How to handle the delimiter unit seperator in Unix
Answers were Sorted based on User's Feedback
Answer / srikanth
create script test.txt with | seperated
The below commands can be used to get the fields values:
awk -F"|" '{print $1}' test.txt
cat sri.txt | cut -d"|" -f1
Is This Answer Correct ? | 4 Yes | 2 No |
Answer / pitambar mishra
Different types of command using delimiter :
1. cut -d "|" -f6 emp.lst
d : delimiter or field separator
2. sort -t "|" -nk6 emp.lst
t : delimiter or field separator
3. awk -F "|" '{print $6}' emp.lst
F : delimiter or field separator
Is This Answer Correct ? | 2 Yes | 0 No |
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Shifting positional parameter in Linux fedora core ? Hi I have written following shell script for display value of positional. But in 11th and 12 field will display without shifting command. May I know is it advance of Linux of programming code error? #!/usr/bin/bash echo "Bellow is the out of ps command" echo "`ps`" echo "The passing Parameter i.e output of \$1,2... value is:==> $11 " echo "Total number of passed argument \$# is:==> $#" echo "Passed argument names (\$*) are:==>$*" echo "This script PID(\$$) is :=>$$" echo "The name of executing script(\$0) is :==>$0" echo "The Parent ID of this script(\$PPID) is:==>$PPID" And my input to this script is ./scriptname arg1 arg2 arg3 arg4 arg5 arg6 arg7 arg8 arg9 arg10 arg11 arg12 Output Is Bellow is the out of ps command PID TTY TIME CMD 2892 pts/0 00:00:00 bash 3172 pts/0 00:00:00 positional_para 3173 pts/0 00:00:00 ps The passing Parameter i.e output of $1,2... value is:==> arg11 Total number of passed argument $# is:==> 13 Passed argument names ($*) are:==>arg1 arg2 arg3 arg4 arg5 arg6 arg7 arg8 arg9 arg10 arg11 arg12 arg13 This script PID($$) is :=>3172 The name of executing script($0) is :==>./positional_parameter The Parent ID of this script($PPID) is:==>2892 After the ps output please see the next line I.e $11 value comes out without shifting the parameter. How is it give me my friends
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