後表單提交沒有發佈數據json_encode

Question

我想從這段代碼得到一些輸出，但我沒有得到我不會。

以下代碼：

<?php 
     if(isset($_POST['send']) && $_POST['send']=="Send"){ 
     $array = $_POST['message']; 
     $var = call_user_func_array("json_encode",$array); 
     $var2 = call_user_func_array("json_encode",array(&$array)); 
     echo "var => ".$var."\n"; 
     echo "var2 => ".$var2."\n"; 
    } 
    ?> 

輸出

var => var2 => "array('a'=>1,'b'=>2)" 

需要

var => 1 var2 => {"a":1,"b":2} 

<div class="container"> 

     <form class="form-horizontal" method="post"> 
     <div class="form-group"> 
      <label class="control-label col-sm-2" for="message">Message:</label> 
      <div class="col-sm-5">   
      <textarea class="form-control" required id="message" name="message" placeholder="Enter Message">array('a'=>1,'b'=>2)</textarea> 
      </div> 
     </div> 
     <div class="form-group">   
      <div class="col-sm-offset-2 col-sm-10"> 
      <input type="submit" name="send" class="btn btn-primary btn-lg" value="Send" /> 
      </div> 
     </div> 
     </form> 
    </div> 

在此先感謝

Answer 1

<?php 
if (isset($_POST['send']) && $_POST['send']=="Send") { 
$array = $_POST['message']; 
$array = eval($array); 
$var = ""; 
if (is_array($array)) { 
    $var = json_encode($array); 
} 
echo "var => ".$var."\n";  
}