extract header if pattern in a column matches - linux

Question

I am trying to extract and print header of a file if the pattern in that particular column matches.
Here is a example :
[user ~]$ cal |sed 's/July 2014//'
Su Mo Tu We Th Fr Sa
1 2 3 4 5
6 7 8 9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31
Expected output :
if input date =31 then print the day on 31st.
Just to be clear, I cannot use date -d flag as its not supported by my OS.Probably would need awk here to crack the question.
[user ~]$ date -d 20140731 +%A
Thursday
I hope I am able to convey my question and concern clearly.

Answer 1

Using awk:
cal | awk -v date=31 'NR == 2 { split($0, header) } NR > 2 { for (i = 1; i <= NF; ++i) if ($i == date) { print header[NR == 3 ? i + 7 - NF : i]; exit } }'
Output:
Th

Answer 2

Here is a gnu awk solution:
cal | awk -v date=31 -v FIELDWIDTHS="3 3 3 3 3 3 3 3" 'NR==2 {split($0,a)} {for (i=1;i<=NF;i++) if ($i==date) print a[i]}'
Th
You set the date that you like to be displayed as a variable, so it can be change to what you like.
Or it could be written like this:
cal | awk 'NR==2 {split($0,a)} {for (i=1;i<=NF;i++) if ($i==date) print a[i]}' FIELDWIDTHS="3 3 3 3 3 3 3 3" date=31
PS FIELDWIDTH was introduced in gnu awk 2.31

Answer 3

Parsing the output of cal isn't really that advisable...
Can your OS's date handle -j?
date -j 073100002014 "+%a"
Thu
How is your OS at perl?
perl -MDateTime -E '$dt=DateTime->new(year=>2014,month=>7,day=>31);say $dt->day_name'
Thursday
Or, if it doesn't do perl -E, you could do
perl -MDateTime -e '$dt=DateTime->new(year=>2014,month=>7,day=>31);print $dt->day_name'
Thursday
How is your OS at php?
php -r '$jd=cal_to_jd(CAL_GREGORIAN,7,31,2014);echo(jdk($jd,2));'
Thu