dynamic array variable with dynamic values

Question

I am in too much trouble. I need below type of array:-

$val = "abc";

$arr1["besk"] = $val
$arr2["besk"] = $val
     .    
     .
$arr15["besk"] = $val

I tried below:-

for($i = 1; $i<16; $i++)
{
   $arr.$i["besk"] = $val
}

here I've $val. so not to worry on that. But array is not properly creating. Any help would be appreciated.

if you don't mind kindly explain what is $var and val in your program. — Kmg Kumar
– Kmg Kumar, Commented Oct 14, 2017 at 5:23
"I need below type of array" - no, more likely you just think that you do ... "numbered" variables names are almost always a sign that you should use an array in that place instead. — C3roe
– C3roe, Commented Oct 14, 2017 at 5:33
@Everybody Thanks for great and quick answer. As everybody have same answer so not able to accept anybody's. Again Thanks for giving me all of Your valuable time. — Ripa Saha
– Ripa Saha, Commented Oct 14, 2017 at 5:52

Qadir Pervez · Accepted Answer · 2017-10-14 05:36:16Z

3

first define the array as string like

$arr = 'arr';

then use the foreach like

for($i = 1; $i<16; $i++)
  {
      ${$arr.$i}["besk"] = $val;
 }

answered Oct 14, 2017 at 5:36

Qadir Pervez

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Comments

Ghassen Rjab · Accepted Answer · 2017-10-14 05:50:48Z

2

You need to use variable variables (not recommended)

for($i = 1; $i<16; $i++)
{
    ${"arr".$i}["besk"] = $val
}

EDIT : @CBroe is right about his comment, you should use an array instead. So the best solution would be to create a two dimensional array like so :

$arr = [];
for($i = 0; $i<15; $i++)
{
    $arr[$i]["besk"] = $val
}

The only difference is your array indexes start from 0 now and if you want to have the third value of your array you need this command $arr[2]["besk"]

edited Oct 14, 2017 at 5:50

answered Oct 14, 2017 at 5:29

Ghassen Rjab

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Comments

Mohammed Ahmed · Accepted Answer · 2017-10-14 05:30:06Z

1

it is very simple use this:

 for($i = 1; $i<16; $i++)
 {
      ${$arr.$i}["besk"] = $val
 }

answered Oct 14, 2017 at 5:30

Mohammed Ahmed

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Comments

Dmitriy Buteiko · Accepted Answer · 2017-10-14 05:31:59Z

1

Use this approach:

for($i = 1; $i<16; $i++)
{
    ${$arr.$i}["besk"] = $val;
}

answered Oct 14, 2017 at 5:31

Dmitriy Buteiko

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Comments

Ronny Sulistio · Accepted Answer · 2017-10-14 05:46:20Z

1

add new variable

$val = "abc";
$arrName = "arr"; //this one
$arr1["besk"] = $val
$arr2["besk"] = $val
     .    
     .
$arr15["besk"] = $val

and to call it

for($i = 1; $i<16; $i++)
 {
      ${$arrName.$i}["besk"] = $val
 }

ps. you did not create array, you just create 15 array variable with 1 index("besk" index)

edited Oct 14, 2017 at 5:46

answered Oct 14, 2017 at 5:41

Ronny Sulistio

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