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# 8  
Old 04-15-2004
Question simple script doesn't work

Smilie red hat linux 8.0
I tried this simple script no lucK
echo "read number:\c:"
read number
echo "read name:\c"
read name
NR == $num{"date"|getline d; printf "%s\t%s%s\n", "profile file", $0, d}NR!=$num{print}
.......
run_file!
awk -f won't work even when I change the script..
and awk 'NR == $num.............. wont work either and the error message says taht fild Zero is name of script file.. please help[!!!!Smilie

Last edited by moxxx68; 04-15-2004 at 06:00 AM..
# 9  
Old 04-15-2004
Pass variable to awk with -v option, e.g...

awk -v num=$number 'NR==num{print "blah"}' file1
# 10  
Old 04-15-2004
PHP didn't work

Smilie not quite the results that I am looking for!
# 11  
Old 04-15-2004
Quote:
Originally posted by Ygor
Pass variable to awk with -v option, e.g...

awk -v num=$number 'NR==num{print "blah"}' file1
Just do it when you want to use these external variables in BEGIN procedure too.

Last edited by home_king; 04-15-2004 at 02:46 PM..
# 12  
Old 04-15-2004
Re: didn't work

Quote:
Originally posted by moxxx68
Smilie not quite the results that I am looking for!
Remember, variable's reference can't be prefixed with $ unless they are position variables like $0~$NF.
Ygor is right. However, you didn't modify your variable reference, so you failed.
Just skip the $, use "num".

By the way, when you want to print patterned line without twice display, just improve your structure like me.

Code:
[root@home root]# cat myawk8
#!/bin/gawk
{
        if(NR==num){
                "date"|getline d;
                printf "%s\t%s%s\n","profile file",$0,d
                next
        }
        print
}
[root@home root]# cat mydata
hi, i am home_king from linuxsir.org.
2003-12-5 Bash it.
I am shell's administrator.
[root@home root]# num=2
[root@home root]# gawk -f myawk8 num=$num mydata
hi, i am home_king from linuxsir.org.
profile file    2003-12-5 Bash it.Îå  4ÔÂ 16 01:44:50 CST 2004
I am shell's administrator.

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