i have a variables like $srange0 , $srange1, $srange2 $srange3.
i am using to declare some value to each value using for loop.
for($i=0;$i<=3;$i++){
$srange.$i = $i;
}
but its not working ?
is there any alternative solution for this
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2That a language provides a feature is not a valid motive to use it.Ignacio Vazquez-Abrams– Ignacio Vazquez-Abrams2011-06-02 04:28:02 +00:00Commented Jun 2, 2011 at 4:28
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4 Answers
for($i=0;$i<=3;$i++){
$var = 'srange'.$i;
$$var = $i;
}
But, whenever I see variables like that, I'd use an array instead.
4 Comments
Matthew
I shouldn't even say this, but you can do
${'srange'.$i} = $i;.
galymzhan
@konforce but sometimes dynamic variable names are very helpful if you know how to use them properly
Matthew
@galymzhan, it is never helpful as a replacement for arrays. And in most cases, arrays suffice. Especially here.
galymzhan
@konforce sure, this code isn't perfect and suitable demonstration of dynamic variables
Use an array:
$srange = array();
for ($i = 0; $i <= 3; ++$i)
$srange[$i] = $i;
For the purpose of this particular task, you can also do this:
$srange = range(0, 3);
That also builds the same array as my first code snippet.
Comments
The properway to add these dynamic variables will be like this
for($i=0;$i<=3;$i++){
$name = 'srange'.$i;
$$name = $i;
}
Comments
This may be helpful to you:
$srange0;
$srange1;
$srange2;
for($i=0;$i<=3;$i++) {
$range = "srange".$i;
$$range = $i;
}
echo $srange2."<br />";
exit;
