如何使用angular.js PHP从mysql显示数据？

发布时间：2020-12-17 08:45:46 所属栏目：安全来源：网络整理

导读：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 我尝试使用node.js mongodb.但想知道php,mysql,angular.js 您好,您可以查看以下示例代码你的HTML页面： html ng-app head titleAngularJs Post Example: DevZone.co.in /t

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

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

您好,您可以查看以下示例代码
你的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);
}
?>

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

（编辑：李大同）

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