converting variable to date format in unix script

Question

if [ -f $FILEPATH3 ]; then
#Will print the Header columns from properties file.
print $header >$CFILEPATH3
#To add rows in output file from input file .
awk -F\" 'BEGIN{OFS=FS;} {for(i=1;i<=NF;i=i+2){gsub(/,/,"~\t",$i);}}1' $FILEPATH3 > $TFILEPATH3

#Removes the footer,header and prints the columns as per mapping by picking column numbers from properties file 
cat $TFILEPATH3| sed '1d' | awk 'BEGIN { FS = "~\t"; OFS = ",";}{ DATE = date -d "$'$ms2'" "+%Y%m%d" } { printf "%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s\n",  $'$ms1', $DATE,$'$ms3',$'$ms4',$'$ms5',$'$ms6',$'$ms7',$'$ms8’,”MS”,$’$ms10',$'$ms11',$'$ms12',$'$ms13',$'$ms14',$'$ms15',$'$ms16',$'$ms17'}' >> $CFILEPATH3

In the above code, I'm trying to copy the data from the input file to output file. ms1, ms2 are the column positions of a CSV input file.

ms2 is the date with format mm/dd/yyyy and which is considered as variable. We need to convert the variable into YYYYMMDD format and write it into output file

In the script I'm trying to change the date format to YYYYMMDD.. but I'm getting an error.

I think the error is from this code

{ DATE = date -d "$'$ms2'" "+%Y%m%d" }

in the properties file i have defined ms2=1 which means it has to pick the column 1 value from the input csv file. In input csv file, the column 1 will have date in the format of mm/dd/YYYY (example.. 6/03/2015).. i want to be formated to YYYYMMDD (20150603) while writing into output file.. — user2954417
– user2954417, Commented Jun 3, 2015 at 9:52

Sriharsha Kalluru · Accepted Answer · 2015-06-03 10:58:09Z

0

You are trying to access the variable in single quotes and which wont happen, So try to use the following syntax.

 DATE=$(echo $ms2|awk -F '/' '{print $3$2$1}')
cat $TFILEPATH3| sed '1d' | awk  -v ms2="$DATE" 'BEGIN { FS = "~\t"; OFS = ",";} { printf "%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s\n",  $'$ms1', $ms2,$'$ms3',$'$ms4',$'$ms5',$'$ms6',$'$ms7',$'$ms8’,”MS”,$’$ms10',$'$ms11',$'$ms12',$'$ms13',$'$ms14',$'$ms15',$'$ms16',$'$ms17'}' >> $CFILEPATH3

edited Jun 3, 2015 at 10:58

answered Jun 3, 2015 at 9:50

Sriharsha Kalluru

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5 Comments

user2954417 Over a year ago

it did not work.. the output was 0$$ms2+%Y%m%d .. but i want the date

Sriharsha Kalluru Over a year ago

Can you please provide an example format and syntax of command that you are using?

user2954417 Over a year ago

ms2 is the date with format mm/dd/yyyy and which is considered as variable. We need to convert the variable into YYYYMMDD format and write it into output file

Sriharsha Kalluru Over a year ago

In that case you can use awk command to convert the variable. echo 'mm/dd/yyyy'|awk -F '/' '{print $3$1$2}' I modified the answer , You can refer that.

user2954417 Over a year ago

i tried.. it did not change the format.. it still prints the date as mm/dd/YYYY

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