If statement syntax error (shell)

Question

if [[ ! -z grep echo "${prof}" | cut -d. -f1 dm_smear.dat ]]

This gives me the following error. I am trying to find a string ${prof} in a file dm_smear.dat and if that exists in the file I will do certain operations

: syntax error in conditional expression   
: syntax error near `1`'   
: `         if [[ ! -z grep `echo "${prof}" | cut -d . -f 1` dm_smear.dat ]]'

What does the variable ${prof} contain? Given the amount if syntax errors and misconceptions in this short piece of code, you'd probably better explain exactly what should happen instead of have us guess. — tripleee
– tripleee, Commented Dec 17, 2013 at 5:23

devnull · Accepted Answer · 2013-12-17 05:09:07Z

1

I am trying to find a string ${prof} in a file dm_smear.dat and if that exists in the file I will do certain operations

You can simply use the -q option for grep. Say:

if grep -q "${prof}" dm_smear.dat; then
   echo "Found the string"
   # Do something here
fi

You could fix your original code by using process substitution correctly:

if [[ ! -z $(grep $(echo "${prof}" | cut -d . -f 1) dm_smear.dat) ]];
   echo "Found the string"
   # Do something here
fi

edited Dec 17, 2013 at 5:09

answered Dec 17, 2013 at 4:32

devnull

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8 Comments

Devansh Agarwal Over a year ago

That just makes it quiet, still the loop doesn't run. I also tried if [[ ! -z cat dm_smear.dat | grep echo "${prof}" | cut -d. -f1 ]] but same error followed.

devnull Over a year ago

@Devansh Why are you executing: if [[ ! -z cat dm_smear.dat | grep echo "${prof}" | cut -d. -f1 ]]; ? Did you even try what was suggested in the answer?

Devansh Agarwal Over a year ago

Yes, I did! The loop doesn't run. I did it with and without -q with -q : loop doesn't run without -q gives the same error

devnull Over a year ago

@Devansh If grep doesn't find the string ${prof} in the file dm_smear.dat then the loop won't run. Isn't that what you wanted? Maybe you should update the question to clarify further if that isn't the case.

Devansh Agarwal Over a year ago

Yes, this is what exactly I want, but I've checked and string ${prof} exits in the file dm_smear.dat.

|

Prakash · Accepted Answer · 2013-12-17 13:26:59Z

0

You can try following code

prof="KEY1"

if ! [ -z cut -d. -f1 dm_smear.dat | grep ${prof} ] ; then

     echo "FOUND"

else

     echo "NOT FOUND"

fi

In above code, we are searching KEY1 in dm_smear.dat file at first column

the output will be FOUND if file contains KEY1 in first column of any line, otherwise it prints NOT FOUND

where dm_smear.dat contains

KEY1.VALUE1

KEY2.VALUE2

KEY3.VALUE2

answered Dec 17, 2013 at 13:26

Prakash

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