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
I want to search
$GLOBALS["\x61\156\x75\156\x61"] using grep but " and / not working perfectly.
grep -rl "$GLOBALS["\x61\156\x75\156\x61"]" <filename>
$GLOBALS["\x61\156\x75\156\x61"] is a virus malware starting code, lots of files are affected. I have a script through I want search effected files and remove top line
Since you are looking for an exact match and you don't want the expression to be interpreted by grep, you have to use -F and single quotes to avoid the variable being expanded:
grep -Frl '$GLOBALS["\x61\156\x75\156\x61"]' <filename>
^ ^ ^
From man grep:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by
newlines, any of which is to be matched. (-F is specified by POSIX.)
See another example of the usage of -F together with single quotes:
We create a file like this:
$ cat a
hello
${myarray[0]}
bye
And an array:
$ myarray=('hello' 'how' 'are' 'you')
Let's use single quotes and look for the value:
$ grep '${myarray[0]}' a
$
Let's use fixed string with double quotes -> it gets interpreted!
$ grep -F "${myarray[0]}" a
hello
Let's use -F and single quotes:
$ grep -F '${myarray[0]}' a
${myarray[0]} #this works!
I am trying to find alphanumeric string including these two characters "/+" with at least 30 characters in length.
I have written this code,
grep "[a-zA-Z0-9\/\+]{30,}" tmp.txt
cat tmp.txt
> array('rWmyiJgKT8sFXCmMr639U4nWxcSvVFEur9hNOOvQwF/tpYRqTk9yWV2xPFBAZwAPRVs/s
ddd73ZEjfy+airfy8DtqIqKI9+dd 6hdd7soJ9iG0sGs/ld5f2GHzockoYHfh
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
KEsAmN4i/+ym8be3wwn KWGYaIB908+7W98pI6qao3iaZB
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
this does not work, Mainly I wanted to have minimum length of the string to be 30
In the syntax of grep, the repetition braces need to be backslashed.
grep -o '[a-zA-Z0-9/+]\{30,\}' file
If you want to constrain the match to lines containing only matches to this pattern, add line-start and line-ending anchors:
grep '^[a-zA-Z0-9/+]\{30,\}$' file
The -o option in the first command line causes grep to only print the matching part, not the entire matching line.
The repetition operator is not directly supported in Basic Regular Expression syntax. Use grep -E to enable Extended Regular Expression syntax, or backslash the braces.
You can use
grep -e "^[a-zA-Z0-9/+]\{30,\}" tmp.txt
grep -e "^[a-zA-Z0-9/+]\{30,\}" tmp.txt
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
man grep
Read up about the difference between between regular and extended patterns. You need the -E option.
I have 70+ strings I need to find and delete in a file. I need to remove the entire line in the file that the string appears in.
I know I can use sed -i '/string to remove/d' fileA.txt to remove them one at a time. However, considering I have 70+, it will take some time doing it this way.
Is there a way I can put these 70+ strings in a file and have sed go through them one by one? Or if I create a file containing the strings, is there a way to compare the two files so it removes any line from fileA that contains one of the strings?
You could use grep:
grep -vf file_with_words.txt file.txt
where file_with_words.txt would be the file containing the list of words, each word being on a different line and file.txt is the file that you want to remove the lines from.
If your list of words contains regex metacharacters, then tell grep to consider those as fixed strings (if that is what you want):
grep -F -vf file_with_words.txt file.txt
Using sed, you'd need to say:
sed '/word1\|word2\|word3/d' file.txt
or
sed -E '/word1|word2|word3/d' file.txt
You could use command substitution to construct the pattern too:
sed -E "/$(paste -sd'|' file_with_words.txt)/d" file.txt
but grep is clearly the tool to use in this case.
If you want to do the job in bash, here's how:
search=fileA.txt
queries=queries.txt
while read query
do
sed -i '' "/$query/d" $search
done < "$queries"
where queries.txt looks like
I
want
to
delete
these
lines
I have a set of 10 CSV files, which normally have a an entry of this kind
a,b,c,d
d,e,f,g
Now due to some error entries in this file have become of this kind
a,b,c,d
d,e,f,g
,,,
h,i,j,k
Now I want to remove the line with only commas in all the files. These files are on a Linux filesystem.
Any command that you recommend that can replaces the erroneous lines in all the files.
It depends on what you mean by replace. If you mean 'remove', then a trivial variant on #wnoise's solution is:
grep -v '^,,,$' old-file.csv > new-file.csv
Note that this deletes just those lines with exactly three commas. If you want to delete mal-formed lines with any number of commas (including zero) - and no other characters on the line, then:
grep -v '^,*$' ...
There are endless other variations on the regex that would deal with other scenarios. Dealing with full CSV data with commas inside quotes starts to need something other than a regex machine. It can be done, within broad limits, especially in more complex regex systems such as PCRE or Perl. But it requires more work.
Check out Mastering Regular Expressions.
sed 's/,,,/replacement/' < old-file.csv > new-file.csv
optionally followed by
mv new-file.csv old-file.csv
Replace or remove, your post is not clear... For replacement see wnoise's answer. For removing, you could use
awk '$0 !~ /,,,/ {print}' <old-file.csv > new-file.csv
What about trying to keep only lines which are matching the desired format instead of handling one exception ?
If the provided input is what you really want to match:
grep -E '[a-z],[a-z],[a-z],[a-z]' < oldfile.csv > newfile.csv
If the input is different, provide it, the regular expression should not be too hard to write.
Do you want to replace them with something, or delete them entirely? Either way, it can be done with sed. To delete:
sed -i -e '/^,\+$/ D' yourfile1.csv yourfile2.csv ...
To replace: well, see wnoise's answer, or if you don't want to create new files with the output,
sed -i -e '/^,\+$/ s//replacement/' yourfile1.csv yourfile2.csv ...
or
sed -i -e '/^,\+$/ c\
replacement' yourfile1.csv yourfile2.csv ...
(that should be entered exactly as is, including the line break). Of course, you can also do this with awk or perl or, if you're only deleting lines, even grep:
egrep -v '^,+$' < oldfile.csv > newfile.csv
I tested these to make sure they work, but I'd advise you to do the same before using them (just in case). You can omit the -i option from sed, in which case it'll print out the results (rather than writing them back to the file), or omit the output redirection >newfile.csv from grep.
EDIT: It was pointed out in a comment that some features of these sed commands only work on GNU sed. As far as I can tell, these are the -i option (which can be replaced with shell redirection, sed ... <infile >outfile ) and the \+ modifier (which can be replaced with \{1,\} ).
Most simply:
$ grep -v ,,,, oldfile > newfile
$ mv newfile oldfile
yes, awk or grep are very good option if you are working in linux platform. However you can use perl regex for other platform. using join & split options.