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Extracting Date from string


 
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Top Forums Shell Programming and Scripting Extracting Date from string
# 8  
Old 08-27-2009

There's still no need for an external command.

Code:
IFS=_
set -- $file
shift $(( $# - 2 ))
echo "date=$1"
echo "version=$2"

# 9  
Old 08-28-2009
this is throwing error shift out of range

line 3: shift: -2: shift count out of range
date=
version=
# 10  
Old 08-28-2009
What is the value of $file?
# 11  
Old 08-28-2009
the file values are

DLR1JPCHASE_All_Trades_Report_20090630_2.csv
GCR_20090630_2.csv
GCR1_20090630_2.csv
b_20090630_2.csv
c_20090630_2
pqr_20090630_2

in each case we need to extract 20090630_2 i.e date 20090630 and version number : 2
# 12  
Old 08-28-2009
Code:
IFS=_
set -- ${file%.csv}
shift $(( $# - 2 ))
echo "date=$1"
echo "version=$2"

# 13  
Old 08-31-2009
Its still throwing

./filename: line 3: shift: -2: shift count out of range


I just need to extract date 20090630 adn version code 2 from the following format ( files with .csv extension and without aswell )

(with extension )
DLR1JPCHASE_All_Trades_Report_20090630_2.csv
GCR_20090630_2.csv
GCR1_20090630_2.csv
b_20090630_2.csv

( without extension)
c_d_e_f20090630_2
pqr_d_20090630_2
pqr_20090630_2

Any help will be highly appreciated
# 14  
Old 08-31-2009

I repeat: What is the value of "$file"?

I don't want the names of the files; I want the actual value of the variable.

(You do realize that the code I posted was to operate on a single file, don't you? To work on all the .csv files, you need to put it in a loop with all of the file names. Use whatever pattern finds them all.)

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