CODEWITHSUNDEEP
bash
bnbfreak
bnbfreak

Reputation: 353

How to change the result when file does not exist in bash?

I have a problem to print out the result when a file does not exist. Let's say I use this code in bash

v1=`cat /ieee80211/phy2/rcstats`
echo $v1

When the file exists, it will display the value like this

0.8
0.6
0.3

But when file doesn't exist, it will just display :

cat: /ieee80211/phy2/rcstats: No such file or directory

How to change that warning into 1.0 ?

What I know, I just do like this

if [! -f $v1]; then
  echo "1.0"
fi

but the result it will just print 1.0 when the file exists, and the warning still appears

What should I do? Thank you

Upvotes: 0

Views: 201

Answers (3)

tamasgal
tamasgal

Reputation: 26259

The best error handling in this case is to check whether the file exist and if so, do the cat, not the other way round.

Like this:

file="/ieee80211/phy2/rcstats"
if [ -f $file ]
then
    v1=$(< $file)
fi

Upvotes: 1

gniourf_gniourf
gniourf_gniourf

Reputation: 46823

If you don't want to handle the case where the file doesn't exist, you might as well just use:

{ v1=$( </ieee80211/phy2/rcstats ) ; } 2>/dev/null
echo "$v1"

Note. Avoid using backticks, use $(...) instead.

Upvotes: 0

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798626

[ is a command. It needs spaces between it and its first argument.

if [ ! ... ]; then
    ^
   here

Upvotes: 4

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