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Posted by Rodney Green on 10/13/05 15:20
Thanks Connor. I changed the line to the following. Is it correct now?
The array $output[] is now being populated. However, when I go to
print the values of the $output array using a foreach loop, it prints
out the word "Array" for each array item.
$output[] = array_slice ($textArray, $int_range[start][$i],
$int_range[end][$i]);
On 10/12/05, Ethilien <spam@ethilien.info> wrote:
> Are you trying to put each return from array_slice into $output as an
> array? You need to replace $output = array_slice... with $output[] =
> array_slice...
>
> Also, you cannot just echo an array (even of characters). It must be a
> string, or else echo will only produce the output 'Array'.
>
> Hope this helps,
> -Connor McKay
>
> Rodney Green wrote:
> > Hello,
> >
> > I'm using a for loop to iterate through an array and slice parts of
> > the array and add into another array.
> >
> > When array_slice is called the variable $i is not being substituted
> > with the current value of
> > $i and the $output array is empty after running the script. Can
> > someone look at the code below and give me a clue as to why this
> > substitution is not happening? If I uncomment the "echo $i;" statement
> > the value of $i is printed just fine.
> >
> > Thanks,
> > Rod
> >
> > Here's my code:
> >
> >
> > $number = count ($int_range[start]);
> > //echo $number;
> > for ($i = 0; $i <= $number; $i++) {
> >
> > //echo $i;
> >
> > $output = array_slice ($textArray, $int_range[start][$i],
> > $int_range[end][$i]);
> >
> > }
> >
> >
> >
> >
> > foreach ($output as $value) {
> >
> > echo $value;
> > }
>
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