如何使用angular.js PHP从mysql显示数据？

How to print data recieved from mysql using angular.js PHP. I want to know how to connect with mysql and angular.js

如何使用angular.js PHP打印从mysql收到的数据。我想知道如何连接mysql和angular.js

I tried with node.js mongodb. but wants to know about php,mysql,angular.js

我尝试使用node.js mongodb。但想知道php，mysql，angular.js

1 个解决方案

#1

Hi Can you check following sample code your HTMLpage :

您好，您可以检查以下示例代码您的HTML页面：

<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);
}
?>

Please checkthe above code you will be understand how it will work. If any problem you are facing just tell me here in comment line

请检查上面的代码，您将了解它将如何工作。如果您遇到任何问题，请在评论栏中告诉我

#1