Shell脚本变量结构

Question

How can I create a variable with a file name format like : FileName pattern: SnapshotIR__somenumber.csv

I tried something like :

TODAY=$(date +"%m%d%Y")    
SNAPSHOT = $(SnapshotIR$TODAY*.csv)

I get error like :

test.sh: line 2: SnapshotIR02122013_2239.csv: command not found
test.sh: line 2: SNAPSHOT: command not found

so, when I want to use with if

if [ -f SnapshotIR$TODAY*.csv]  -> works 
if [ -f ${SNAPSHOT} ]           -> does not work (I get the above error)

Answer 1

SNAPSHOT=SnapshotIR${TODAY}\*.csv

when you give $(.....) it says shell to execute the command within the braces. i guess you are just forming the file name.

also remove the spaces:

SNAPSHOT<space>=<space>

i would also like to add wildcard "*" will not work for -f flag.

for file in $SNAPSHOT
do
        if [ -f "$file" ]
        then
                FOUND="$file"
                break
        fi
done