Adding in php question

[TheServant]

I have the following code which is part of a for loop. There are 2 arrays previously defined ($var, $bar):
[php]echo ( "<tr><td><input name=\"".$var[$i]."_guild_upgrade\" type=\"submit\" value=\"Level ".++$bar[$i]."\" /></td></tr>" ) ;[/php]
That works 100%. However that is not what I tired first. I was tring to do this:
[php]echo ( "<tr><td><input name=\"".$var[$i]."_guild_upgrade\" type=\"submit\" value=\"Level ".$bar[$i]+1."\" /></td></tr>" ) ;[/php]
This was coming up with a parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING?
Does anyone know why I can't use +1 instead of ++?

[hsriat]

Enclose the adding thing in brackets.

($bar[$i]+1)

[TheServant]

That's it! Thanks for that. I will continue using the ++ though!

[hsriat]

That's it! Thanks for that. I will continue using the ++ though!

Yeh, that's better to use ++ instead of doing increment separately.

[Atli]

It should be noted that by using ++ instead of +1 you are in fact changing the value of your array element.

Consider this:

$value = 1;
echo "Value + 1 = ". ++$value ."<br />";
echo $value; # Prints 2
 

$value = 1;
echo "Value + 1 = ". ($value + 1) ."<br />";
echo $value; # Prints 1
 

[Markus]

It should be noted that by using ++ instead of +1 you are in fact changing the value of your array element.

Consider this:

$value = 1;
echo "Value + 1 = ". ++$value ."<br />";
echo $value; # Prints 2
 

$value = 1;
echo "Value + 1 = ". ($value + 1) ."<br />";
echo $value; # Prints 1
 

So $value will remain at 2.

Suarve.

[Atli]

So $value will remain at 2.

Suarve.

Yes, but if you are only trying to print the value + 1, incrementing it like this may cause problems.

For example, if your trying to print all elements of an array:

$someArray = array(1, 2, 4, 8, 16, 32, 64);
for($i = 0; $i < count($someArray); $i++) {
  echo "Element ". ++$i ." = ". $someArray[$i] ."<br />";
}
 

This will skip every other element in the array and cause an undefined index error on the last loop.

You would either have to skip the incrementation in the for loop or just do ($i + 1).

[TheServant]

Hmm, that's good to know, it works in my case but I will have to keep that in mind!