时间：2021-07-01 10:21:17 帮助过：8人阅读
【HTML 代码】
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width,user-scalable=no, initial-scale=1">
<link rel="stylesheet" type="text/css" href="" />
<title>angularjs实现 ajax</title>
</head>
<body ng-app="HelloAjax">
<div ng-controller="HelloAjax">
<form>
<input type="text" ng-model="username" />
<input type="text" ng-model="email" />
</form>
<table>
<tr ng-repeat="user in users">
<td>{{user.username}}</td>
<td>{{user.email}}</td>
</tr>
</table>
<button ng-click="get_more();">get more</button>
</div>
</body>
<script type="text/javascript" src="./js/angular.min.js" charset="utf-8"></script>
<script type="text/javascript" src="ajax.js" charset="utf-8"></script>
</html>
【js代码 ajax.js】
var myModule = angular.module("HelloAjax",[]);
myModule.controller("HelloAjax",["$scope","$http",function HelloAjax($scope,$http){
/*
$scope.users=[{'username':"zhangsan","email":"zs@11.com"},
{'username':"zhangsan2","email":"zs@22.com"},
{'username':"zhangsan3","email":"zs@33.com"}];
*/
$scope.get_more = function(){
$http({
method: "POST",
url: "./ajax.php",
data:{'username':$scope.username,
'email':$scope.email
}
}).
success(function(data, status) {
//$scope.status = status;
$scope.users = data;
}).
error(function(data, status) {
//$scope.data = data || "Request failed";
//$scope.status = status;
});
}
}]);
【PHP代码 ajax.php】
<?php
//获取参数
$data = file_get_contents("php://input");
$user = json_decode($data);
//查询数据库
$conn = mysql_connect("localhost","root","");
mysql_select_db("test");
$sql ="select username,email from users ";
$res = mysql_query($sql,$conn);
$users = array();
while($row = mysql_fetch_assoc($res)){
$users[] = $row;
}
//当然这里简化了插入数据库
$users[] = array('username'=>$user->username,
'email'=>$user->email);
//返回数据库
echo json_encode($users);