2

I want to merge 3 arrays.

My first one is:

Array ( [0] => Leaves-19 [1] => Shifts-1 [2] => Shifts-1 [3] => Shifts-1 [4] => Shifts-1 [5] => Shifts-1 [6] => Leaves-19 [7] => Leaves-19 [8] => Shifts-1 [9] => Shifts-1 [10] => Shifts-1 [11] => Shifts-1 [12] => Shifts-1 [13] => Leaves-19 [14] => Leaves-19 [15] => Shifts-1 [16] => Shifts-1 [17] => Shifts-1 [18] => Shifts-1 [19] => Shifts-1 [20] => Leaves-19 [21] => Leaves-19 [22] => Shifts-1 [23] => Shifts-1 [24] => Shifts-1 [25] => Shifts-1 [26] => Shifts-1 [27] => Leaves-19 [28] => Leaves-19 [29] => Shifts-1 [30] => Shifts-1 [31] => Shifts-1 [32] => Shifts-1 [33] => Shifts-1 [34] => Leaves-19 [35] => Leaves-19 [36] => Shifts-1 [37] => Shifts-1 [38] => Shifts-1 [39] => Shifts-1 [40] => Shifts-1 [41] => Leaves-19 )

My second is:

Array ( [0] => 2013-04-28 [1] => 2013-04-29 [2] => 2013-04-30 [3] => 2013-05-01 [4] => 2013-05-02 [5] => 2013-05-03 [6] => 2013-05-04 )

The third one is:

Array ( [0] => 13 [1] => 10 [2] => 12 [3] => 9 [4] => 14 [5] => 11 )

I want:

  • 2013-04-28 / 13 / Leaves-19
  • 2013-04-29 / 13 / Shifts-1
  • 2013-04-30 / 13 / Shifts-1
  • 2013-05-01 / 13 / Shifts-1
  • 2013-05-02 / 13 / Shifts-1
  • 2013-05-03 / 13 / Shifts-1
  • 2013-05-04 / 13 / Leaves-19
  • 2013-04-28 / 10 / Leaves-19
  • 2013-04-29 / 10 / Shifts-1
  • 2013-04-30 / 10 / Shifts-1
  • 2013-05-01 / 10 / Shifts-1
  • 2013-05-02 / 10 / Shifts-1
  • 2013-05-03 / 10 / Shifts-1
  • 2013-05-04 / 10 / Leaves-19
  • ...

Thansk for help.

What I tryed:

echo print_r($_POST['dayType'])."<hr />";
echo print_r($_POST['dayArr'])."<hr />";
echo print_r($_POST['userArr'])."<hr />";

//echo count($_POST['dayType'])." --- ".count($_POST['dayArr'])." --- ".count($_POST['userArr']);

// 2013-05-04 / 13 / Leaves-19
$loopNb1 = count($_POST['dayType']);
$loopNb2 = count($_POST['dayType'])/7;

for($a=0; $a<$loopNb1; $a++) {
    echo $_POST['dayType'][$a]."<br />";
}

echo "<hr />";

for($b=0; $b<$loopNb2; $b++) {
    echo $_POST['userArr'][1]."<br />";
}
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5
  • 2
    It would be nice to post what you have tried. Commented May 4, 2013 at 19:08
  • Your third array has only 6 elements, not 7. Commented May 4, 2013 at 19:09
  • Thanks. I put an example. Think it's better for understanding now. Commented May 4, 2013 at 19:11
  • You might be looking for php.net/array_map, see Example #4 Creating an array of arrays Commented May 4, 2013 at 19:11
  • Thanks for links but every examples shows egual arrays. Mines are not eqal. Commented May 4, 2013 at 19:21

3 Answers 3

Reset to default
1

First/second/third arrays are the same that you pasted (Same order).

