ajax-返回值 - php返回json数据的demo如下，请问是什么原因导致该问题？-php教程-PHP中文网

首页

后端开发

php教程

ajax-返回值 - php返回json数据的demo如下，请问是什么原因导致该问题？

WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWB

Sep 12, 2016 pm 05:44 PM

jsonphp

<code><?php // Include ezSQL core
include_once('../include/ez_sql_core.php');

// Include ezSQL database specific component
include_once('../include/ez_sql_postgresql.php');
$username='';
$password='';
$result='';
if(!empty($_POST)){
    $username = $_POST["username"];
    $password = $_POST["password"];
}else{
    
}

$db = new ezSQL_postgresql('i2soft', 'i2s@Shanghai', 'sdata', '127.0.0.1');


///$user = $db->query("select id from user2 where name = 'admin' and password ='admin'");
//$user_details = $db->get_row(null, OBJECT, 0);
//var_dump($user_details) ;

$sql = sprintf("select * from user2 where name='%s' and password='%s'",$username,$password);
//echo $sql;
//select * from user2 where name='admin' and password='admin'
$user=$db->get_results($sql);

//var_dump($user);
//array(1) { [0]=> object(stdClass)#5 (3) { ["id"]=> string(1) "1" ["name"]=> string(5) "admin" ["password"]=> string(5) "admin" } }
if($user){
  $result='0';
   
}else{
   $result='-1';
  
}

$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo json_encode($arr);
?>

</code>

php返回json数据，若含有

<code>$user=$db->get_results($sql);  </code>

这行代码ajax就返回error。

若去掉这行代码，则ajax结果是success
这行代码是没问题的，返回的数据我打印过，注释的那个数组就是打印出出来，我不用ajax请求服务器代码的话，直接提交表单的时候测过代码。

html代码如下：

<code>


    
        <meta charset="utf-8">
        <meta http-equiv="X-UA-Compatible" content="IE=edge">
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <meta name="description" content="">
        <meta name="author" content="">
        <link rel="icon" href="../media/image/favicon.ico">
        <title>英方I2Active</title>
        <!-- Bootstrap core CSS -->
        <link href="../media/bootstrap/css/bootstrap.min.css" rel="stylesheet">
        <!-- Custom styles for this template -->
        <link href="../media/css/login.css" rel="stylesheet">
        <script type="text/javascript" src="../media/js/jquery-1.9.1.js"></script>
    

    
        <div class="container">
            <form class="form-signin" role="form" action="">
                <img  src="/static/imghwm/default1.png" data-src="../media/image/logo.png" class="lazy" alt="ajax-返回值 - php返回json数据的demo如下，请问是什么原因导致该问题？" >
                <h2 id="I-Active登录">I2Active登录</h2>
                <input id="username" type="text" class="form-control" placeholder="User name" required autofocus>
                <input id="password" type="password" class="form-control" placeholder="Password" required>
                <div class="checkbox">
                    <label>
                        <input type="checkbox" value="remember-me"> 记住密码
                    </label>
                </div>
                <button id="login" class="btn btn-lg btn-primary btn-block">登录</button>
            </form>
        </div>
        <!-- /container -->
        <script type="text/javascript">  
            
            $("#login").click(function () {
                var username = $("#username").val();
                var password = $("#password").val();
                if(username&&password){
                    $.ajax({
                        url: '../controller/login2.php', // 跳转到 action  
                        data: {
                            'username': username,
                            'password': password
                        },
                        type: 'post',
                        cache: false,
                        dataType: 'json',
                        success: function (data) {
                            alert('ajax success');
                        },
                        error: function () {
                          alert('ajax error');
                        }
                    })
                }
            });
        
        </script>
    

</code>

回复内容：

<code><?php // Include ezSQL core
include_once('../include/ez_sql_core.php');

// Include ezSQL database specific component
include_once('../include/ez_sql_postgresql.php');
$username='';
$password='';
$result='';
if(!empty($_POST)){
    $username = $_POST["username"];
    $password = $_POST["password"];
}else{
    
}

$db = new ezSQL_postgresql('i2soft', 'i2s@Shanghai', 'sdata', '127.0.0.1');


///$user = $db->query("select id from user2 where name = 'admin' and password ='admin'");
//$user_details = $db->get_row(null, OBJECT, 0);
//var_dump($user_details) ;

$sql = sprintf("select * from user2 where name='%s' and password='%s'",$username,$password);
//echo $sql;
//select * from user2 where name='admin' and password='admin'
$user=$db->get_results($sql);

//var_dump($user);
//array(1) { [0]=> object(stdClass)#5 (3) { ["id"]=> string(1) "1" ["name"]=> string(5) "admin" ["password"]=> string(5) "admin" } }
if($user){
  $result='0';
   
}else{
   $result='-1';
  
}

$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo json_encode($arr);
?>

</code>

php返回json数据，若含有

<code>$user=$db->get_results($sql);  </code>

这行代码ajax就返回error。

html代码如下：

<code>


    
        <meta charset="utf-8">
        <meta http-equiv="X-UA-Compatible" content="IE=edge">
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <meta name="description" content="">
        <meta name="author" content="">
        <link rel="icon" href="../media/image/favicon.ico">
        <title>英方I2Active</title>
        <!-- Bootstrap core CSS -->
        <link href="../media/bootstrap/css/bootstrap.min.css" rel="stylesheet">
        <!-- Custom styles for this template -->
        <link href="../media/css/login.css" rel="stylesheet">
        <script type="text/javascript" src="../media/js/jquery-1.9.1.js"></script>
    

    
        <div class="container">
            <form class="form-signin" role="form" action="">
                <img  src="/static/imghwm/default1.png" data-src="../media/image/logo.png" class="lazy" alt="ajax-返回值 - php返回json数据的demo如下，请问是什么原因导致该问题？" >
                <h2 id="I-Active登录">I2Active登录</h2>
                <input id="username" type="text" class="form-control" placeholder="User name" required autofocus>
                <input id="password" type="password" class="form-control" placeholder="Password" required>
                <div class="checkbox">
                    <label>
                        <input type="checkbox" value="remember-me"> 记住密码
                    </label>
                </div>
                <button id="login" class="btn btn-lg btn-primary btn-block">登录</button>
            </form>
        </div>
        <!-- /container -->
        <script type="text/javascript">  
            
            $("#login").click(function () {
                var username = $("#username").val();
                var password = $("#password").val();
                if(username&&password){
                    $.ajax({
                        url: '../controller/login2.php', // 跳转到 action  
                        data: {
                            'username': username,
                            'password': password
                        },
                        type: 'post',
                        cache: false,
                        dataType: 'json',
                        success: function (data) {
                            alert('ajax success');
                        },
                        error: function () {
                          alert('ajax error');
                        }
                    })
                }
            });
        
        </script>
    

</code>

SQL执行出错了，PHP抛出错误，建议你用Web浏览器直接访问看看原因

看一下 ezSQL_postgresql 类的代码

声明

本文内容由网友自发贡献，版权归原作者所有，本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容，请联系admin@php.cn