Suppose we want to write a text file containing

user001
user002
user003
....
user999

Is there any smart way to build it instead of writing a program? Of course it would be easy by writing a C program or an awk script.

I am trying to find some automatic way in vi/vim or in a shell to add lines containing a constant text pattern and the updated value of a counter.

CodePudding user response：

There are a couple of interesting ways to do this in a short amount of characters

For bash:

echo user{000..999} | tr ' ' '\n' > file

or

seq 000 999 | sed 's/^/user/' > file

For vim:

You could just invoke one of the bash solutions using :r!

:r!seq 000 999 | sed 's/^/user/'

You could also just run seq from vim, and prepend "user" to all the lines

:r!seq 000 999
:%s/^/user

or do it in a completely vim-native way:

• generate a 1000 lines containing just "user000": i user000 CRESC999.
• select all text: ggctrl-vG$
• tell vim to increment them one by one: gctrl-a

CodePudding user response：

for counter in {001..999}; do echo "user${counter}" >> outputfile; done

For vim, there's a good answer here: https://vi.stackexchange.com/a/5600

CodePudding user response：

A bash one-liner, without using an explicit loop nor an external utility:

printf 'user%s\n' {001..999} >> outputfile

This uses printf's implicit loop and requires bash version 4.0 or newer (for leading zeros in brace expansions).

CodePudding user response：

You can use loops in shell. If you just paste this snippet into your terminal and press enter, it will create the file at your current location:

for i in {1..999}; do echo $(printf "userd" $i) >> test.txt; done

the loop iterates the variable i. Note that it's not enough to just concatenate "user" with i because we need leading 0s. Thus we need to format the number with d, meaning it is a decimal that contains at least 3 digits. A number smaller than 100 will be prefixed with 0s. With source >> target you can take the result from the echo command and write it into the file.