Question

我有一个发布到PHP文件的基本表单。

<form action="index.php" method="POST">
<input name="operation" id="operation" placeholder="operation" />
<br>
<input id="name" name="name" placeholder="Name" />
<br>
<input id="email" name="email" placeholder="Email"/>
<br>
<input id="password" name="password" placeholder="Password"/>
<br>
<button type="submit" >POST</button>
</form>

问题是操作是通过下面的索引文件发布NULL或空。我使用基本的php：//输入来通过json进行编码。

if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
    $data = json_decode(file_get_contents('php://input'));

    if(isset($data -> operation)){

      $operation = $data -> operation;
      echo $operation;
       if(!empty($operation)){

       }else{

           //$operation is empty ...

       }

     }else{

          //$operation is not set ...

     }

  }

但是，回显file_get_contents（'php：// input'）会从发布的表单中显示正确的值。

为什么$ operation返回始终为空？

Answer 1

在转换为Json格式之前，您的数据应该准备好试试这个代码:) 我测试这个代码，它工作。祝你好运



    if ($_SERVER['REQUEST_METHOD'] == 'POST')
{

    $data = file_get_contents('php://input');
    $data = str_replace('=','":"',$data);
    $data = str_replace('&','","',$data);
    $data = '{"'.$data.'"}';

    $data = json_decode($data);

    if(isset($data->operation)){

      $operation = $data -> operation;
      echo $operation;
       if(!empty($operation)){
            echo "NOT EMPTY";
       }else{

            echo "IS EMPTY";
           //$operation is empty ...

       }

     }else{
            echo "NO OPERATION";
          //$operation is not set ...

     }

  }

通过JSON的PHP POST是空的

1 个答案: