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如何使用angular.js PHP顯示來自mysql的數據？

2014-10-07 88 views

回答

2

嗨您能查看下面的示例代碼 您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); 
} 
?>

請checkthe上面的代碼，你將瞭解它是如何工作的。如果您遇到任何問題，請在評論行

來源

2014-10-07 06:38:09

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