php loop with variable changing simplified

Question

pretty sure my title needs to be fixed but anyhow, i have this loop. background info, i have 36 input "types" that need to be inserted 1 by 1. if the value has something.. set to a variable, if not set it to NULL.

MY question is, is there a loop i can perform to do this without listing 36 of these things.

using php version 5.2.17

    if (!empty($_POST['type1'])){
        $type1 = $_POST['type1'];
    }
    else 
        $type1 = NULL;

    if (!empty($_POST['type2'])){
        $type2 = $_POST['type2'];
    }
    else 
        $type2 = NULL;

    if (!empty($_POST['type3'])){
        $type3 = $_POST['type3'];
    }
    else 
        $type3 = NULL; // to 36...

php/html

<?php 
    for ($i = 1; $i < 37; $i++){
    echo "Type$i:<input name='type$i' type='text' size='20' maxlength='35' /><br />";
    }
 ?>

edit: I don't want to use an array.

This would be a whole lot easier if you defined the input names in your markup as an array: type[] — Mark Baker
– Mark Baker, Commented Jan 15, 2014 at 23:50
Or you could do the same thing: loop trough your post variables. — putvande
– putvande, Commented Jan 15, 2014 at 23:50
Starting question: why do you need them as seperate variables? Seems like an array would really do... Of course is can be done, but it's nasty to debug later on (you can't find easily where a variable was defined). — Wrikken
– Wrikken, Commented Jan 15, 2014 at 23:50
thats fine, i'm just looking for an answer, and i don't want to use an array. — Albert D
– Albert D, Commented Jan 15, 2014 at 23:52
with a for loop ? but still @wrikken 's comment is true, You dont WANT to use an array now, but debugging will become hell LATER — Gert B.
– Gert B., Commented Jan 15, 2014 at 23:56

grebneke · Accepted Answer · 2014-01-16 00:16:29Z

1

To actually define all those variables $type1, $type2 etc, use "variable variables":

for ($i = 1; $i < 37; $i++) {
    $varname = "type$i";
    if (! empty($_POST[$varname])) {
        $$varname = $_POST[$varname];
    }
    else {
        $$varname = NULL;
    }
}

Others suggest using an array instead (and in principle I agree), but this is an actual answer to the question.

edited Jan 16, 2014 at 0:16

answered Jan 15, 2014 at 23:59

grebneke

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3 Comments

Albert D Over a year ago

YES! indeed it does work, thanks for this. I know an array would work, there is 5000 ways of doing anything really, i wanted a simple solution to my problem without having to redo my entire script. thanks.

Albert D Over a year ago

if you could just explain to me why you use two $$ in $$varname ?

grebneke Over a year ago

It's all explained in the link to PHP-documentation on variable variables. Think of it this way: Let's say $i = 6. Then $varname = "type6", and $$varname is the same as saying $type6.

MitMaro · Accepted Answer · 2014-01-16 00:02:37Z

Question was updated to not use an array. Will keep it here for someone looking for an answer that can use an array.

You could add the $_POST array to a predefined array of values storing the result in another array. For example:

<?php
$defaults = array(
    "type1" => NULL,
    "type2" => NULL,
    "type3" => NULL,
    "type4" => NULL
    // etc
);
$_POST = array(
    "type1" => 1,
    "type2" => "foo"
);
$types = $_POST + $defaults;
print_r($types);

Which results in the array:

Array
(
    [type1] => 1
    [type2] => foo
    [type3] =>
    [type4] =>
    ...
)

Then for your html loop:

for ($i = 1; $i < 37; $i++){
    echo "Type$i:<input name='" . $types["type" . $i] . "' type='text' size='20' maxlength='35' /><br />\n";
}

It is important to note that this is slightly different than using a check with empty.

3dgoo · Accepted Answer · 2014-01-15 23:56:16Z

0

Use an array on both the php and form inputs.

$type = array();

for ($i = 1; $i <= 36; $i++) {
    if (isset($_POST['type'][$i])){
        $type[$i] = $_POST['type'][$i];
    }
}

HTML/PHP

<?php 
    for ($i = 1; $i <= 36; $i++){
        echo "Type$i:<input name='type[$i]' type='text' size='20' maxlength='35' /><br />";
    }
?>

answered Jan 15, 2014 at 23:56

3dgoo

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Comments

accurate respondetur · Accepted Answer · 2014-01-16 00:00:52Z

0

Just an example with an array named $post.

$post = array(
'type1' => 'one',
'name' => 'test',
'type3' => 'three'

);
$matches  = preg_grep ('^type[0-9]^', array_keys($post));

foreach ($matches as $val) {
$$val = $post[$val];
}

var_export($type1);

answered Jan 16, 2014 at 0:00

accurate respondetur

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Comments

Mando Madalin · Accepted Answer · 2014-01-16 00:10:23Z

0

Use array for this

$i = 0;
$myinputs = 36;

while($i<$myinputs){
 $i++;
 if(empty($_POST['type'.$i.''])){
    $type[$i] = $_POST['type'.$i.''];
 }else{
    $type[$i] = null;
 }



}

print_r($type);

answered Jan 16, 2014 at 0:10

Mando Madalin

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