how to extract a part of file name in unix scripting and check if it exists

Question

I need to write a script where I have to extract part of the name from a file abc_def_xxx_yyy_20210515.txt and check if abc_def_xxx_yyy.txt exists. If not then create the file.

FILE="abc_def_xxx_yyy_20210515.txt"
DLS=echo $FILE | awk -F_ '{print $1,$2,$3,$4".txt"}'
DLL=\home\xxx\$DLS
if [-f "$DLL"]; then
echo "$DLL exists"
else 
touch "$DLL"
fi

I get the error saying "abc_def_xxx_yyy_20210515.txt" not found

but if I just run with

FILE="abc_def_xxx_yyy_20210515.txt"
echo $FILE | awk -F_ '{print $1,$2,$3,$4".txt"}'

I get abc_def_xxx_yyy.txt.

Can someone please help me with this?

There are a bunch of syntax errors here; run it through [shellcheck.net](https://www.shellcheck.net) and fix what it points out. Also, it looks you're using DOS/Windows paths (with `/` rather than `\`) -- is this on Windows or Unix/Linux? — Gordon Davisson, Aug 15 '21 at 08:15
You forgot the `$(...)` around the command substitution. You are assigning the string `echo` to the variable `DLS` and then attempting to execute `$FILE` as a command. — tripleee, Aug 15 '21 at 10:22

score 1 · Accepted Answer · answered Aug 15 '21 at 09:38

1

Try:

FILE="abc_def_xxx_yyy_20210515.txt"
DLS="${FILE%_*}.txt"
DLL="$HOME/$DLS"
if [ -f "$DLL"]; then
  echo "$DLL exists"
else 
  touch "$DLL"
fi

answered Aug 15 '21 at 09:38

Pierre François

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how to extract a part of file name in unix scripting and check if it exists

1 Answers1