Javascript 将变量指定给操作输入

我正在尝试一个POST方法，根据输入使用不同的操作

我的方法是这样的

<form enctype="multipart/form-data" method="POST" action="upload.php">
<p>Upload your .json file</p>

}

使用单选按钮的onclick方法调用check，例如

<input onclick="check(0)" type="radio" name="options" id="pie" aria-label="Radio button for choosing Pie chart" autocomplete="off" checked>

document.myForm.action=上传

}试试看

HTML

---

嘿，谢谢你的建议，不幸的是由于某种原因，行动没有改变。我测试了upload变量，它记录了choosen值，当提交时，它仍然总是上传。php@kukgStone当然，因为上面的代码不会改变表单操作。谢谢，它现在可以用于document.myForm.action=upload；而不是document.getElementById（“myForm”）.formAction=upload；是的，代码实际上应该是“document.getElementById（“myForm”）.action=upload；”。我已经在上面的示例中更正了它。

<input onclick="check(0)" type="radio" name="options" id="pie" aria-label="Radio button for choosing Pie chart" autocomplete="off" checked>

  function check(x) {

  if (x===0){
    upload = "upload.php";
    console.log(upload);

  } else if (x===1){
    upload = "upload_line.php";
    console.log(upload);

  } else if (x===2){
    upload = "upload_pie.php";
    console.log(upload);

  }else if (x===3){
    upload = "upload_round.php";
    console.log(upload);

  }else if (x===4){
    upload = "upload_bar2.php";
    console.log(upload);

  }else{
    upload = "upload.php";

  }

<form id="myForm" enctype="multipart/form-data" method="POST" action="upload.php">
<p>Upload your .json file</p>

function check( x ) {

  if ( x === 0 ) {
    upload = "upload.php";
    console.log(upload);
  } 
  else if ( x === 1 ) {
    upload = "upload_line.php";
  } 
  else if ( x === 2 ) {
    upload = "upload_pie.php";
  } else if ( x === 3 ) {
    upload = "upload_round.php";
  } else if ( x === 4 ) {
    upload = "upload_bar2.php";
  }
  else {
    upload = "upload.php";
  }

  // set the form's POST action.

  // Whoops! This should be
  // document.getElementById( "myForm" ).formAction = upload; 
  document.getElementById( "myForm" ).action = upload; 

}