Finding max value of a specific date awk

Question

I have a file with several rows and with each row containing the following data-

name 20150801|1 20150802|4  20150803|6  20150804|7  20150805|7  20150806|8  2015         


        

Answer 1

I think you mean this:

awk -v date=20150823 '{for(f=2;f<=NF;f++){split($f,a,"|");if(a[1]==date&&a[2]>max){max=a[2];name=$1}}}END{print name,max}' YourFile

So, you pass the date you are looking for in as a variable called date. You then iterate through all fields on the line, and split the date and value of each into an array using | as separator - a[1] has the date, a[2] has the value. If the date matches and the value is greater than any previously seen maximum, save this as the new maximum and save the first field from this line for printing at the end.