PHP Passing Variable to Variable

Question

I posed anther thread about variable variables. I person responded but his code is not working. I asked him to clarify and he never responded. So I'm posting a follow up question.

I have the below code. When I put the dynamic $ct in the code it fails and says undefined variable. But when I remove the $ct, it works fine. Would someone clarify how I can add the $ct variable to the end of my variable so that I can have dynamic variables?

$ct = 1;
${"var".$ct} = "String";
echo  $var1;

The above fails.

$ct = 1;
${"var"} = "String";
echo $var1;

This works fine.

So something is not working right with adding the dynamic $ct.

He said it should work fine. I have also tried $$ using a variable then setting that variable and still fails the same.

/////////Edit

so my code looks like this:

$ct=1;
foreach ($Emp1 as $e){
    if ($e['Status']!="Primary"){



     ${"B1_E1_Info$ct"} = $e['Employer_Name']." ".$e['Address_Unit']." ".$e['City']." ".$e['State']." ".$e['Zip'];
     $B1_E1_SE.$ct  = (strpos($e['Self_Employed'], "| Y") !== false) ? true : false;
     $B1_E1_YrsOn.$ct  = $e['Years_on_Job'];
     $B1_E1_YrsIn.$ct  = $e['Years_in_Line_of_Work'];
     $B1_E1_Pos.$ct  = $e['Position_Title'];
     $B1_E1_Ph.$ct  = $e['Business_Phone'];
    }
    $ct= $ct+1;

}
 echo  $B1_E1_Info1;

When called iside the loop it doesn not work. If i take it out of the loop it works. Likewise if i change the code inside the loop to not inlcude the $ct it works as well. The $ct just doesn't work in the loop.

Your first example works fine for me, although generally when you have variable variables, what you should be doing is using an array. — Niet the Dark Absol
– Niet the Dark Absol, Commented Sep 24, 2019 at 16:08
Use arrays, not variable variables. Variable variables are confusing, difficult to use, and can create errors that are virtually impossible to find. — Sammitch
– Sammitch, Commented Sep 24, 2019 at 16:10
The first code works! The second code does not work as you have NOT created a $var1 !! Notice: Undefined variable: var1 — RiggsFolly
– RiggsFolly, Commented Sep 24, 2019 at 16:14
Maybe you can provide php version, the latest update are strictic types. Yu can use: ${"var$ct"} = "String"; — jashk
– jashk, Commented Sep 24, 2019 at 16:21

AmigoJack · Accepted Answer · 2019-09-24 17:46:23Z

0

Welcome to the internet - it works thru links. If your question is a follow-up then link to the answer you refer to. If it was https://stackoverflow.com/a/58083537/4299358 then that code might "fail" because $record could be empty, never ever setting $var1.

Everything else has been said in the comments already: your first code does work - you either have a very outdated PHP version or you haven't tested that code yourself (in its own PHP file with nothing before). Also comments here and there suggest you to use arrays instead.

answered Sep 24, 2019 at 17:46

AmigoJack

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user982853 · Accepted Answer · 2019-09-24 20:04:36Z

0

Ugh!!! Thanks Amigojacks. Also i figured out my error. I was calling $ct = $ct+1, outside the condition loop so it was actually setting $var2 instead of $var1 because the loop already triggered once. When i moved the $ct + 1, inside the loop, it now works fine. Thanks and sorry to waste your time. Sometimes is just takes talking through to find your own mistake.

answered Sep 24, 2019 at 20:04

user982853

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