BATCH > manipuler les chaines de caractères

ALIGNER DU TEXTE

aligner à droite

l’astuce est d’ajouter un grand nombre d’espace, puis de n’afficher que les 8 derniers caractères :

 

set x=3000
set y=2
set x=        %x%
set y=        %y%
echo.X=%x:~-8%
echo.Y=%y:~-8%

 

résultat :

 

X=    3000
Y=       2

AFFICHER DES CARACTÈRES

afficher les premiers caractères

:~debut,longueur

 

set str=abcdefghi
set str=%str:~0,4%

 

résultat :

 

abcd

afficher les derniers caractères

 

Map and Lookup - Use Key-Value pair list to lookup and translate values

name of a month into it`s corresponding two digit number. The key-value pairs are listed in the map variable separated by semicolon. Key and value itself are separated by one dash character. Same can be used to tranlate a day-of-the-week short string into a day-of-the-week long string by changing the map content only.

 

REM ---- Ex 1: name of month into two digit number ----
SET v=Mai 
SET map=Jan-01;Feb-02;Mar-03;Apr-04;Mai-05;Jun-06;Jul-07;Aug-08;Sep-09;Oct-10;Nov-11;Dec-12
CALL SET v=%%map:*%v%-=%%
SET v=%v:;=&rem.%
ECHO.%v%

REM ---- Example 2: Translate abbreviation into full string ----
SET v=sun
set map=mon-Monday;tue-Tuesday;wed-Wednesday;thu-Thursday;fri-Friday;sat-Saturday;sun-Sunday
CALL SET v=%%map:*%v%-=%%
SET v=%v:;=&rem.%
ECHO.%v%

05
Sunday

Mid String - Extract a Substring by Position

:~  The example here shows how to extract the parts of a date.

 

echo.Date   : %date%
echo.Month  : %date:~4,2%
echo.Day    : %date:~7,2%
echo.Year   : %date:~10,4%

 

résultat :

 

Date   : 26/03/2018
Month  : 26
Day    : 03
Year   : 2018

supprimer des caractères par substitution

supprimer tous les foo :

 

set str=abcfoodeffoo
set str=%str:foo=%

 

résultat :

 

abcdef

SUPPRIMER DES CARACTERES

Enlever les Premier et le Dernier caractères

:~1,-1 enlève le premier et le dernier caractère.

 

set str=abcdef
set str=%str:~1,-1%

 

résultat :

 

bcde

Enlever tous les espaces

set str=      toto       &rem
set str=%str: =%

REMPLACER UNE SOUS-CHAINE

Remplacer tous les foo par bar :

 

set str=abcdefoo123456foo
set str=%str:foo=bar%

 

résultat :

 

abcdebar123456bar

EXTRAIRE DES CARACTÈRES

derniers caractères

récupère les 4 derniers caractères :

 

set str=%str:~-4%

Découper une chaine

echo.-- Jour de la semaine
FOR /f %%a IN ("%date%") DO set d=%%a
echo.Date   : %date%
echo.d      : %d%
echo.

echo.-- couper la date grace aux slash et espace comme  delimiters

for /f "tokens=1,2,3,4 delims=/ " %%a in ("%date%") do set wday=%%a&set month=%%b&set day=%%c&set year=%%d
echo.Weekday: %wday%
echo.Month  : %month%
echo.Day    : %day%
echo.Year   : %year%

résultat :

 

-- Jour de la semaine
Date   : Thu 12/02/2005
d      : Thu

-- couper la date grace aux slash et espace comme  delimiters
Weekday: Thu
Month  : 12
Day    : 02
Year   : 2005

Concaténer

set "str1=Hello"
set "str2=World" 

set "str3=%str1%%str2%"
set "str4=%str1% %str2%"

echo.%str3%
echo.%str4%

 

résultat :

 

HelloWorld
Hello World

Enlever les espaces du début d’une chaine

Utilser FOR pour enlever les espaces au début d’une variable: str.

