why does not this work, shouldnt this output 30:84:A9:9B:2A:67 from my textfile?
grep [A-F0-9]\:{5}[A-F0-9] textfile.txt
try this:
$ echo 30:84:A9:9B:2A:67 | grep -P "([A-F0-9]{2}:){5}[A-F0-9]{2}"
30:84:A9:9B:2A:67
In your question "[A-F0-9]:{5}" was trying to match an alpha numeric character plus colon five times: X:X:X:X:X:
Also, grep accepts basic regular expressions (BRE) so you need to escape brackets and parenthesis.
Related
This question already has answers here:
grep with regexp: whitespace doesn't match unless I add an assertion
(2 answers)
Closed 3 years ago.
I have a text file that contains quotes, comma and spaces.
"'x','a b c'"
"'x','a b c','1','2 3'"
"'x','a b c','22'"
"'x','a b z'"
"'x','s d 2'"
However, when I try using grep to pull the exact match, it doesn't display the results. Below is the command I'm trying to use.
grep -E "\"\'x\'\,\'a\s\+b\s\+c\'\"" test.txt
Expected output: "'x','a b c'"
Am I missing anything? Any help would be really appreciated.
You were close! Couple of notes:
Don't use \s. It is a gnu extension, not available everywhere. It's better to use character classes [[:space:]], or really just match a space.
The \+ may be misleading - in -E mode, it matches a literal +, while without -E the \+ matches one or more preceding characters. The escaping depends on the mode you are using.
You don't need to escape everything! When in " doublequotes, escape doublequotes "\"", don't escape singlequotes and commas in doublequotes, "\'\," is interpreted as just "',".
If you meant only to match spaces with grep -E:
grep -E "\"'x','a +b +c'\""
This is simple enough without -E, just \+ instead of +:
grep "\"'x','a \+b \+c'\""
I like to put things in front of + inside braces, helps me read:
grep "\"'x','a[ ]\+b[ ]\+c'\""
grep -E "\"'x','a[ ]+b[ ]+c'\""
If you want to match spaces and tabs between a and b, you can insert a literal tab character inside [ ] with $'\t':
grep "\"'x','a[ "$'\t'"]\+b[ "$'\t'"]\+c'\""
grep -E "\"'x','a[ "$'\t'"]+b[ "$'\t'"]+c'\""
But with grep -P that would just become:
grep -P "\"'x','a[ \t]+b[ \t]+c'\""
But the best is to forget about \s and use character classes [[:space:]]:
grep "\"'x','a[[:space:]]\+b[[:space:]]\+c'\""
grep -E "\"'x','a[[:space:]]+b[[:space:]]+c'\""
This question already has answers here:
Using different delimiters in sed commands and range addresses
(3 answers)
Closed 6 years ago.
I'm trying to replace text in a file with the output of another command. Unfortunately, the outputted text contains characters bash expands. For example, I'm running the following script to change the file (somestring references output that would break the sed command):
#!/bin/bash
somestring='$6$sPnfj/lnXwZVrec7$fCnL9uy1oWIMZduInKTHBAxhsQxGCsBpm2XfVFFqDPHKidrd93yfjbYvKgYexXHVcvkKdu9lbfy16Ek5GvKy/1'
sed '0,/^title/s/^title*/'"$somestring"'\n&/' $HOME/example.txt
sed fails with this error:
sed: -e expression #1, char 30: unknown option to `s'
I think bash is substuting the contents of $somestring when building the sed command, but is then trying to expand the resulting text. I can't put the entire sed script in single quotes, I need bash to expand it the first time, just not the second. Any suggestions? Thanks
here the forward slash / is the problem. If it's the only issue you can set sed to use a different delimiter.
for example
$ somestring="abc/def"; echo xxx | sed 's/xxx/'"$somestring"'/'
sed: -e expression #1, char 11: unknown option to `s'
$ somestring="abc/def"; echo xxx | sed 's_xxx_'"$somestring"'_'
abc/def
you also need to worry about & and \ chars and escape them if can appear in the replacement text.
