Menu
I am trying to pass the argument as max limit for the for loop like this:
#!/bin/bash for i in {1..$1} do echo $i done This however returns {1..2} when called with argument 2, instead of executing the script and giving me
1 2 
746
bash [filename] runs the commands saved in a file. $@ refers to all of a shell script's command-line arguments. $1 , $2 , etc., refer to the first command-line argument, the second command-line argument, etc. Place variables in quotes if the values might have spaces in them.
Arguments can be passed to the script when it is executed, by writing them as a space-delimited list following the script file name. Inside the script, the $1 variable references the first argument in the command line, $2 the second argument and so forth. The variable $0 references to the current script.
Save this answer. Show activity on this post. && and || are boolean operators. && is the logical AND operator, and bash uses short cut evaluation, so the second command is only executed if there is the chance that the whole expression can be true.
Just like && , || is a bash control operator: && means execute the statement which follows only if the preceding statement executed successfully (returned exit code zero). || means execute the statement which follows only if the preceding statement failed (returned a non-zero exit code).
Variable substitutions are not done inside of curly braces. You can use fixed numbers but not variables.
Brace Expansion
A sequence expression takes the form {x..y}, where x and y are either integers or single characters. ...
Brace expansion is performed before any other expansions, and any characters special to other expansions are preserved in the result. It is strictly textual. Bash does not apply any syntactic interpretation to the context of the expansion or the text between the braces.
A correctly-formed brace expansion must contain unquoted opening and closing braces, and at least one unquoted comma or a valid sequence expression. Any incorrectly formed brace expansion is left unchanged.
Try one of these alternatives:
for ((i = 1; i <= $1; i++)); do echo $i done # Not recommended with large sequences. for i in $(seq 1 $1); do echo $i done This will cycle through all true arguments (a.k.a. "testo mesto" is one argument)
#cycle through all args for (( i=1; i<=$1; i++ )); do eval arg=\$$i echo "$arg" done OR
#cycle through all args for (( i=1; i<=$1; i++ )); do echo "${!i}" done If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
