1

CORS(Cross Origin Request Method)のエラーが発生したため、angularjsファイル用のWebプロキシスクリプトを作成する必要があり、サーバーエンドに変更を加えることができないため、Access Control Allow Originを使用するオプションがありません。私のバックエンド データは Java です。私のangularjsアプリケーション用のWebプロキシを作成する方法を教えてください。

または、ブラウザからのcorsリクエストをバイパスする方法はありますか?

4

3 に答える 3

1

パラメータjson_decodeを持つユーザーtrue

$data = "{"studentid":"5","firstame":"jagdjasgd","lastname":"kjdgakjd","email":"dgahsdg@em.com"}";
$d = json_decode($data,true); // true means it will result in aaray
print_r($d);
$stdId = $d['studentid'];
$fname = $d['firstname'];
$lname = $d['lastname'];
$mail = $d['email'];

編集: 複数の json データの場合:

$data = '[
    {
        "0": "1",
        "studentid": "1",
        "1": "David",
        "firstname": "David",
        "2": "Beckham",
        "lastname": "Beckham",
        "3": "1",
        "gender": "1",
        "4": "david123@gmail.com",
        "email": "david123@gmail.com",
        "5": "Beckham",
        "fathername": "Beckham",
        "6": "Beckhamii",
        "mothername": "Beckhamii",
        "7": "2016-03-13",
        "birthday": "2016-03-13",
        "8": "dgasdhghasd\nkajsdgjaksdh\nkahdgjaksgdas",
        "address": "dgasdhghasd\nkajsdgjaksdh\nkahdgjaksgdas",
        "9": "58.25",
        "tenth": "58.25",
        "10": "62.25",
        "twelfth": "62.25"
    },
    {
        "0": "3",
        "studentid": "3",
        "1": "Chris",
        "firstname": "Chris",
        "2": "Gayle",
        "lastname": "Gayle",
        "3": "1",
        "gender": "1",
        "4": "chrisgayle@email.com",
        "email": "chrisgayle@email.com",
        "5": "Chris Potters",
        "fathername": "Chris Potters",
        "6": "Christine",
        "mothername": "Christine",
        "7": "2016-04-20",
        "birthday": "2016-04-20",
        "8": "adhafsdh\njgadahksgdkjas\njagdjahsdlkajsld\nkajsgdjlahsdlkas",
        "address": "adhafsdh\njgadahksgdkjas\njagdjahsdlkajsld\nkajsgdjlahsdlkas",
        "9": "87.587",
        "tenth": "87.587",
        "10": "98.256",
        "twelfth": "98.256"
    },
    {
        "0": "5",
        "studentid": "5",
        "1": "jagdjasgd",
        "firstname": "jagdjasgd",
        "2": "kjdgakjd",
        "lastname": "kjdgakjd",
        "3": "1",
        "gender": "1",
        "4": "dgahsdg@em.com",
        "email": "dgahsdg@em.com",
        "5": "hashsdh",
        "fathername": "hashsdh",
        "6": "djhavshd",
        "mothername": "djhavshd",
        "7": "2016-03-21",
        "birthday": "2016-03-21",
        "8": "gafdhfadhs\nagdkjashdas\ndjkahsdklsaj",
        "address": "gafdhfadhs\nagdkjashdas\ndjkahsdklsaj",
        "9": "45.235",
        "tenth": "45.235",
        "10": "56.25",
        "twelfth": "56.25"
    }
]';

$json = json_decode($data, true);
echo '<pre>'; 
foreach ($json as $key => $value) {
    echo "StudentID: ".$value['studentid']."<br>";
}

出力:

StudentID: 1
StudentID: 3
StudentID: 5
于 2016-04-01T10:53:22.037 に答える
0

以下のように配列をデコードします。

$newarr= json_decode('urjsonstring');

extract($newarr);

$query="insert into stud values($studentid, $firstname,$lastname...)";
于 2016-04-01T11:26:02.507 に答える