$datesCount        = count( $secondArray );
$firstArrayLength  = count( $firstArray );
$thirdArrayLength  = count( $thirdArray );

for( $i=0 ; $i < $thirdArrayLength ; $i++ )
{
    $currentThirdArrayValue = $thirdArray[$i];

    for( $inner=0, $firstArrayIndex=0 ; $inner < $datesCount ; $inner++, $firstArrayIndex++ )
    {
        if( $firstArrayIndex == $firstArrayLength )
                $firstArrayIndex = 0;

        echo "{$secondArray[$inner]} / {$currentThirdArrayValue} / {$firstArray[$firstArrayIndex]}<br/>\n";
    }
}

Will give you:

2013-04-28 / 13 / Leaves-19<br/>
2013-04-29 / 13 / Shifts-1<br/>
2013-04-30 / 13 / Shifts-1<br/>
2013-05-01 / 13 / Shifts-1<br/>
2013-05-02 / 13 / Shifts-1<br/>
2013-05-03 / 13 / Shifts-1<br/>
2013-05-04 / 13 / Leaves-19<br/>
2013-04-28 / 10 / Leaves-19<br/>
2013-04-29 / 10 / Shifts-1<br/>
2013-04-30 / 10 / Shifts-1<br/>
2013-05-01 / 10 / Shifts-1<br/>
2013-05-02 / 10 / Shifts-1<br/>
2013-05-03 / 10 / Shifts-1<br/>
2013-05-04 / 10 / Leaves-19<br/>

etc... ending with:

2013-05-01 / 11 / Shifts-1<br/>
2013-05-02 / 11 / Shifts-1<br/>
2013-05-03 / 11 / Shifts-1<br/>
2013-05-04 / 11 / Leaves-19<br/>
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9 Comments

I really need to make like in my example. :(
Hm, I see what you mean, the fix should be to loop over the dates array and just check if the other two are going to be above their last item and reset the array. -- Want me to modify it?
It would be so appreciated please. Thanks.
Sure, a simple clarification, before your first "..." is 2013-05-04 / 13 / Leaves-19, which is the next value you expect?
Yes, I just put ... to separate them. :) Thanks.
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1

Get an idea-

//Your array
$a = array(1, 2, 3, 4, 5, 6,7,8);
$b = array("one", "two", "three", "four", "five",'123', 'asdfsadf');
$c = array("uno", "dos", "tres", "cuatro", "cinco", 'jina');


function myFunc($first, $second, $third) {
    if (!empty($first) && !empty($second) && !empty($second)) {
        return $first . '/' . $second . '/'. $third;
    }
}

// get the lowest size of array
$minimumSize = min(count($a), count($b), count($c));

//modify array in terms of lowest size
$firstArray = array_slice($a, 0, $minimumSize);
$secondArray = array_slice($b, 0, $minimumSize);
$thirdArray = array_slice($c, 0, $minimumSize);

$d = array_map('myFunc', $firstArray, $secondArray, $thirdArray);
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1 Comment

Thanks for your help but my arrays aren't equal.
1

If you want all combinations from those 3 arrays (call them $first, $second, $third), you can do something like this:

foreach($first as $a){
 foreach($second as $b){
   foreach($third as $c{
     //do whatever you want with $a,$b,$c
     echo "$a/$b/$c";
   }
 }  
}
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6 Comments

Thanks but I need to keep relation between informations. Elvena code work like a charm but not for the 1st array if datas are not the same between users, Do your understand? Could your please help? Thanks.
does this data by any chance come from a database? Isn't it better to solve the problem in the database instead of PHP?
This is to save this sort of table send-picture.com/upload/…
I really don't understand the connection between the first array (which has indices from [0] => Leaves-19 [1] => Shifts-1 ... [40] => Shifts-1 [41] => Leaves-19) what does the index 40 and 41 mean for example?
Ok. First array array the day the user need to make (for example Shifts-41 is from 9am to 5pm). Second array is to know wich date are in the current view/table. The third one is for user. I know that in one week there's 7 days. My first array got 42 datas. 42/7=6. I have 6 users. I want for each date, for each user, each type of day he needs to make. With it solution Elvena was so close but there a trouble with day type if there are not the same for users.
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