php variable after a variable in for loop

Question

I need a way to do this

for($i=1;$i<=10;$i++){
$id$i = "example" . $i;
}

notice the second line has $id$i so for the first loop $id1 will equal example1 in the second loop $id2 will equal example2 and so on... Thank you very much!

That's what arrays are for, is there a reason you can't use an array? — jeroen
– jeroen, Commented Jul 11, 2012 at 21:26
i have tryed $id[1]= "example" . $i; but its not working. i think the syntax is wrong — EtayRock
– EtayRock, Commented Jul 11, 2012 at 21:28
It is possible using variable variables, but not recommended: ideone.com/gjvIX. — mellamokb
– mellamokb, Commented Jul 11, 2012 at 21:29
Just initialize the array before your loop and use $id[$i] = "example" . $i in the loop — jeroen
– jeroen, Commented Jul 11, 2012 at 21:30
the syntax would be $id[$i] instead of $id[1]. What you tried would be putting each string in position 1 of the array — Borophyll
– Borophyll, Commented Jul 11, 2012 at 21:31

Palladium · Accepted Answer · 2012-07-11 21:28:31Z

6

You can use variable variable names to do this; however, it's probably much more convenient if you just used an array:

for($i = 1, $i <= 10, $i++) {
    $id[] = "example" . $i;
}

answered Jul 11, 2012 at 21:28

Palladium

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4 Comments

EtayRock Over a year ago

what should be inside the []?

Palladium Over a year ago

Nothing. $id[] means append to the end of array $id.

Palladium Over a year ago

Note that array indices start from 0; if you REALLY want the indices to start from 1, you can use $id[$i] in this particular case instead.

EtayRock Over a year ago

I know about arrays i just aint knew that [] means the end of the array. thank you! it works.

Stefan Fandler · Accepted Answer · 2012-07-11 21:28:50Z

0

You can convert a string into a variable (the name of the variable), if you put another $ in front of it:

$str = "number";
$number = 5;
$$str = 8;
echo $number;  // will output 8

So in your example, you could do it like that:

for($i = 1; $i <= 10; $i++) {
    $str_var = "id".$i;
    $$str_var = "example".$i;
}

answered Jul 11, 2012 at 21:28

Stefan Fandler

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Comments

Andrew · Accepted Answer · 2012-07-11 21:30:31Z

0

It would be much better to use an array, but you could do this:

for($i=1; $i<=10; $i++){
    $var ="id$i";
    $$var = "example" . $i;
}

Here's what I would recommend doing instead:

$ids = array;
for($i = 1; $i <= 10; $i++) {
    $ids[$i] = "example" . $i;
}

answered Jul 11, 2012 at 21:30

Andrew

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Comments

Brian J · Accepted Answer · 2012-07-11 21:30:37Z

0

You could create an array of size $i with a name of $id, and insert each element into a different index.

for($i=1;$i<=10;$i++){
    $id[$i] = "example" . $i;
}

answered Jul 11, 2012 at 21:30

Brian J

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Comments

Federkun · Accepted Answer · 2012-07-11 21:35:57Z

0

$var = array();
for($i=1; $i<=10; $i++) {
$var['id' . $i] = 'example' . $i;
}
extract($var, EXTR_SKIP);
unset($var);

but why not use a simple array?

answered Jul 11, 2012 at 21:35

Federkun

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