有没有办法检查$ _POST是否存在？ [重复]

Question

This question already has an answer here:

• PHP - proper check if $_POST['variable'] is posted 4 answers

I am repeatedly calling the same page to perform the same action an x amount of times. In the page before this loop, I have a form asking for input. Only the first time when the loop runs I want to use the given value, but when I have the $_POST in the code, it gives an error.

So I have two questions:

1. Does PHP go on with executing the code when 'erroring', if not, is there a way to let it do so, because that would solve my problem.
2. Can I skip a piece of the code? Something like skip lines 12-14 or something...

I could post my code, but I don't think that will make much sense, since I have some other pages too and txt files and such and if you don't know what it is for, it is rather hard to understand.

I will summarize though:

page1.html:

<form action="page2.php" method="post">
<input type="text" name="input">
<input type="submit" value="Submit">
</form>

page2.php:

for($i = 0; $i < $_POST['text']; $i++){
print "<form action=\"page2.php\" method=\"post\">";
}

As you can see, the loop calls page2.php again, but the $_POST['text'] will not exist anymore. So, is there a way to go around this?

I have tried this:

if($another_counter_from_another_page > 1){
//Do nothing
}
else
{
$_SESSION['counter']=$_POST['text'];
}

And then replace the $_POST['text'] in the for-loop with $_SESSION['counter'], but I still get an error even when $another_counter_from_another_page > 1 is true, so that is why I asked for a way to skip a piece of code.

Thank you in advance, I hope my problem is clear from the example I gave...

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Answer 1

For a single key:

 if(isset($_POST['myKey'])) {
     echo "myKey exists"
 }

If you want to know if there is any POST data:

if(!empty($_POST)) {
    echo "We have POST data"
}