How to appending variable to a variable name

Question

I have a similar problem with this topic Appending a value of a variable to a variable name?

but the answer not solve my problem, i have variable,$item->ans_cont1 ..... to $item->ans_cont34 tried

for($i=1;$i<35;$i++) {      
  $inputval = 'item->ans_cont'.$i;
  echo $$inputval;  //output is empty
  echo ${'item->ans_cont'.$i};  //output is empty
}

but not work,is it "->" arrow problem?how to fix it?

Thanks!

`$$inputval` consider as variable. which is undefined. in short you are echoing the 34 variables without assigning. — devpro, Sep 21 '16 at 15:44
for($i=1;$i<35;$i++) { $inputval = $item->ans_cont.$i;echo $inputval; echo $$inputval;} — Alive to die - Anant, Sep 21 '16 at 15:45
FYI... You will save yourself grief by learning to use arrays instead: `echo $item->ans_cont[$i];` — AbraCadaver, Sep 21 '16 at 15:55

score 2 · Accepted Answer · edited Sep 21 '16 at 15:53

2

$item is an Object so you have to call its attribute:

$inputval = $item->{'ans_cont'.$i};
// Or 
$inputval = 'ans_cont'.$i;
echo $item->$inputval;

edited Sep 21 '16 at 15:53

Don't Panic

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answered Sep 21 '16 at 15:47

Gectou4

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How to appending variable to a variable name

1 Answers1