Bash Script – for i in {x..y} [Duplicate]

bashprogrammingscripts

i am writing a script in bash and I would like to give the two numbers in the brackets, x and y to be precise, as variables.

I wrote the script this way:

echo "x?"
read x
echo "y"
read y

for i in {$x..$y}

but it does not work.

How could I set it up to make it function?

Thanks a lot in advance.

Best Answer

Use this simple bash solution:

for ((i=$x;i<=$y;i++))

Before using the following functions, read below !

For a sequence of numbers, you might use seq:

for i in $(seq $x $y)

If you really need brace expansion, you can use eval:

for i in $(eval echo {$x..$y})

Disclaimer:
eval is evil. This case is a perfect example: If the user writes $(do_something_evil) instead of a number, do_something_evil will be executed. So before using eval, make sure your input is not evil. The same applies to seq $x $y.

If you still want to use one of these functions, test the user input, which is generally a good idea: How do I test if a variable is a number in Bash?

Related Solutions

Ubuntu – making a bash script executable programatically

There are two things you need to do:

Reference the script interpreter at the beginning of the script:
```
#!/bin/bash
```
Set the permissions to make it executable:
```
chmod +x myscript.sh
```

Ubuntu – How to set the output of a function as a variable value in a bash script

Bash functions are not like functions on other programming languages, they are more like commands. This means they have no classical return value, but

an exit/return code. This is an integer number in the range 0-255, where 0 means "success" and every other value represents an error. If you try to specify a number outside this range, it will be taken modulo 256 (add or subtract 255 from your number until it fits the range 0-255).

This code is automatically set to the return code of the last statement that got executed inside the function, unless you manually set it using the return command, like this:
```
return 1
```
output streams. Each Bash function can write arbitrary strings to the output streams (STDOUT and STDERR), just like normal scripts. The output can either be directly from the commands you run in the function, or set manually by using echo.

But instead of letting this output get displayed in the console, you can capture it when you run the function, e.g. using Bash's command substitution syntax, and store it in a variable in your main script:
```
example_function() {
    # do something useful
    echo "return this message"
}

returned_value="$(example_function)"
```

So your code would have to look like this:

file_count() {
    find "$1" -maxdepth 1 -type f -printf '.' | wc -c
    # the line above already prints the desired value, so we don't need an echo
}

files_in_somedir="$(file_count "$somedir")"

See Return value in bash script (on Stack Overflow) for more info.

Best Answer

Related Solutions

Ubuntu – making a bash script executable programatically

Ubuntu – How to set the output of a function as a variable value in a bash script

Related Question