Fix formatting and copy-paste fail in dialplan expression docs

Author: Xiretza

docs/Configuration/Dialplan/Expressions/Expressions-Examples.md

@@ -7,13 +7,13 @@ pageid: 4620379
 
 | Expression | Result | Note |
 | --- | --- | --- |
-| "One Thousand Five Hundred" =~ "(T[Expressions Examples^ ])" | Thousand |  |
-| "One Thousand Five Hundred" =~ "T[Expressions Examples^ ]" | 8 |  |
-| "One Thousand Five Hundred" : "T[Expressions Examples^ ]" | 0 |  |
+| "One Thousand Five Hundred" =~ "(T[^ ])" | Thousand |  |
+| "One Thousand Five Hundred" =~ "T[^ ]" | 8 |  |
+| "One Thousand Five Hundred" : "T[^ ]" | 0 |  |
 | "8015551212" : "(...)" | 801 |  |
 | "3075551212":"...(...)" | 555 |  |
-| ! "One Thousand Five Hundred" =~ "T[Expressions Examples^ ]" | 0 | Because it applies to the string, which is non-null, which it turns to "0", and then looks for the pattern in the "0", and doesn't find it |
-| !( "One Thousand Five Hundred" : "T[Expressions Examples^ ]+" ) | 1 | Because the string doesn't start with a word starting with T, so the match evals to 0, and the ! operator inverts it to 1 |
+| ! "One Thousand Five Hundred" =~ "T[^ ]" | 0 | Because it applies to the string, which is non-null, which it turns to "0", and then looks for the pattern in the "0", and doesn't find it |
+| !( "One Thousand Five Hundred" : "T[^ ]+" ) | 1 | Because the string doesn't start with a word starting with T, so the match evals to 0, and the ! operator inverts it to 1 |
 | 2 + 8 / 2 | 6 | Because of operator precedence; the division is done first, then the addition |
 | 2+8/2 | 6 | Spaces aren't necessary |
 | (2+8)/2 | 5 |  |

docs/Configuration/Dialplan/Expressions/Operators.md

@@ -20,12 +20,9 @@ Operators are listed below in order of increasing precedence. Operators with equ
 * ! expr1  
  Return the result of a logical complement of expr1. In other words, if expr1 is null, 0, an empty string, or the string "0", return a 1. Otherwise, return a 0. It has the same precedence as the unary minus operator, and is also right associative.
 * expr1 : expr2  
- The `:' operator matches expr1 against expr2, which must be a regular expression. The regular expression is anchored to the beginning of the string with an implicit `'.
-
-If the match succeeds and the pattern contains at least one regular expression subexpression `', the string corresponing to `\1' is returned; otherwise the matching operator returns the number of characters matched. If the match fails and the pattern contains a regular expression subexpression the null string is returned; otherwise 0.
-
-Normally, the double quotes wrapping a string are left as part of the string. This is disastrous to the : operator. Therefore, before the regex match is made, beginning and ending double quote characters are stripped from both the pattern and the string.
-
+ The `:` operator matches expr1 against expr2, which must be a regular expression. The regular expression is anchored to the beginning of the string with an implicit `^`.
+ If the match succeeds and the pattern contains at least one regular expression match group `()`, the string corresponing to `\1` is returned; otherwise the matching operator returns the number of characters matched. If the match fails and the pattern contains a regular expression match group the null string is returned; otherwise 0.
+ Normally, the double quotes wrapping a string are left as part of the string. This is disastrous to the : operator. Therefore, before the regex match is made, beginning and ending double quote characters are stripped from both the pattern and the string.
 * expr1 =~ expr2  
  Exactly the same as the ':' operator, except that the match is not anchored to the beginning of the string. Pardon any similarity to seemingly similar operators in other programming languages! The ":" and "=~" operators share the same precedence.
 * expr1 ? expr2 :: expr3