How do I create an array for a range of values with keys using php?

Question

I am using CodeIgniter. I want to create an array to add to a dropdown which contains the numbers 1 -> 1000.

I have tried the php range() function like so

$arr = range(1,1000);

It worked and create a dropdown from 1 to 1000.

I do have one problem though.

When select text 1 from my drop down and post, the posting value is 0. Because by default the keys are starting from 0 and the key is set to the dropdown value

Here is part of my drop down HTML

<select id="user-day" class="dropdown-small Verdana11-424039" tabindex="123456" name="days_of_month">
<option value="0">1</option>
<option value="1">2</option>
<option value="2">3</option>
<option value="3">4</option>
<option value="4">5</option>
<option value="5">6</option>
<option value="6">7</option>

Is there any way I can define range() with key values?

Such that the values will become

<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>

nickb · Accepted Answer · 2011-11-22 13:51:05Z

11

This will form your array correctly using array_combine:

$array = array_combine( range(1,1000), range(1,1000));

answered Nov 22, 2011 at 13:51

nickb

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Comments

WWW · Accepted Answer · 2011-11-22 13:50:58Z

4

You could create an array the same way you're doing it, just expand the range a bit then unset the 0th element.

$arr = range(0,1000);
unset($arr[0]);

answered Nov 22, 2011 at 13:50

WWW

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4 Comments

manny Over a year ago

this will destroy first array, so instead 1000 list option you will get only 999

WWW Over a year ago

Yes it will destroy the 0th element of the array, but range(0,1000) is inclusive - there would be 1001 elements to the array it generates, therefore unsetting the 0th element would leave you with 1000 elements, 1-1000.

Wallace Vizerra Over a year ago

but if the range starting with 50, then will be wrong.

WWW Over a year ago

@WallaceMaxters Umm, yes, this answer would be wrong then, as what you're saying is not the original problem. Nice 4+ year necro comment, btw.

Rooneyl · Accepted Answer · 2011-11-22 14:05:12Z

1

Could use an old-fashioned for loop;

for ($i=1; $i <= 1000; $i++) {
   $arr[$i] = $i; 
}
print_r($arr);

Or just adjust the form population

$arr = range(1,10);
print_r($arr);
echo '<select>';
foreach ($arr as $a) {
   $value = $a+1;
   echo '<option value=\"'.$value.'">'.$a.'</option><br />';
}
echo '</select>';

answered Nov 22, 2011 at 14:05

Rooneyl

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Comments

David · Accepted Answer · 2011-11-22 13:54:31Z

0

Another way is to create the array through a loop:

for ($i = 1; $i <= 1000; $i++)
    $arr[$i] = $i;

answered Nov 22, 2011 at 13:54

David

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Comments

Your Common Sense · Accepted Answer · 2011-11-22 14:01:46Z

0

Just don't use value attribute at all.
So, the form will send you option instead.

answered Nov 22, 2011 at 14:01

Your Common Sense

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Comments

user1058988 · Accepted Answer · 2011-11-23 07:31:27Z

-2

Use the following code

<select id="user-day" class="dropdown-small Verdana11-424039" tabindex="123456" name="days_of_month">
<?php for ($i=0; $i <= 1000; $i++) { ?>
 <option value="<?php echo $i; ?>"><?php echo $i; ?></option>
<? } ?>

answered Nov 23, 2011 at 7:31

user1058988

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