Replace number in variable with another variable in for loop

Question

I'm trying to produce 25 variables from $day1Txt through to $day2Txt.

How do I turn the '1' in the variable $day1Txt into the variable $i?

for ($i = 1; $i < 25; $i++) { 

        $day1Txt = echo call_user_func('Day_'.$i.'_Offer', 'EMAIL_OFFER'); 

}

I've made an attempt myself but it white screens...

  ${'day'.$i.'Txt'} = echo call_user_func('Day_'.$i.'_Offer', 'EMAIL_OFFER');

Please expand on your question, what are you trying to achieve? If it is in a loop it will be keep changing until the last iteration of the loop. — Script47
– Script47, Commented Nov 6, 2015 at 11:19
I just removed the echo and it hasn't errored, but I dont know if the variables are stored yet... so this probably isn't an answer. — Amesey
– Amesey, Commented Nov 6, 2015 at 11:32
It’s possible to get the variable name and then you could split the string (or you could hardcode the ints by using a for as you tried to), but for all sanity, you really really should use arrays instead. There is no reason not to. — Smar
– Smar, Commented Nov 6, 2015 at 15:16

Niranjan N Raju · Accepted Answer · 2015-11-06 11:33:07Z

1

This should help you(remove echo).

for ($i = 1; $i < 25; $i++) {

        ${"day".$i."Txt"} =  call_user_func('Day_' . $i . '_Offer', 'EMAIL_OFFER');
    }

If you want to see what all variables are there in your function use get_defined_vars().

See deme here

answered Nov 6, 2015 at 11:33

Niranjan N Raju

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Comments

Harikrishnan · Accepted Answer · 2015-11-06 11:34:41Z

0

Try this

for ($i = 1; $i < 25; $i++) { 

        ${"day".$i."Txt"} = call_user_func('Day_'.$i.'_Offer', 'EMAIL_OFFER'); 

}

It's called Variable variables ie a variable name which can be set and used dynamically

answered Nov 6, 2015 at 11:22

Harikrishnan

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that echo is ruining the party

@NiranjanNRaju Its Ok. Time to delete comments