Print first field in awk


 
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# 1  
Old 05-14-2014
Print first field in awk

Hi,

I have below text file

Code:
01.02.2014,asdas,arse,aere,4tfsd
12.03.2014,sdte,45gf,8iuj,qw343w
01.02.0214,aetre,sdfgter,asfrwe

I have writen below code to print only first field that is only date field from text file

Code:
#!/bin/ksh
echo "enter week"
read week
while  read -r line
do
ffld=${line}|/usr/xpg4/bin/awk 'BEGIN { FS = "," } ;{print $1}'
echo $ffld
done < inputfile >> file

Want to print like below

Code:
output file
 
01.02.2014
12.03.2014
01.02.0214

but the ffld taking the complete row
Code:
12.03.2014,sdte,45gf,8iuj,qw343w

and o/p is not printing, the file is empty.
# 2  
Old 05-14-2014
Hello,

Could you please use the following code.

Code:
awk -F, '{print $1}' print_first_files_check12112

Output will be as follows.

Code:
01.02.2014
12.03.2014
01.02.0214


Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
# 3  
Old 05-14-2014
I tried below code

Code:
#!/bin/ksh
echo "enter week"
read week
while  read -r line
do
sdate=${line}|/usr/xpg4/bin/awk -F, '{print $1}'
echo $sdate
done< inputfile >> file

still its printing empty file
# 4  
Old 05-14-2014
Hello,

Could you please let us know why you are getting week as input here ?


Thanks,
R. Singh
# 5  
Old 05-14-2014
try
Code:
#!/bin/ksh
echo "enter week"
read week
while  read -r line
do
sdate=`echo ${line}| nawk -F, '{print $1}'`
echo $sdate
done< inputfile >> file

Also no need to write while loop
Code:
nawk -F, '{print $1}' inputfile > file

# 6  
Old 05-14-2014
Use parameter substitution instead:
Code:
while read line
do
        echo "${line%%,*}"
done < inputfile > outputfile

# 7  
Old 05-14-2014
I'm a bit confused. From your input file, do you just want to grab the first column into another file?
Code:
cut -f1 -d, infile > outfile

...or do you want to process them in a loop:-
Code:
for val in `cut -f1 -d, infile`
do
   # whatever you like
done > outfile

.... and what do you do with the value read in as week anyway?




Regards,
Robin
 
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