shell script variable for loop
Method 2: Bash For Loop using C like syntax Before the first iteration, expr1 is evaluated. This is usually used to initialize variables for the loop. All the statements between do and done are executed repeatedly until the value of expr2 is TRUE. After ,It looks like you're trying to dynamically create variable names using $i . Firstly, I would recommend against this if possible - it would be better to use an array if ... ,Either use another shell like zsh or ksh93 or use an alternative syntax: ... Using loops in a shell script are often an indication that you're not doing it right. , The meaning of for i in "$test1" "$test2" "$test3" is: $i will hold the content of the three variables test1, test2, test3 - which will actually holds: 1, 2, ..., or you can use the c-style for loop, which is also safe: for (( i = 0 ... though that may give you a large command line if length is massive. Another ..., How can I iterate bash for loop using a variable range of numbers in Unix or Linux or BSD or Apple OS X operating systems? You can use the ...,The reason for this is the order in which things occur in bash. Brace expansion occurs before variables are expanded. In order to accomplish your goal, you ... , Use an array. #! /bin/bash servers=( 192.xxx.xxx.2 192.xxx.xxx.3 192.xxx.xxx.4 192.xxx.xxx.5 192.xxx.xxx.6 192.xxx.xxx.7 ) for server in ...,hi, my first question is :- i would like to know how do i loop through the output of a variable. for ex:- if i have a variable called x and echo --$x gives the output like ... , Your mistake is $x=$var . It should be x=$var . You can think of it as $x is the value contained the in variable x so you don't want to assign to ...
| 相關軟體 PuTTY 資訊 | |
|---|---|
 shell script variable for loop 相關參考資料
12 Bash For Loop Examples for Your Linux Shell Scripting
Method 2: Bash For Loop using C like syntax Before the first iteration, expr1 is evaluated. This is usually used to initialize variables for the loop. All the statements between do and done are execu... https://www.thegeekstuff.com Assigning value to a variable in for loop in shell scripting ...
It looks like you're trying to dynamically create variable names using $i . Firstly, I would recommend against this if possible - it would be better to use an array if ... https://stackoverflow.com bash variables in for loop range - Unix & Linux Stack Exchange
Either use another shell like zsh or ksh93 or use an alternative syntax: ... Using loops in a shell script are often an indication that you're not doing it right. https://unix.stackexchange.com echo variable name in for loop - Unix & Linux Stack Exchange
The meaning of for i in "$test1" "$test2" "$test3" is: $i will hold the content of the three variables test1, test2, test3 - which will actually holds: 1, 2, ... https://unix.stackexchange.com How to use variables in a bash for loop - Stack Overflow
or you can use the c-style for loop, which is also safe: for (( i = 0 ... though that may give you a large command line if length is massive. Another ... https://stackoverflow.com HowTo: Iterate Bash For Loop Variable Range Under Unix ...
How can I iterate bash for loop using a variable range of numbers in Unix or Linux or BSD or Apple OS X operating systems? You can use the ... https://www.cyberciti.biz In bash, is it possible to use an integer variable in the loop ...
The reason for this is the order in which things occur in bash. Brace expansion occurs before variables are expanded. In order to accomplish your goal, you ... https://unix.stackexchange.com Loop Through Variables - Unix & Linux Stack Exchange
Use an array. #! /bin/bash servers=( 192.xxx.xxx.2 192.xxx.xxx.3 192.xxx.xxx.4 192.xxx.xxx.5 192.xxx.xxx.6 192.xxx.xxx.7 ) for server in ... https://unix.stackexchange.com looping through a variable in a shell script - UNIX and Linux Forums
hi, my first question is :- i would like to know how do i loop through the output of a variable. for ex:- if i have a variable called x and echo --$x gives the output like ... https://www.unix.com Shell script variable & loop - Stack Overflow
Your mistake is $x=$var . It should be x=$var . You can think of it as $x is the value contained the in variable x so you don't want to assign to ... https://stackoverflow.com |