Variable variables with $_POST

Question

I have a long list of variables with a number at the end. For example address1 and I have all the way up top address14. I want to post these from a form but rather than type out $address1 = $_POST[address1] I would like to create a loop that loops round 14 times and changes the number at the end of address in both the variable name and in the $_POST section...

I am struggling to do this. I have a loop that creates the varaibles, but I keep getting errors as it isn't doing the $_POST bit.

Can someone please assist? Thank you.

This is what I currently have:

        $x = 0;
        while($x < 14) {
            $address = "address" . $x;
            $address = $$address;

            $string = "<p>Address$x:" . $address[0] . "</p>";
            echo $string;
            $x = $x + 1;

}

what do you mean by the $_POST - bit? You have no form? Are you trying to create a form with names and adress that user has to type in? And then you want to get these values with $_POST. Is that what you want? — bestprogrammerintheworld, Apr 20 '13 at 14:12
I have a form in another page. It posts the fields across as it should, but I have 14 of the same field, but the names change just by one digit, so address1, address2, address3, all the way up to address14. I need to create a loop to receive these from the form that has posted them. So where at the top of the page you create a new variable with the $_POST function I need that but 14 times. For example: $address1 = $_POST [address1], $address2 = $_POST [address2] and so on up until address 14. — Ade, Apr 20 '13 at 14:16
I understand I could use an array but I am trying to use variable variables. — Ade, Apr 20 '13 at 14:18
why do you need to use variable instead of an array? Can you show the code of how you retrieve the $_POST-variables then? — bestprogrammerintheworld, Apr 20 '13 at 14:19
Because on the form they are individual entities, they aren't set up as an array and I have dynamically created them on my form too using a counter, which is easier as it is just the name given to the input fields, but when trying to append the end of 'address' with a number for a variable it becomes more complex. — Ade, Apr 20 '13 at 14:22
You could easily create dynamically created elements in a form (add brackets after name). Like this or you could solve it like user2301283 answered. — bestprogrammerintheworld, Apr 20 '13 at 14:23

score 2 · Accepted Answer · answered Apr 20 '13 at 14:24

2

why you don't do:

for ($i=0; $i < 14; $i++) {
    $address[$i] = $_POST['address'.$i];
}

answered Apr 20 '13 at 14:24

Fosfor

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2
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score 0 · Answer 2 · answered Apr 20 '13 at 14:26

Try this:

$addresses = array();
for( $x = 0; $x <= 14; $x++ )
{
    $address_field = "address" . $x;
    if( array_key_exists( $address_field, $_POST ) ) {
        $addresses[$x] = $_POST[$address_field];
        echo '<p>Address', $x, ': ', $addresses[$x], "</p>\n";
    }
}

score 0 · Answer 3 · answered Apr 20 '13 at 14:56

0

You could also use the array syntax in your form :

<input type="text" name="address[1]" />
<input type="text" name="address[2]" />
<input type="text" name="address[3]" />
... up to 14

In PHP :

$address = $_POST['address'];

$address is now an array containing all the addresses

answered Apr 20 '13 at 14:56

Frédéric Clausset

1,086
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in name you don't need to specify the element. just name="address[]" for all elements will work fine. – Aris Apr 20 '13 at 15:06
Yes it works without specifying the keys. Just added for clarity, but you don't need them. – Frédéric Clausset Apr 20 '13 at 15:21

Variable variables with $_POST

3 Answers3