CODEWITHSUNDEEP
tcl
velpandian
velpandian

Reputation: 471

How to delete a character in a string using TCL

I have the following string

"TCL is known as "tool command language", TCL is known as "tool command language", TCL is known as "tool command language""

from the above input I want a output like below

"TCL is known as tool command language, TCL is known as tool command language, TCL is known as tool command language"

i.e. only first and last double quotes should be displayed on output, and all other should be deleted, Could someone let me know the different methods to accomplish this

Upvotes: 0

Views: 21563

Answers (3)

Dinesh
Dinesh

Reputation: 16436

There can be many ways. I have tried with regsub

set str {"TCL is known as "tool command language", TCL is known as "tool command language", TCL is known as "tool command language""}
puts "Input : $str"
regsub -all {(.)"} $str {\1} output
puts "Output : $output"

which will produce the following 

Input : "TCL is known as "tool command language", TCL is known as "tool command language", TCL is known as "tool command language""
Output : "TCL is known as tool command language, TCL is known as tool command language, TCL is known as tool command language"

The pattern I have used is (.)". In regular expressions, the atom . will match any single character. (Will talk about the parenthesis usage at the bottom). Then a single quote. So, basically, this will match any single char and having a single quote next to it as shown below.

enter image description here

As you can see, we have a total of 6 matches. Let us take the 2nd match which is e". Our main intention is to remove the quotes. But, we have matched 2 characters. This is the reason why we have grouped it with parenthesis.

With Tcl, we can access 1st subgroup with the help of \1 and 2nd subgroup with \2 and so on. Finally, we are substituting the 2 characters with one character which is nothing but the first letter other than quote. i.e. e" is substituted with character e.

Notice the use of -all flag at the beginning which is responsible for matching all the occurrence of this pattern.

Note : \1 should be used with braces like {\1} as I have mentioned. In case, if you want to access it without braces, you have to use \\1

Reference : Non-capturing subpatterns

Upvotes: 4

Donal Fellows
Donal Fellows

Reputation: 137787

You seek to remove all " characters from the string that have at least one character on each side of them. That leads to this regular expression substitution:

set transformedString [regsub -all {(.)[""]+(.)} $inputString {\1\2}]

The " is doubled up and in [brackets] just to make the highlighting here work. You could use {(.)"+(.)} instead.

Upvotes: 1

glenn jackman
glenn jackman

Reputation: 247210

You could remove all quotes and re-add the outer ones. One way:

set new [format {"%s"} [string map {{"} {}} $str]]

Upvotes: 2

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