if（isset（$ _ POST [“myvar”]）&& isset（$ _ POST [“myvar2”]）失败

Question

If i use this code to take as input a number it works fine

<form  method = "post">
    Hours: <input type="text" name="myvar">
</form>

<?php
require 'connect.php';
if (isset($_POST["myvar"])) {

But the following code to get 2 numbers doesn't go pass if. How can i make this work?

<form  method = "post">
    Hours: <input type="text" name="myvar">
</form>
<form  method = "post">
    Minutes: <input type="text" name="myvar2">
</form> 

<?php
require 'connect.php';
if (isset($_POST["myvar"]) && isset($_POST["myvar2"])) {

Answer 1

You just need to use one form for both fields as:

<form method="post">
    Hours: <input type="text" name="myvar">
    Minutes: <input type="text" name="myvar2">
    <input type="submit" name="submit">
</form>

And in PHP you can use !empty() for form input.

<?
if (!empty($_POST["myvar"]) && !empty($_POST["myvar2"])) {
...
?>

Second solution, if you still want to use isset() than also add !empty() for checking as:

<?
if (isset($_POST["myvar"]) && !empty($_POST["myvar"]) && 
    (isset($_POST["myvar2"]) && !empty($_POST["myvar2"]))) {
    echo "<pre>";
    print_r($_POST);
}
?>