Array shows its pointer instead of its value, Part 2

Question

Well, I know there was already a similar question before, but I think, this one is a bit different and/or specific. The following Code always returns the correct value-positions within my "$targetArray" (1,2,3,4,5,...), but not its value (*,::,:::,-,...).

function BlaBla($text, $chars = 40) {
$targetArray = array(" * ", " :: ", " ::: ", " - ", " # ", " ", " ", " ", " / ", " \\ ", " ++ ", " ~ ");
$rand = array_rand($targetArray);
 $text = $text." ";
 $text = substr($text,0,$chars);
 $text = substr($text,0,strrpos($text,' '));
 $text = $text." ".$rand;
 return $text;
 }

So: Where is the hook? Any hints for this?

When picking only one entry, array_rand() returns the __key__ for a random entry. — Mark Baker
– Mark Baker, Commented Aug 4, 2016 at 11:46
Well, I didn´t catch your sentence completely, because I´ve already used "array_rand()" in the same way before (I thought so) - anyway @apokryfos answer was perfect for me. Thanks to both of you! — NinjaPopeye
– NinjaPopeye, Commented Aug 4, 2016 at 12:05

apokryfos · Accepted Answer · 2016-08-04 11:47:31Z

1

From the array_rand manual page

Picks one or more random entries out of an array, and returns the key (or keys) of the random entries.

In your case you need:

$randInd = array_rand($targetArray);
$rand = $targetArray[$randInd];

answered Aug 4, 2016 at 11:47

apokryfos

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1 Comment

NinjaPopeye Over a year ago

Wow - it works, great! Two lightning fast answers within three minutes - perhaps this question was too easy... ;-D .....Thanks a lot!

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