Bash grep function for strings in array, can't get the quotation marks correct

Question

I'm trying to create a function that will do a grep on strings from an array that are sent as arguments but I can't get the quotation marks to work as I want.

LOG=/tmp/log.log
function grepCheck () {
    if grep $1 ; then
    STATE="true"
    else
    STATE="false"
    fi }

possibleErrors=("string1 string1 string1" "string2 string2 string2")

for checkError in ${possibleErrors[*]}
do
grepCheck "${possibleErrors[$checkError]} $LOG"
done

Output example:

+ LOG=/tmp/log.log
+ possibleErrors=("string1 string1 string1" "string2 string2 string2")
+ for checkError in '${possibleErrors[*]}'
+ grepCheck 'string1 string1 string1 /tmp/log.log'
+ grep string1 string1 string1 /tmp/log.log grep: string1: No such file or directory grep: string1: No such file or directory

How do I get the function to grep the string that it recieves?

Check: How to iterate over list which contains whitespaces in bash — fedorqui
– fedorqui, Commented Dec 2, 2014 at 11:58

tripleee · Accepted Answer · 2014-12-02 13:09:08Z

2

What makes sense is for your function to accept the same arguments grep does.

LOG=/tmp/log.log
grepCheck () {
    # Edit: use "$@"; add -q option to avoid spurious output
    if grep -q "$@"; then
        STATE="true"
    else
        STATE="false"
    fi
}

possibleErrors=("string1 string1 string1" "string2 string2 string2")

for checkError in ${possibleErrors[*]}
do
    # Edit: fixed quotes; pass two quoted arguments
    grepCheck "${possibleErrors[$checkError]}" "$LOG"
done

If you simply care whether any of these strings matched, it would be simpler to pass them all to grep in one go.

regex=$(IFS='|'; echo "${possibleErrors[*]}")
grep -Eq "$regex" "$LOG"

edited Dec 2, 2014 at 13:09

answered Dec 2, 2014 at 12:50

tripleee

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2 Comments

Bitdiot Over a year ago

I think you want to use egrep in the second example with '|' ?

tripleee Over a year ago

@Bitdiot Meh, I was trying to wiggle around that, but you're right, it's the easiest solution.

Bitdiot · Accepted Answer · 2014-12-02 12:55:42Z

0

Your code looks a little funny. grep usually wants a regular expression and a file to look for the expression. However, in your example, grepCheck is given an error string to-find, and a log to look for it.

Here you have:

if grep $1 ; then

Because $1 is unquoted, the shell expands this to grep string1 string1 string1. So grep wants to look for string1 in the two files called string1 and string1 (basically the same two files.).

I think you want the following:

function grepCheck () {
    needle=$1
    haystack=$2
    if [[ -n $(grep "$needle" "$haystack") ]] ; then
        STATE="true"     # Error found
    else
        STATE="false"
    fi }

Then in your calling code:

grepCheck "${possibleErrors[$checkError]}" $LOG

Notice the change in quoting.

edited Dec 2, 2014 at 12:55

answered Dec 2, 2014 at 12:49

Bitdiot

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6 Comments

tripleee Over a year ago

That's a Useless Use of [[. grep tells you perfectly well whether or not it succeeded just by itself, and the OP's code has this part exactly right.

Bitdiot Over a year ago

Ah yes, thank you for correcting my bad habit. Let me change that. If you don't mind.

Bitdiot Over a year ago

Hey triplee. I changed it back to the original. The reason is, and I just remembered, he may not want to see the results of the grep, but to set the flags.

tripleee Over a year ago

Use grep -q for that. See my answer.

Bitdiot Over a year ago

Wow, that's nice. I didn't know of the -q option. Thanks.

|

Geoff Nixon · Accepted Answer · 2014-12-02 13:07:57Z

0

I believe you need something like:

... for checkError in seq 0 $((${#possibleErrors[*]} - 1)) ...

But this is also a useless use of an array:

for error in "string1 string1 string1" "string2 string2 string2"; do grepCheck "$error" "$LOG" done

answered Dec 2, 2014 at 13:07

Geoff Nixon

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