Any ideas why this wont print out

Question

Revising for php and cant seem to get this to print the values out that i want

Any ideas?

Thanks

<form action="revision.php" method="GET">
<input type=“text” name=“number[]”/>
<input type=“text” name=“number[]”/>
<input type=“text” name=“number[]”/>
<input type=“text” name=“number[]”/>
<input type=“text” name=“number[]”/>
<input type="Submit" name="Calcuate"/>
</form>

<?php 
if(isset($_GET['number'])){
    $amount = count($number);

    for($i=0; $i < $amount; $i++){
        echo $number[$i];
    }
}
?>

Can you put `var_dump($_GET)` somewhere in your code and show us the result? — Samir Talwar, Apr 23 '10 at 13:45

score 2 · Accepted Answer · answered Apr 23 '10 at 14:20

2

I think the actual problem with your code is that the quotation marks " are wrong you are using “ and ” instead of ". Replace those and everything will work.

answered Apr 23 '10 at 14:20

jondro

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lol thanks very much, thats what you get when you copy sample code from powerpoint lecture slides :D – stan Apr 23 '10 at 14:28

Dolph · Answer 2 · 2010-04-24T10:12:50.013

1

EDIT: My answer is completely wrong. See @rmarimon in comments below.

Text fields can't be mapped to an array. You'll have to name them something ugly like "number1", "number2", etc and add them up with $_GET['number1'] + ...

edited Apr 24 '10 at 10:12

answered Apr 23 '10 at 13:46

Dolph

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Actually text fields can be mapped to arrays in the way @stan is doing. PHP maps all request variables ending in [] to an array and discards multiple values of variables not ending in []. You can check this here http://docs.php.net/manual/en/faq.html.php#faq.html.arrays – Ricardo Marimon Apr 23 '10 at 14:08

score 0 · Answer 3 · answered Apr 23 '10 at 13:45

0

 <form action="revision.php" method="GET" enctype="multipart/form-data">

Change form to this. The multipart tag must be used for this

you also need this for file uploads

and for printing use this

foreach ($_GET['number'] AS $key => $value)
{
    echo "$key => $value";
}

because the array can be number[1] -> number[3]

answered Apr 23 '10 at 13:45

Marco

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Does that let them be mapped to an array? If so, I wasn't aware of that! – Dolph Apr 23 '10 at 13:47
my lecture slides appear to say that we can create the form how i have done tho? – stan Apr 23 '10 at 13:47
1

Try it and print the get print_r($_GET); and you will see the data – Marco Apr 23 '10 at 13:57

score 0 · Answer 4 · edited May 23 '17 at 11:47

0

It is not in your code but do you have

$number = $_GET["number"]

What you are doing is the correct way. This is similar to this other question.

edited May 23 '17 at 11:47

Community

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1

answered Apr 23 '10 at 13:52

Ricardo Marimon

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score 0 · Answer 5 · answered Apr 23 '10 at 14:56

The way I see it, there's a couple things you should change in your code, first, the names of the fields, you're trying to name them number[0], number[1], number[2] from the looks of it but it won't work that way, try naming them differenty or try to make a FOR cicle to create the fields with those custom names. Second, in order to save the array coming in the $_GET variable into the $number variable you need something like this:

if(isset($_GET['number']))
{
    $number = $_GET['number'];
    $amount = count($number);
    for( $i = 0 ; $i < $amount ; $i++ )
        echo $number[$i];
}

Hope this helps, if you're still having problems try posting or describing the context and what you have in mind for the form and the array.

Any ideas why this wont print out

5 Answers5