沒有jQuery的

Question

阿賈克斯JSON請求我使用一個文件backend.php

echo json_encode(array('post' => $_POST)); 

和JavaScript Ajax請求：

function ajax_(requst) { 
    requst = requst || {}; 
    var result = {}; 
    (function() { 
     var xhttp = window.XMLHttpRequest ? new XMLHttpRequest() : 
       new ActiveXObject("Microsoft.XMLHTTP"); 
     xhttp.onreadystatechange = function() { 
      if (xhttp.readyState === 4 && xhttp.status === 200) { 
       result = JSON.parse(xhttp.responseText); 
       //result = (xhttp.responseText); 
      } else { 
       result = { 
        error: 'status="'+xhttp.statusText+' ('+xhttp.status+')", state="'+xhttp.readyState+'"' 
       }; 
      } 
     }; 
     xhttp.open(requst.type || 'POST', 'backend.php', false); 
     xhttp.setRequestHeader("Content-Type", "application/json;charset=UTF-8"); 
     //xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); 
     xhttp.send(requst.data); 
    })(); 
    return result; 
} 

而且還要求使用

console.log('before'); 
console.log(ajax_({ 
    data: { 
     a: 1, 
     b: { 
      b1: [], 
      b2: {}, 
      b3: 'b3' 
     } 
    } 
})); 
console.log('after'); 

如何做到這一點控制檯顯示

before 
Object { post: a: 1, b: {b1: [], b2: [], b3: 'b3'} 
after 
? 

問題是json發送不起作用。只有json從後端得到作品。

Answer 1

您無法通過AJAX發送對象。

<script> 
    function ajax_(requst) { 
     requst = requst || {}; 
     var result = {}; 
     var xhttp = window.XMLHttpRequest ? new XMLHttpRequest() : 
     new ActiveXObject("Microsoft.XMLHTTP"); 
     xhttp.onreadystatechange = function() { 
      if (xhttp.readyState === 4 && xhttp.status === 200) { 
       result = JSON.parse(xhttp.responseText); 
       //result = (xhttp.responseText); 
      } else { 
       result = { 
        error: 'status="'+xhttp.statusText+' ('+xhttp.status+')", state="'+xhttp.readyState+'"' 
       }; 
      } 
     }; 
     xhttp.open(requst.type || 'POST', 'backend.php', false); 
     //xhttp.setRequestHeader("Content-Type", "application/json;charset=UTF-8"); 
     xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); 
     xhttp.send('data='+JSON.stringify(requst.data)); 
     return result; 
    } 

    console.log('before'); 
    console.log(ajax_({ 
     data: { 
      a: 1, 
      b: { 
       b1: [], 
       b2: {}, 
       b3: 'b3' 
      } 
     } 
    })); 
    console.log('after'); 
</script> 

而對於backend.php

<?php echo $_POST['data'];