如何在循环中提交没有提交按钮的选择菜单？

I have a code that submits a form when an option in a select box is clicked, but my problem is when I put it inside a loop, it doesn't work. Can you help me with it?

Here's the code:

<?php 
  $y1=5;
  for($x1=1; $x1<=$y1; $x1++){
?>
    <form name="myform[]" action="test.php" id ="myform" method="post">
      <select id="sel_id" name="sel_name[]"  onchange="submitform();">
      <?php 
        $y=5;
        for($x=1; $x<=$y; $x++){
      ?>
          <option value="<?php echo $x;?>"><?php echo $x;?></option>
      <?php
        }
      ?> 
      </select>
    </form>

    <script type="text/javascript">
      function submitform()
      {
        document.myform.submit();
      }
    </script>

<?php
  }
?>  

Now, I want every value of it be printed:

<?php
  echo $_POST['sel_name'];
?>

You are using multiple form with same id="myform". this should be unique for every form.

Example:

<?php 
$y1=5;
For($x1=1; $x1<=$y1; $x1++){
?>
<form name="myform[]" action="test.php" id ="myform_<?=$x1?>" method="post">
<select id="sel_id" name="sel_name[]"  onchange="submitform(<?=$x1?>);">
<?php 
$y=5;
For($x=1; $x<=$y; $x++){
?>
<option value="<?php echo $x;?>"><?php echo $x;?></option>
<?php
}
?> 
</select>
</form>
<?php
}
?>  

Your Java Script:

<script type="text/javascript">
function submitform(id)
{
    document.getElementById("myform_"+id).submit();
}
</script>

Your test.php:

<?php
if(isset($_POST['sel_name']))
    print_r($_POST['sel_name']);
?>

I am using different form ids as myform_<?=$x1?> and also using this id into submitform() function.

As per @Mr.Engineer, no need to use java script function submitform() inside the for() when you are using the unique IDs for form submission.

Just use this inside your function submitform() :

document.getElementById("myform").submit();

EDIT : I didnt notice your for loop

You are using form in for loop, so you need separate id for each form.

So try this :

<?php 
$y1=5;
for($x1=1; $x1<=$y1; $x1++){
?>
<form name="myform[]" action="test.php" id ="myform<?php echo $x1;?>" method="post">
<select id="sel_id<?php echo $x1;?>" name="sel_name[]"  onchange="submitform(<?php echo $x1;?>);">
<?php 
$y=5;
for($x=1; $x<=$y; $x++){
?>
<option value="<?php echo $x;?>"><?php echo $x;?></option>
<?php
}
?> 
</select>
</form>
<?php } ?>

And outside of your for loop write your JS function :

function submitform(id)
{
document.getElementById("myform"+id).submit();
}

Please try below code for submit form using selectbox.

<select id="sel_id" name="sel_name[]" onchange="this.form.submit()">
    ...
</select>

Try to this may be it's help for you

$(field).closest("form").submit();

For example, to handle the onchange event, you would have this:

$(select your fields here).change(function() {
    $(this).closest("form").submit();
});