角桩到MySQL误差

Question

我正在开发一个离子型应用程序，用户需要将数据发布到MySQL数据库。由于一些非常有用的答案，所以我已经绕过了CORS的问题，然而，我现在遇到了另一个问题。submit.php文件如下所示：

<?php
    if (isset($_SERVER['HTTP_ORIGIN'])) {
        header("Access-Control-Allow-Origin: {$_SERVER['HTTP_ORIGIN']}");
        header('Access-Control-Allow-Credentials: true');
        header('Access-Control-Max-Age: 86400');    // cache for 1 day
    }

    // Access-Control headers are received during OPTIONS requests
    if ($_SERVER['REQUEST_METHOD'] == 'OPTIONS') {

        if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_METHOD']))
            header("Access-Control-Allow-Methods: GET, POST, OPTIONS");         

        if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']))
            header("Access-Control-Allow-Headers:        {$_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']}");

        exit(0);
    }

    $data = json_decode(file_get_contents("php://input"));
    $celeb = $data->celeb;
    $camp = $data->camp;
    $spirit = $data->spirit;

    $sport = $data->sport;
    $bizs = $data->bizs;
    $entrep = $data->entrep;

    $young = $data->young;
    $conser = $data->conser;
    $saty = $data->saty;

    $name = $data->name;
    $surname = $data->surname;
    $email = $data->email;

    $contacts = $data->contacts;

    $con = mysql_connect('localhost', 'root', 'thePssWrd');
    mysql_select_db('theDB', $con);

    $qry = 'INSERT INTO test 
                    (celeb,camp,spirit,sport,bizs,
                     entrep,young,conser,saty,name,
                     surname,email,contacts) 
             values ("' . $celeb . '","' . $camp . '",' .$spirit . 
                     ','.$sport. ','.$bizs. ','.$entrep. ','.$young. 
                     ','.$conser. ','.$saty. ','.$name. ','.$surname. 
                     ','.$email. ','.$contacts. ')';
    $qry_res = mysql_query($qry);

    if ($qry_res) {

        if ($qry_res) {
            $arr = array('msg' => "Submitted Successfully!!!", 'error' => '');
            $jsn = json_encode($arr);
            print_r($jsn);
        } else {
            $arr = array('msg' => "", 'error' => 'Error In Submit');
            $jsn = json_encode($arr);
            print_r($jsn);
        }
    } else {
        $arr = array('msg' => "", 'error' => 'This is the big error thing...');
        $jsn = json_encode($arr);
        print_r($jsn);
    }
?>

我的控制器：

.controller('FrmController', function ($scope , $http) {
    $scope.errors = [];
    $scope.msgs = [];

    $scope.vote = function() {

        $scope.errors.splice(0, $scope.errors.length); 
        $scope.msgs.splice(0, $scope.msgs.length);

        $http.post('http://www.ann7.com/saty/submit.php', {

            'celeb'     : $scope.celeb, 
            'camp'      : $scope.camp, 
            'spirit'    : $scope.spirit,
            'sport'     : $scope.sport,
            'bizs'      : $scope.bizs, 
            'entrep'    : $scope.entrep, 
            'young'     : $scope.young,
            'conser'    : $scope.conser,
            'saty'      : $scope.saty, 
            'name'      : $scope.name, 
            'surname'   : $scope.surname,
            'email'     : $scope.email,
            'contacts'  : $scope.contacts

        }
        ).success(function(data, status, headers, config) {
            if (data.msg != '')
            {

              console.log(data.msg);
                $scope.msgs.push(data.msg);
            }
            else
            {
              console.log(data.error);
                $scope.errors.push(data.error);
            }
        }).error(function(data, status) { 
            $scope.errors.push(status);
        });
    }
});

如果您看一下PHP，它会输出错误消息，而我收到的是“这是一个大错误.”在我的控制台中的错误消息，我认为它是一个成功的连接到DB，但是在插入值时出错.我的html文件看起来像一个例子：

<form  name="nominationForm" ng-controller="FrmController" class="falecomigo novalidate form-manager" >
    <div ng-controller="VoteCtrl" class="list list-inset" >
    <div class="styled-select">
           <select ng-model="bizs">
              <option value="e">option1</option>
              <option value="1">option1</option>
              <option value="2">option1</option>
              <option value="3">option1</option>
              <option value="4">option1</option>
              <option value="5">option1</option>
           </select>
        </div>
        <label class="item item-input">
            <input ng-model="name" name="fieldName" type="text" placeholder="Name" required ng-minlength="2" ng-maxlength="70">
        </label>

        <label class="item item-input">
            <input ng-model="surname" name="fieldSurname" type="text" placeholder="Surname" required ng-minlength="2" ng-maxlength="70">
        </label>
        <label class="item item-input">
            <input ng-model="email" name="fieldEmail" type="email" placeholder="E-mail" required ng-maxlength="50">
        </label>

        <label class="item item-input">
            <input ng-model="contacts" name="fieldNumber" type="text" placeholder="Contact No." required ng-minlength="2" ng-maxlength="70">
        </label>
    </div>

    <div class="padding container">
        <div class="row">
            <button ng-click="vote()" class="button button-balanced  button-small col col-100"> Vote </button>
        </div>
    </div>
</form>

我不知道我哪里出了问题..。

Answer 1

不要给自己发送一条毫无意义的错误消息，而是始终这样做。

} else {
    $arr = array('msg' => "", 'error' => mysql_error());
    $jsn = json_encode($arr);
    //print_r($jsn);
    echo $jsn;
}

这个错误实际上会告诉你出了什么问题。

当然，在一个活动的系统中，您不会想这样做，因为用户会看到很可能帮助黑客的错误消息。所以你会做这样的事

} else {
    // log error for admins to check
    error_log( mysql_error(),3, '/some/path/to/error.log');
    $arr = array('msg' => "", 'error' => 'An error occured, see the log');

    $jsn = json_encode($arr);
    //print_r($jsn);
    echo $jsn;
}

额外检查：

我现在看到您没有对任何mysql_函数调用执行错误检查，可能错误在连接过程中，所以请添加

$con = mysql_connect('localhost', 'root', 'thePssWrd');
if (!$con) {
    // log error for admins to check
    //error_log( mysql_error(),3, '/some/path/to/error.log');
    die('Could not connect: ' . mysql_error());
}


if (! mysql_select_db('theDB', $con) ) {
   die ('Can\'t select the database: ' . mysql_error());
}

Ahhhh

看起来，您实际上并没有将json编码的数据作为一个简单的字符串发送回浏览器。与print_r($jsn)不同，做一个简单的echo $jsn;，它适用于json数据字符串的所有重调。此时，$json实际上不是一个数组，它是一个简单的字符串变量。

还可以简化查询的编码。

请记住，双引号允许您展开$variables，因此它可能解决了底层问题，只需使查询的编码更加直接。

$qry = "INSERT INTO test 
                (celeb,camp,spirit,sport,bizs,
                 entrep,young,conser,saty,name,
                 surname,email,contacts) 
         values ('$celeb','$camp','$spirit','$sport','$bizs',
                 '$entrep','$young','$conser','$saty','$name',
                 '$surname','$email','$contacts')";

$surname失踪了

我觉得我应该加上这个，好像我没有别人会那样。如果您只是学习PHP及其数据库访问机制，请不要浪费时间学习mysql_扩展。它被废弃了，在即将发布的新PHP7中不再可用。看看这个，学习mysqli_扩展或PDO。Have a read on this to help you pick one.