How to enter keys into array using defined ranges?

Question

So I have code that outputs the following:

$test = array('test1.txt' => '1 June 2015',
               'test2.txt' => '1 June 2015',
               'test3.txt' => '1 June 2015',
               'test4.txt' => '1 June 2015',
               'test5.txt' => '1 June 2015');

But instead of 5 files there are hundreds. How would I make it so the I can define the date outputted depending on the line number i.e.

If file 1-5 (first 5 files), then date = "1 June 2015". If file 6-8 (next 2 files), then date = "5 June 2015. Etc. Etc. How would I go about doing so?

How are you defining your range. or on what basis you have your range — Meenesh Jain
– Meenesh Jain, Commented Jun 23, 2015 at 4:47
The number of files are fixed or they may vary? Or how the number will be defined? — Sougata Bose
– Sougata Bose, Commented Jun 23, 2015 at 4:54
The number of files may vary. The dates are defined manually by me for range (i.e. i define 1 date for first 5 files, define another for next 2, etc. etc.) — Ted
– Ted, Commented Jun 23, 2015 at 4:56
Then how one can decide the number of files? There must be some way that tell about that!! — Sougata Bose
– Sougata Bose, Commented Jun 23, 2015 at 4:57

Sougata Bose · Accepted Answer · 2015-06-23 05:29:41Z

1

You can try this -

$date = "1 June 2015"; // The initial date
$fileNum = array(5, 2, 3); // The number of files to be considered
$total = array_sum($fileNum); // Total number of files
$array = array(); // Array to be filled
$j = 1; // The number for increment
foreach($fileNum as $num) {
    $i = 1; // The number to compare the file numbers 
    while($i <= $num) {
         $array['test' . $j . '.txt'] = $date; // Store the values
         $j++;
         $i++;
    }
    $date = date('j F Y', strtotime('+ 1 DAY', strtotime($date))); // Increment the date
}

var_dump($array);

Output

array(10) {
  ["test1.txt"]=>
  string(11) "1 June 2015"
  ["test2.txt"]=>
  string(11) "1 June 2015"
  ["test3.txt"]=>
  string(11) "1 June 2015"
  ["test4.txt"]=>
  string(11) "1 June 2015"
  ["test5.txt"]=>
  string(11) "1 June 2015"
  ["test6.txt"]=>
  string(11) "2 June 2015"
  ["test7.txt"]=>
  string(11) "2 June 2015"
  ["test8.txt"]=>
  string(11) "3 June 2015"
  ["test9.txt"]=>
  string(11) "3 June 2015"
  ["test10.txt"]=>
  string(11) "3 June 2015"
}

Update

$fileNum = array(5 => "1 June 2015", 2 => "2 June 2015", 3 => "3 June 2015"); // Set the array with file number and corresponding dates
$total = array_sum($fileNum);
$array = array();
$j = 1;
foreach($fileNum as $num => $date) {
    $i = 1;
    while($i <= $num) {
         $array['test' . $j . '.txt'] = $date;
         $j++;
         $i++;
    }
}

edited Jun 23, 2015 at 5:29

answered Jun 23, 2015 at 5:11

Sougata Bose

31.8k8 gold badges52 silver badges89 bronze badges

Sign up to request clarification or add additional context in comments.

10 Comments

Ted Over a year ago

Hey, thanks for the reply. This is my actual code, can it integrate with this? $path = "/home/files/"; $date = "1 June 2015"; $files = array();foreach (glob($path . "*.txt") as $filename) { $files[$filename] = ""; var_export($files);

Ted Over a year ago

What does array(5, 2, 3) do? How would I go about doing that? Sorry, I am quite a beginner when it comes to PHP.

Sougata Bose Over a year ago

It is the number of files.

Ted Over a year ago

I don't want to increment dates by 1 I want to define using something like this: $files1to5 = "1 June 2015"; $files6to8 = "2 June 2015"; $files9to50 = "11 June 2015";

Ted Over a year ago

I ran the code but it had no output. How would I add my "$files" array in the code $path = "/home/files/"; $date = "1 June 2015"; $files = array();foreach (glob($path . "*.txt") as $filename) { $files[$filename] = ""; var_export($files);

|

Dinuka Dayarathna · Accepted Answer · 2015-06-23 05:50:27Z

0

Try This

  $numberOfFiles = 100;
//$startDate = '1 June 2015';

$startDate = '1 June 2015';
for($i=1;$i<=$numberOfFiles;$i++)
{
    if($i%5==0)
    {
        $test['test'.$i.'.txt'] = $startDate;
        $startDate = date('j F Y',strtotime($startDate.' + 5 days'));
    }
    else
    {
        $test['test'.$i.'.txt'] = $startDate;
    } 
}

print_r($test);

Updated to requirement. Note You have to define the date against file number.

$numberOfFiles = 50;
//$startDate = '1 June 2015';

for($i=1;$i<=$numberOfFiles;$i++)
{
    if($i<=5)
    {
        $startDate ='1 June 2015';
        $test['test'.$i.'.txt'] = $startDate;
    }
    elseif((5<$i)&&($i<=8))
    {
        $startDate ='2 June 2015';
        $test['test'.$i.'.txt'] = $startDate;
    }
    elseif((8<$i)&&($i<=50))
    {
        $startDate ='11 June 2015';
        $test['test'.$i.'.txt'] = $startDate;
    }
    else
    {
        // Define the date value if does not match above criterian
    } 
}

print_r($test);

edited Jun 23, 2015 at 5:50

answered Jun 23, 2015 at 4:56

Dinuka Dayarathna

1691 gold badge3 silver badges9 bronze badges

3 Comments

Sougata Bose Over a year ago

Are you sure the number of files will be 5?

Dinuka Dayarathna Over a year ago

Need a specific pattern to increment date against file number.

Ted Over a year ago

Thanks! I'm testing it out right now, will comment back results

Collectives™ on Stack Overflow

How to enter keys into array using defined ranges?

2 Answers 2

10 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

10 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related