array check ahead loop

[calmchess]

I need to run a for loop on an array and read ahead 1 index  so that if that spot is null then i can break out of the for loop something like this but it doesn't work.....

$whopro = array("slot0","slot1","slot2");

for($l=17; $l < 34; $l++){

if($whopro[$i+1]==null){
break;
}
echo $whopro[i]; 
}

now obviously the loop loops mor times than there are array indexes.The whole point of this is to stop getting php notice errors without changing the for loop integers.

 

 

[Mark Baker]

Why can't you change the array integers.... especially as they're the whole cause of your problem

[calmchess]

I can't change the array integers because they are part of a dynamic array that changes length

[calmchess]

Bear with me and assume that there are above 17 items in the array

[laffin]

what is $i?

why is $l way out of the array bounds

 

$whopro has three elements, starting with 0, 1, 2

so trying to reference $l=17 gives ya an notice, about a variable not being declared.

 

php dusn assign non existant variables to null.

if the variable dusn exist why not just use isset?

 

 

[calmchess]

I know my code is bad above just fix it up and get on with the "look ahead" and break part

[laffin]

ya can perform a check within the for loop

for($i=0;$i<=20;$i++) $slot[$i]="Slot{$i}";

for($l=17;$i<34 && issset($slot[$i]);$i++)
  echo $slot[$l];

 

but if u really need it within the loop, the lookahead (note that by program flow, the last valid entry wont display, as the lookahead breaks out of the loop)

 

[/code]

for($l=17;$i<34;$i++)

{

  if(i!sset($slot[$i+1])) break;

  echo $slot[$l];

}

[/code]

 

[calmchess]

Thankyou so much that worked perfectly and pardon my dirty post.

[laffin]

No Probs here

 

There are a lot of php functions, some that even surprise me

 

Good luck on yer project