I am trying grep some string in a script.
I don't know what string i am finding or what is in file where I am trying to find it. I just need grep exact string from file.
So, my problem is, sometimes greping string contains square brackets and as I found out, grep consider them special characters.
string='some [text]'
grep "$string" file
I can escape them with sed
string='some [text]'
grep "$(sed -e 's/\[/\\[/g' <<< "$string")" file
I need grep to match exact string no matter what input can be. So is there a nicer way to do it? some way to tell grep to consider every character in string as regular character? If no, are there any other special characters like [ I need to worry about?
You can use -F option, to interpret it as fixed string:
grep -F "$string" file
Related
I have a file, the content of file has a string like this:
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
I want to check the file has this string. But when I use grep to check, It always return false. I try some ways:
grep "'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'" foo.txt
grep "'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'" foo.txt
str="'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'"
grep "$str" foo.txt
Can you help me? Maybe, another command line.
This is my case:
while read str; do
if [ ! -z "$str" ]; then
if grep -Fxq "$str" "$file_path"; then
do somthing
fi
fi
done < <(cat /usr/local/caotoc/db.dat)
Thank you so much!
First, you need to make sure the string is quoted properly. This is a bit of an art form, since your string contains both single and double quotes.
One thought would be to use read and a here-document to avoid having to escape anything.
Second, you need to use -F to perform exact string matching instead of more general regular-expression matching.
IFS= read -r str <<'EOF'
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
EOF
grep -F "$str" foo.txt
Based on the update, you can use a simple loop to read them one at a time.
while IFS= read -r str; do
grep -F "$str" foo.txt
done < /usr/local/caotoc/db.dat
You may be able to simply use the -f option to grep, which will cause grep to output lines from foo.txt that match any line from db.dat.
grep -f /usr/local/caotoc/db.dat -F foo.txt
Instead of trying to workaround regexes, the simplest way is to turn off regular expressions using -F (or --fixed-strings) option, which makes grep act like a simple string search
-F, --fixed-strings PATTERN is a set of newline-separated strings
like this:
grep -F "'/ad/e','#'.base64_decode(\"ZXZhbA==\").'(\$zad)', 'add'" test
Note: because of the shell, you still need to escape:
double quotes
dollar sign or else $zad is evaluated as an environment variable
Below is the string I am trying to grep for this in the bash shell:
'#Hostname=sometext.company.com, sometext.company.com' filename
I want to only find the string if it matches that exact pattern. I already tried the command below and a few others.
grep -Fx "#Hostname=sometext.company.com, sometext.company.com" filename
Did you specify the -xoption on purpose?
grep -F '#Hostname=sometext.company.com, sometext.company.com' filename
most likely is what you want. Also, it's better to put single quotes instead of double quotes, just in case your search pattern happens to contain special shell characters.
I wanted to search for this particular string $string['site:config'] in my folder. But when I use a normal grep function grep -r "$string['site:config']" it gives me random results.
The problem with your pattern is that characters like $ [ ] are characters used to define regular expressions and you have either to escape them:
grep "\$string\['site:config'\]"
or instruct grep to look for the given string as is:
grep -F "$string['site:config']"
without attempting to interpret it as a regular expression.
Is there a way to grep (or use another command) to find exact strings, using NO regex?
For example, if I want to search for (literally):
/some/file"that/has'lots\of"invalid"chars/and.triggers$(#2)[*~.old][3].html
I don't want to go through and escape every single "escapable". Essentially, I want to pass it through, like I would with echo:
$ echo "/some/file\"that/has'lots\of\"invalid\"chars/and.triggers$(#2)[*~.old][3].html"
/some/file"that/has'lots\of"invalid"chars/and.triggers$(#2)[*~.old][3].html
Use fgrep, it's the same as grep -F (matches a fixed string).
Well, you can put the information you want to match, each in a line, and then use grep:
grep -F -f patterns.txt file.txt
Notice the usage of the flag -F, which causes grep to consider each line of the file patterns.txt as a fixed-string to be searched in file.txt.
I am using sed -e "s/foo/$bar/" -e "s/some/$text/" file.whatever to replace a phrase in a certain file. The problem is that the $bar string contains multiple special characters like /. So when I try to replace something in a text file using the following code...
#!/bin/bash
bar="http://stackoverflow.com/"
sed -e "s/foo/$bar/" -e "s/some/$text/ file.whatever
...then I get an error saying : sed: unknown option to s is there anything I can do about it?
You can use any delimiter. s#some#SOME# for example. Another good delimiter is vertical-bar. Other chars can work but have special significance for some contexts such as regular expressions.
You can get this difficulty in sed regardless of what delimiters you use, especially if you don't know what the string contains. I'd pick a different method for passing the shell variables into the helper interpreter.
awk -v rep1="$bar" -v rep2="$text" '{sub(/foo/, rep1); sub(/some/, rep2); print}'
or
perl -spe 's/foo/$rep1/; s/some/$rep2/' -- -rep1="$bar" -rep2="$text"
Correctness trumps brevity in this case.
(reference for Perl example)