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up vote 4 down vote favorite
1

I do have variables like

<?php
$num1 = 'txt1';
$num2 = 'txt2';
$num3 = 'txt3';   
...

and I do have a loop 

for ($i=1; $i<100; $i++){  
   echo 'This is textNr'.$i.':'   .$num.$i  ;
}

I need to produce a result like:

This is textNr1: txt1

Of course with this code i would get something like:

This is textNr1: 'undefined'

,because in $num.$i the $num is not defined;

so $num.$i should be parsed once and become $num1 and then $num1 should be parsed a second time like:

 echo 'This is textNr'.$i.':'   .$num1  ;

Does anybody know how to handle this, or could give me a link about this problem and maybe the technical term of this problem (solved :) variable variables)?

Please feel free to edit my question, into something which will fit this case better!

share|improve this question
Why not just use an array and loop that? – elclanrs 14 hours ago
I don't get what you're trying to do inside the loop. – Lance 14 hours ago
@Lance he thought using $num.$i can get $num1 if $i=1 – pinkpanther 14 hours ago
He's trying to loop the variables with $i like $num.$i == $num1 etc. But this is the wrong approach... – elclanrs 14 hours ago
Ohhh. I see. Well, for starters define $num inside the loop. – Lance 14 hours ago
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6 Answers

up vote 1 down vote accepted

Using an array (Antoine's answer) is better practice. But if you want/need to use your solution

for ($i=1; $i<100; $i++){  
   echo 'This is textNr'.$i.':' . ${"num$i"}  ;
}

PHP does first evaluate "num$i" and use that result to find a variable with that name,
thanks to ${ }.

Note that you can use a deeper level of variable names

$a = "b";
$b = "c";
${${$a}}  = 7; 
echo $c; // echoes 7

Via an array: you could also declare an array

$array = ('text1', 'text2', 'text3', ... , 'textN');
echo count ( $array ); // echoes N, size of array

To display its items

// Loop on array
foreach ($array as $text) {
  echo "$text\n";
}

displays text1, text2, ... textN (one per line)

You could also use a for loop

$n = count($array);
for($i=0 ; $i<$n ; $i++) echo $array[$i] . "\n";

that produces the same output.

share|improve this answer
thx for the detailed answer – AddingColor 27 mins ago
up vote 1 down vote

Try this: 

<?php
$nums = array();
  $nums[0] = 'txt1';
  $nums[1] = 'txt2';

  for ($i=0; $i<count($nums); $i++){  
    echo 'This is textNr'.$i.':'   .$nums[$i]  ;
  }
?>

Is an improved version of the answer you have choosed. It will only show the values you have set to the array.

share|improve this answer
up vote 0 down vote

You can use an array , like this:

$nums = array('txt1','txt2','txt3');

for ($i=0; $i<sizeof($nums); $i++){  
    echo 'This is textNr'.$i.':'   .$nums[$i]  ;
}

The upper bound of array should be the size of the array, otherwise you may get empty values at $nums[x]

share|improve this answer
thx for the sizeof($nums) this comes quite handy , know this from javascript as array.length – AddingColor 13 hours ago
Sure, no problem :) – Jacob Amerz 13 hours ago
up vote 1 down vote

The php term you look for is "variable variable". php its one of very few languages that support it, and for a good reason - the approach smells wrong.

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up vote 2 down vote

php variable can be constructed with curly brackets like example below:

echo 'This is textNr'.$i.':'   .${'num'.$i}  . "\r\n" ;

you can put anything that returnrs string between brackets

share|improve this answer
up vote 4 down vote

You should do something like:

  $nums = array();
  $nums[0] = 'txt1';
  $nums[1] = 'txt2';

  for ($i=0; $i<100; $i++){  
    echo 'This is textNr'.$i.':'   .$nums[$i]  ;
  }

Or start the index at 1 in the array, if you want to start it at one in the loop,

share|improve this answer
Array seems to be the magic here. Thx, this should work. I'm quite new to PHP, but I guess arrays are often quite handy. – AddingColor 14 hours ago

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