So zeigen Sie Daten aus mysql mit angular.js PHP?

So drucken Sie Daten empfangen von mysql verwenden angular.js PHP.
Ich möchte wissen, wie Sie mit mysql und angular.js

Habe ich versucht mit node.js mongodb. aber wissen will über php,mysql,angular.js

InformationsquelleAutor Digant Shah | 2014-10-07

  1. 2

    Hallo Können Sie überprüfen Sie folgenden Beispiel-code
    Ihre HTMLpage :

    <html ng-app>
    <head>
        <title>AngularJs Post Example: DevZone.co.in </title>
        <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/angularjs/1.0.7/angular.min.js"></script>
        <style>
            #dv1{
                border:1px solid #DBDCE9; margin-left:auto;
                margin-right:auto;width:220px;
                border-radius:7px;padding: 25px;
            }

            .info{
                border: 1px solid;margin: 10px 0px;
                padding:10px;color: #00529B;
                background-color: #BDE5F8;list-style: none;
            }
            .err{
                border: 1px solid;  margin: 10px 0px;
                padding:10px;  color: #D8000C;
                background-color: #FFBABA;   list-style: none;
            }
        </style>
    </head>
    <body>
        <div id='dv1'>
            <form ng-controller="FrmController">
                <ul>
                    <li class="err" ng-repeat="error in errors"> {{ error}} </li>
                </ul>
                <ul>
                    <li class="info" ng-repeat="msg in msgs"> {{ msg}} </li>
                </ul>
                <h2>Sigup Form</h2>
                <div>
                    <label>Name</label> 
                    <input type="text" ng-model="username" placeholder="User Name" style='margin-left: 22px;'> 
                </div>
                <div>
                    <label>Email</label>
                    <input type="text" ng-model="useremail" placeholder="Email" style='margin-left: 22px;'> 
                </div>
                <div>
                    <label>Password</label>
                    <input type="password" ng-model="userpassword" placeholder="Password">
                </div>
                <button ng-click='SignUp();' style='margin-left: 63px;margin-top:10px'>SignUp</button>
            </form>
        </div>

        <script type="text/javascript">
            function FrmController($scope, $http) {
                $scope.errors = [];
                $scope.msgs = [];

                $scope.SignUp = function() {

                    $scope.errors.splice(0, $scope.errors.length); //remove all error messages
                    $scope.msgs.splice(0, $scope.msgs.length);

                    $http.post('post_es.php', {'uname': $scope.username, 'pswd': $scope.userpassword, 'email': $scope.useremail}
                    ).success(function(data, status, headers, config) {
                        if (data.msg != '')
                        {
                            $scope.msgs.push(data.msg);
                        }
                        else
                        {
                            $scope.errors.push(data.error);
                        }
                    }).error(function(data, status) { //called asynchronously if an error occurs
//or server returns response with an error status.
                        $scope.errors.push(status);
                    });
                }
            }
        </script>
        <a href='http://devzone.co.in'>Devzone.co.in</a>
    </body>
 </html>

////////////////////////////////////////////////////////////////////////
Your php code

<?php

$data = json_decode(file_get_contents("php://input"));
$usrname = mysql_real_escape_string($data->uname);
$upswd = mysql_real_escape_string($data->pswd);
$uemail = mysql_real_escape_string($data->email);

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

$qry_em = 'select count(*) as cnt from users where email ="' . $uemail . '"';
$qry_res = mysql_query($qry_em);
$res = mysql_fetch_assoc($qry_res);

if ($res['cnt'] == 0) {
    $qry = 'INSERT INTO users (name,pass,email) values ("' . $usrname . '","' . $upswd . '","' . $uemail . '")';
    $qry_res = mysql_query($qry);
    if ($qry_res) {
        $arr = array('msg' => "User Created Successfully!!!", 'error' => '');
        $jsn = json_encode($arr);
        print_r($jsn);
    } else {
        $arr = array('msg' => "", 'error' => 'Error In inserting record');
        $jsn = json_encode($arr);
        print_r($jsn);
    }
} else {
    $arr = array('msg' => "", 'error' => 'User Already exists with same email');
    $jsn = json_encode($arr);
    print_r($jsn);
}
?>

    Bitte checkthe oben genannten code ein, den Sie werden verstehen, wie es funktioniert. Wenn Sie ein problem konfrontiert sind, Sag mir einfach hier in der Kommentar-Zeile

    InformationsquelleAutor Rahul_Dabhi

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