本文介绍了未定义JSONP跨域的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个基于PHP的超薄框架的API,并希望生成JSONP为我的网站。当我呼吁网站:。它返回JSON回调()上写一个空白页。当登录到控制台是不确定的。
I have an API based on Php Slim Framework and want to generate JSONP for my website. When I call the website on: 'http://api.mangayurdu.com/users?callback=JSON_CALLBACK'. It returns a blank page with JSON CALLBACK() write on it. When logged to the console it is undefined.
API的index.php文件:
API's index.php:
<?php
require 'vendor/autoload.php';
$app = new \Slim\Slim();
$app->contentType('application/json');
$app->get('/users', 'getUsers');
$app->run();
function getConnection() {
$dbhost="localhost";
$dbuser="";
$dbpass="";
$dbname="";
$dbh = new PDO("mysql:host=$dbhost;dbname=$dbname;mysql:charset=utf8mb4", $dbuser, $dbpass);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
return $dbh;
}
function getUsers() {
$sql = "select * FROM manga";
try {
$db = getConnection();
$stmt = $db->query($sql);
$users = $stmt->fetchAll(PDO::FETCH_OBJ);
$db = null;
echo $_GET['callback'] . '('.json_encode($users).')';
}
catch(PDOException $e) {
echo $_GET['callback'] . '('.json_encode($e->getMessage()).')';
}
}
使用Javascript:
Javascript:
.factory('MY', function($http){
var fax= {};
var url = 'http://api.mangayurdu.com/users?callback=JSON_CALLBACK';
fax.isimler = $http.jsonp(url);
return fax;
})
.controller('indexCtrl', function($scope, MY) {
MY.isimler.success(function(alHemen){
$scope.mangas = alHemen;
});
})
推荐答案
不要为它服务为应用程序/ JSON
。服务为应用程序/ JavaScript的
。 JSONP需要被执行
Don't serve it as application/json
. Serve it as application/javascript
. JSONP needs to be executed.
此外,它似乎与修身你不应该使用回声
。尝试调用 $ APP-GT&;响应 - 方式&gt; setBody
Also it appears with Slim you shouldn't use echo
. Try calling $app->response->setBody
.
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