generate dynamic variable name in for loop php

Question

I am trying to generate variable name dynamically in php for loop. I want call a function within for loop and want to pass iterated values. What I tried is:

for ($i=1, $j=1; $i<=16 ; $i++, $j++) {
    $a='arrtreelevel_1';
    $b='0';
    $newname = $a.$j.$b;
    echo $newname,"<br>";
    echo $arrtreelevel_1[0];
    $arrtreelevel_2.$i = profiledetails($newname);
}

My problem is '$newname' variable printing $arrtreelevel_1[0] as a text not taking the value, and $arrtreelevel_11[0] returns the actual value. I tried with echo ${'arrtreelevel_1$i'} also which printing nothing. Assist me with any solution please.

Thanks in advance.

you need to use double quotes instead of single quotes ${"arrtreelevel_2$i"} — JYoThI
– JYoThI, Commented Jul 7, 2017 at 11:10

JYoThI · Accepted Answer · 2017-07-07 11:12:33Z

6

you need to use double quotes instead of single quotes

${"arrtreelevel_2$i"}=profiledetails($newname);

answered Jul 7, 2017 at 11:12

JYoThI

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Comments

Niklesh Raut · Accepted Answer · 2017-07-07 11:10:40Z

1

Use $$ to make runtime php variable. Replace this line

$arrtreelevel_2.$i = profiledetails($newname);

To

$tempvar = $arrtreelevel_2.$i;
$$tempvar = profiledetails($newname);

answered Jul 7, 2017 at 11:10

Niklesh Raut

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2 Comments

Shashikala Over a year ago

I am getting values for $arrtreelevel_2.$i properly as arrtreelevel_21,arrtreelevel_22 and son on.But $newname is printing $arrtreelevel_21 but actual $arrtreelevel_21value is 61 which i wanted.not the name of concatinated variable but the value of it

Niklesh Raut Over a year ago

@Webbie : $tempvar will give you name $arrtreelevel_21 and $$arrtreelevel_21 using $$ will give you value of $arrtreelevel_21

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