Question

我必须为我的angularjs文件创建一个web-proxy脚本，因为我收到了CORS（Cross Origin Request Method）的错误，我没有任何选项来使用Access Control Allow Origin，因为我无法对我的服务器进行任何更改结束。我的后端数据是在java中。所以请有人告诉我如何为我的angularjs应用程序制作一个web代理。

或者无论如何都要绕过我浏览器的cors请求。

Answer 1

用户json_decode true参数

$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'];

修改对于多个json数据：

$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>"; }

<强>输出：

StudentID: 1 StudentID: 3 StudentID: 5

Answer 2

快速解决使用foreach和json_decode的问题。

如果你的print_r（$ json）有这种格式：

Array
(
    [0] => Array
        (
            [studentid] => 5
            [firstame] => jagdjasgd
            [lastname] => kjdgakjd
            [gender] => 1
            [email] => dgahsdg@em.com
            [fathername] => hashsdh
            [mothername] => djhavshd
            [birthday] => 2016-03-21
            [address] => gafdhfadhs
            [tenth] => 45.235
            [twelfth] => 56.25
        )

)

这样可以解决问题：

 if ($_SERVER['REQUEST_METHOD'] == 'POST')
 {
    $json = json_decode(file_get_contents("php://input"), true);
  //print_r($json);

    $data =array();//Open blank array for student data
    $num = array();//Open Blank array for number of student
    foreach($json as $k => $v):
        $num [] = $v; //number of student
        if(is_array($v)){
           foreach($v as $key=>$val):
             $data[$key] = $val;//Student data
           endforeach;
       }    
    endforeach;

$row= count($num);//Put number of student in $row    
for($i=1; $i<=$row; $i++){
     $q = 'INSERT INTO table (`col1`) 
           VALUES($data['studentid'])';//Looping through sql statement
}

希望这会有所帮助。

Answer 3

如下所示解码你的数组..

$newarr= json_decode('urjsonstring');

extract($newarr);

$query="insert into stud values($studentid, $firstname,$lastname...)";

angularjs应用程序的web-proxy

3 个答案: