Bash - binary operator expected

Question

I've got bash script for counting rows in the reports. I have one array where all reports names are stored and in the loop I'm counting rows. However for some files my script receives binary operator expected error. Do anyone have a solution?

for i in ${ARRAY[@]}; do
if [ ! -f "$BASE_DIR/$i"* ];
then
        echo "File not generated yet"
else
        ARRAY2=$(wc -l < "$BASE_DIR/$i"*.tab | awk '{print $1-2}')
        echo ${ARRAY2[$i]} $i
fi

You might be aware of the fact that * in this context will do globbing — sjsam
– sjsam, Commented Jul 1, 2016 at 11:47
globbing is used on purpose, cause there is also text before filename and extension — ulsterboy
– ulsterboy, Commented Jul 1, 2016 at 11:56

Rohit Verma · Accepted Answer · 2016-07-01 11:52:56Z

2

Use double straight braces instead of ones as follows since you r using extended expressions.

if [[ ! -f "$BASE_DIR/$i"* ]];

Need to check with array contents. Special characters as ' ' (spaces) in file names must be escaped.

edited Jul 1, 2016 at 11:52

answered Jul 1, 2016 at 11:34

Rohit Verma

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chepner · Accepted Answer · 2016-07-01 11:59:04Z

2

-f takes just one argument, so the error occurs when the pattern matches more than one file.

It seems to work with [[, although I can't find any documentation as to why it does.

The bigger problem is you can also only use one file with the < operator; if the pattern matches multiple files, you'll get an ambiguous redirect error. To fix that, you'll need to use cat:

cat "$BASE_DIR/$i"*.tab | wc -l

However, it's not clear what you are expecting from the output; ARRAY2 will not actually be an array.

answered Jul 1, 2016 at 11:59

chepner

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1 Comment

ulsterboy Over a year ago

yeah, as @sjsam has mentioned the problem is with globbing, I have found out that I have some files with .bak in the end

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