String Length Issue

Question

I have four string length statements I am trying to validate,but only two are working.

$name = "mike";
$age = "54";
$city = "Chicago";
$building = "Madadnock";
$lengthOfString1 = strlen ( $name );
echo "String length of 'name' " . $lengthOfString1  . "</br>";

$lengthOfString2 = strlen ( $age );
echo "String length of 'age' " . $lengthOfString2  . "</br>";
$lengthOfString3 = strlen ( $city );
echo "String length of 'city' " . $lengthOfstring3  . "</br>";

$lengthOfString4 = strlen ( $building );
echo "String length of 'building' " . $lengthOfstring4  . "</br>";

The top two are the only ones that work. Any help would be great!

add more details about how those last two statements not working — RomanPerekhrest, Apr 16 '17 at 20:49
This question should have been closed as a duplicate with [PHP: “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset”](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-notice-undefined-index-and-notice-undef) — Funk Forty Niner, Apr 16 '17 at 21:10

score 3 · Answer 1 · answered Apr 16 '17 at 20:51

3

variables in PHP are case sensitive

$lengthOfString4 vs $lengthOfstring4
         ^                   ^

answered Apr 16 '17 at 20:51

Peter

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same goes for `$lengthOfstring3` – Funk Forty Niner Apr 16 '17 at 21:11
@Fred-ii- great observation fred :) – Peter Apr 16 '17 at 21:19
I worked in a typesetting/graphics-related environment for a long time ;-) One picks up on those after a while. – Funk Forty Niner Apr 16 '17 at 21:21

score 0 · Answer 2 · answered Apr 16 '17 at 20:56

The problem lies in variable names. The variable names in PHP are case-sensitive.

$lengthOfString3 = strlen ( $city );
echo "String length of 'city' " . $lengthOfString3  . "</br>";

$lengthOfString4 = strlen ( $building );
echo "String length of 'building' " . $lengthOfString4  . "</br>";

Fixed it.

String Length Issue

2 Answers2