 

set str=               15 espaces
echo."%str%"
FOR /f "tokens=* delims= " %%a IN ("%str%") DO set str=%%a
echo."%str%"

 

résultat :

 

"               15 espaces"
"15 espaces"

Enlever les guillemets

set str="cmd politic"
echo.%str%
for /f "useback tokens=*" %%a in ('%str%') do set str=%%~a
echo.%str%

 

résultat :

 

"cmd politic"
cmd politic

Enlever les espaces à la fin d’une chaine

Attention: le module Delayed Expansion doit être activé.

 

set str=15 espaces               &rem
echo."%str%"
FOR /l %%a IN (1,1,31) DO if "!str:~-1!"==" " set str=!str:~0,-1!
echo."%str%"

 

résultat :

 

"15 espaces               "
"15 espaces"

substitution

set str=15 espaces               &rem
set str=%str%##
set str=%str:  ##=##%
set str=%str: ##=##%
set str=%str:##=%
echo."%str%"

—

https://www.dostips.com/DtTipsStringOperations.php

 

https://ss64.com/nt/syntax-substring.html

 

https://stackoverflow.com/questions/13805187/how-to-set-a-variable-inside-a-loop-for-f

 

https://stackoverflow.com/questions/10456801/batch-file-equivalent-of-currentdir-pwd

 

—

 

 

Replacing a Substring With Another String in Bash

Replace substring natively in bash (good for a single line)

Bash has some built-in methods for string manipulation. If you want to replace part of a string with another, this is how you do it:

${main_string/search_term/replace_term}

Create a string variable consisting of the line: “I am writing a line today” without the quotes and then replace today with now:

line="I am writing a line today"
echo "${line/today/now}"
I am writing a line now

Did you understand what just happened? In the syntax "${line/today/now}", line is the name of the variable where I’ve just stored the entire sentence. Here, I’m instructing it to replace the first occurrence of the word today with now. So instead of displaying the contents of the original variable, it showed you the line with the changed word.

Hence, the line variable hasn’t actually changed. It is still the same:

echo $line
I am writing a line today

But you can definitely replace the word you want to and modify the same variable to make the changes permanent:

line="${line/today/now}"
echo $line
I am writing a line now

Now the changes have been made permanent and that’s how you can permanently replace the first occurrence of a substring in a string.

You can also use other variables to store specific substrings that you wish to replace:

replace="now"
replacewith="today"
line="${line/${replace}/${replacewith}}"
echo $line
I am writing a line today

Here, I stored the word to be replaced in a variable called replace and the word that it would be replaced with inside replacewith. After that, I used the same method as discussed above to “revise” the line. Now, the changes I had made in the beginning of this tutorial have been reverted.

Let us look at another example:

hbday="Happy Birthday! Many Many Happy Returns!"
hbday="${hbday/Many/So Many}"
echo $hbday
Happy Birthday! So Many Many Happy Returns!

Replacing all occurrences of a substring

You can also replace multiple occurrences of substrings inside strings. Let’s see it through another example:

hbday="${hbday//Many/So Many}"
echo $hbday
Happy Birthday! So Many So Many Happy Returns!

That extra / after hbday made it replace all occurrences of Many with So Many inside the sentence.

Replace string using sed command (can work on files as well)

Here’s another way for replacing substrings in a string in bash. Use the sed command in this manner:

Replace first occurrence:

line=$(sed "s/$replace/s//$replacewith/" <<< "$line")

If I take the first example, it can be played as:

You can also replace all occurrences by adding g at the end:

line=$(sed "s/$replace/$replacewith/g" <<< "$line")

Now, you may think this is a more complicated than the native bash string method. Perhaps, but sed is very powerful and you can use it to replace all the occurrences of a string in a file.

sed -i 's/$replace/$replacewith/' filename

 

Sed is a very powerful tool for editing text files in Linux. You should at least learn its basics.

 

line=$(sed "s/$replace/s//$replacewith/" <<< "$line")

and it should be:

line=$(sed "s/$replace/$replacewith/" <<< "$line")

just like in the screenshot below that.