angularjs應用程序的web代理

Question

我必須爲我的angularjs文件創建一個web代理腳本，因爲我得到了CORS（交叉源請求方法）的錯誤，而且我沒有任何選項可以使用訪問控制允許源，因爲我不能對我的服務器端做任何更改。 我的後端數據是用java編寫的。所以請有人告訴我如何爲我的angularjs應用程序創建一個web代理。

或者無論如何繞過我的瀏覽器cors請求。

Answer 1

快速解決使用foreach和json_decode的問題。

如果您的print_r（$ JSON）自帶的格式如下：

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 
     ) 

) 

這將這樣的伎倆：

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 
} 

希望這會有所幫助。

Answer 2

用戶json_decode與true PARAMS

$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']; 

編輯： 對於多個JSON數據：

$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>"; 
} 

輸出：

StudentID: 1 
StudentID: 3 
StudentID: 5 

Answer 3

解碼烏爾陣列像下面..

$newarr= json_decode('urjsonstring'); 

extract($newarr); 

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