方法中的Jquery.Get请求？_Javascript_Php_Jquery_Ajax

Javascript 如何将数据返回到格式化方法中的Jquery.Get请求？

javascript php jquery ajax

Javascript 如何将数据返回到格式化方法中的Jquery.Get请求？,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,刚刚从Asp.net迁移到php 我正在学习一些与php mvc相关的教程我目前面临的问题是将数据返回到ajax调用。数据即将到来，但仪表板视图也已附加到它我的路线是： Options -MultiViews RewriteEngine On Options -Indexes RewriteCond %{REQUEST_FILENAME} !-d RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-l

刚刚从Asp.net迁移到php

我正在学习一些与php mvc相关的教程

我目前面临的问题是将数据返回到ajax调用。数据即将到来，但仪表板视图也已附加到它

我的路线是：

Options -MultiViews
RewriteEngine On
Options -Indexes
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-l

RewriteRule ^(.+)$ index.php?url=$1 [QSA,L]

我正在发出jquery请求

$.get('http://localhost:44500/phpmvc/dashboard/xhrGetListItem',function(o){
       console.log(o);
    });

答复如下：

[{"id":"1","text":"hassaan khan"},{"id":"2","text":"we are working fine by the way and all is good hope you all are good too"},{"id":"3","text":"\r\n    how day how days are going hope you all working fine"},{"id":"4","text":"\r\n    how day how days are going hope you all working fine"},{"id":"5","text":"i am doing some thing extremely good. know that\r\ni am working on php.\r\n    "}]
<html>
    <head>
        <title>Test</title>
        <link rel="stylesheet" href="http://localhost:44500/phpmvc/public/css/default.css" />
        <script type="text/javascript" src="http://localhost:44500/phpmvc/public/js/jquery-1.11.3.min.js"></script>
        <script type="text/javascript" src="http://localhost:44500/phpmvc/public/js/custom.js"></script>
        <script type="text/javascript" src="http://localhost:44500/phpmvc/views/dashboard/js/default.js"></script>
        <script type="text/javascript">
        $(document).ready(function(){
         //Custom Functions here
        })
    </script>
    </head>
    <body>
        <div id="header">
        Header

            <br />
            <a href="http://localhost:44500/phpmvc/index">Index</a>
            <a href="http://localhost:44500/phpmvc/help">Help</a>
            <a href="http://localhost:44500/phpmvc/dashboard/logout">Logout</a>
        </div>
        <div class="content">
        Content



dashboard ... logged in


            <br/>
            <form id="dashForm" action="http://localhost:44500/phpmvc/dashboard/xhrInsert" method="post">
                <textarea rows="4" cols="50" name="text"></textarea>
                <input type="submit"/>
            </form>
            <br/>
            <hr/>
            <div id="listInsert"></div>
        </div>
        <hr />
        <div id="footer">
    &copy; Footer
</div>
    </body>
</html>

在仪表板模型中，我有方法定义

function xhrGetListComment(){
        //$this->db get the database object
        $sth = $this->db->prepare('SELECT * FROM data');
        $sth->setFetchMode(PDO::FETCH_ASSOC);    
        $sth->execute();  
        $data = $sth->fetchAll();
        //print(json_encode($data));
        echo json_encode($data);            
    }

我如何只返回json对象？

如果以后需要进行一些处理，可能不是最干净的方法，但在没有处理的情况下可以工作：add

exit（）在回显JSON数据之后。否则，您可能会对其感兴趣或与之相近的内容，以“结束”输出。如果您使用它，请小心：“输出缓冲区必须由ob_start（）启动”（在您开始回显任何内容之前）。@blex exit（）有效。我需要关闭连接并完成响应。谢谢，伙计
function xhrGetListComment(){
        //$this->db get the database object
        $sth = $this->db->prepare('SELECT * FROM data');
        $sth->setFetchMode(PDO::FETCH_ASSOC);    
        $sth->execute();  
        $data = $sth->fetchAll();
        //print(json_encode($data));
        echo json_encode($data);            
    }