$ _POST和代码操作

Question

I want to grab the value of a field using $_POST, manipulate it, then pass the value back to the same page to the same field before the PHP code manipulates it.

If I put the PHP code after the field, it manipulates the code, reloads the page but doesn't put the manipulated code back into the field.

    if (!isset($input)) {
        $input = '';
    }   

    echo '<form id="testform" method="post" action="">';
    echo '<input type="text" name="inputText" value="' . $input . '">';
    echo '<button type="submit" name="button"> Button </button>';
    echo '</form>';

    $input = $_POST['inputText'];

    if(isset($_POST['inputText'])) {
        $input = $input . ' manipulated';
    }

    echo $input; //test

If I put the PHP code before the field, it can't find the field to manipulate the value...

    if (!isset($input)) {
        $input = '';
    }

    $input = $_POST['inputText'];

    if(isset($_POST['inputText'])) {
        $input = $input . ' manipulated';
    }

    echo $input; //test

    echo '<form id="testform" method="post" action="">';
    echo '<input type="text" name="inputText" value="' . $input . '">';
    echo '<button type="submit" name="button"> Button </button>';
    echo '</form>';

Obviously the first approach is more correct, but how do I pass the $input variable to the field before the rest of my PHP manipulation code executes?

I tried $_POST['inputText'] = $input as a desperate attempt but nothing..

Answer 1

Well, from what I've understood in your explanation, you want to change the input value to something else and show it in he same field. If that's correct, you may want to do this:

<form id="testform" method="post" action="">
    <input type="text" name="inputText" value="<?php echo ( isset($_POST['inputText']) ) ? sprintf( '%s manipulated', $_POST['inputText'] ) : ''; ?>">
    <button type="submit"> Send </button>
</form>

Let me know if that's what you wanted. Regards !