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How to Process input files from folder in shell script?

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# 1  
Old 10-01-2010
How to Process input files from folder in shell script?
Hi,

I want to process all input files available into folder (C:\ShellPrg\InputFile\) Input files are abc.CSV , XYZ.zip (zip of CSV file), PQR.gz (zip of CSV file).

I want to check the extension of file, If its .zip/.gz then need to unzip the file as .CSV
I want to parse line by line of each .CSV file.

Here is my code,
Code:
FILE=$1
echo $FILE
EXTN=`echo $FILE | cut -d "." -f 2`
 
if [ $EXTN == "zip" ]; then
    input_file=`unzip $FILE`
    echo "input file is $input_file"
elif [ $EXTN == "gz" ]; then
    input_file=`gzip -d $FILE`
fi
EXTN1=`echo $input_file | cut -d "." -f 2`
if [ $EXTN1 == "CSV" ]; then
    // code of line parse
fi

where $1 = <PATH> a_b_c_d_1234.CSV / a_b_c_d_1234.zip / a_b_c_d_1234.gz

How can I put the for loop to process each file from folder?
Also help me to unzip the file and store the unzipped file name into a variable?

Moderator's Comments:
Mod Comment Time to start using code tags.
# 2  
Old 10-01-2010
Code:
for FILE in *; do
echo $FILE
EXTN=`echo $FILE | cut -d "." -f 2`
 
if [ $EXTN == "zip" ]; then
    input_file=`unzip $FILE`
    echo "input file is $input_file"
elif [ $EXTN == "gz" ]; then
    input_file=`gzip -d $FILE`
fi
EXTN1=`echo $input_file | cut -d "." -f 2`
if [ $EXTN1 == "CSV" ]; then
    // code of line parse
fi
done

# 3  
Old 10-01-2010
OK.
Help me to unzip the file and store the unzipped file name into a variable.
From my code i am getting output as,

Code:
input file is Archive:  a_b_c_d_1234.zip

Thanks in advance.
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