If you can't control the the replacement string, either you have to sanitize with another sed script or, alternatively use r command to read it from a file. For example,
$ seq 5 | sed -e '/3/{r replace' -e 'd}'
1
2
3slashes///1ampersand&and2backslashes\\end
4
5
where
$ cat replace
3slashes///1ampersand&and2backslashes\\end
You have several errors here:
the string somestring has characters that are significative for sed command (the most important being '/' that you are using as a delimiter) You can escape it, by substituting it with a previous
somestring=$(echo "$somestring" | sed -e 's/\//\\\//g')
that will convert your / chars to \/ sequences.
you are using sed '0,/^title/s/^title*/'"$somestring"'\n&/' $HOME/example.txt which is looking to substitute the string titl followed by any number of e characters by that $somestring value, followed by a new line and the original one. Unfortunately, sed(1) doesn't allow you to use newline characters in the pattern substitution side of the s command, but you can afford the result by using the i command with a text consisting of you pattern (preceding any new line by a \ to interpret it as literal):
Finally the script leads to:
#!/bin/bash
somestring='$6$sPnfj/lnXwZVrec7$fCnL9uy1oWIMZduInKTHBAxhsQxGCsBpm2XfVFFqDPHKidrd93yfjbYvKgYexXHVcvkKdu9lbfy16Ek5GvKy/1'
somestring=$(echo "$somestring" | sed -e 's/\//\\\//g')
sed '/^title/i\
'"$somestring\\
" $HOME/example.txt
If your shell is Bash, you can use parameter substitution to replace the problematic /:
somestring="{somestring//\//\\/}"
That looks scary, but is easier to understand if you look at the version that replaces x with __:
somestring="${somestring//x/__}"
It might be easier to use (say) underscore as the delimiter for your sed s command, and then the substitution above would be
somestring="${somestring//_/\\_}"
If you already have backslashes, you'll need to first replace those:
somestring="${somestring//\\/\\\\}"
somestring="{somestring//\//\\/}"
If there were other characters that needed escaping (e.g. on the search side of s///), then you could extend the above appropriately.
This URL provides the cleanest answer:
Command to escape a string in bash
printf "%q" "$someVariable"
will escape any characters you need escaped for you.
I am trying to find alphanumeric string including these two characters "/+" with at least 30 characters in length.
I have written this code,
grep "[a-zA-Z0-9\/\+]{30,}" tmp.txt
cat tmp.txt
> array('rWmyiJgKT8sFXCmMr639U4nWxcSvVFEur9hNOOvQwF/tpYRqTk9yWV2xPFBAZwAPRVs/s
ddd73ZEjfy+airfy8DtqIqKI9+dd 6hdd7soJ9iG0sGs/ld5f2GHzockoYHfh
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
KEsAmN4i/+ym8be3wwn KWGYaIB908+7W98pI6qao3iaZB
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
this does not work, Mainly I wanted to have minimum length of the string to be 30
In the syntax of grep, the repetition braces need to be backslashed.
grep -o '[a-zA-Z0-9/+]\{30,\}' file
If you want to constrain the match to lines containing only matches to this pattern, add line-start and line-ending anchors:
grep '^[a-zA-Z0-9/+]\{30,\}$' 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 a file which contains Inverted exclamation mark, I want to count number of occurrences of these inverted exclamation marks using Linux grep command.
I have tried hex representation of this character as follows. but it is returning complete file , not exactly the lines which are matching this text.
grep -v "["$'\xA1'"]" K2345061.005
Thanks in advance for sharing any idea on this issue.
If your grep supports the -P flag for PCRE regex syntax, you can use that:
$ echo -e '\xa1Ay caramba!' > /tmp/a1.dat
$ grep -P '\xa1' /tmp/a1.dat
¡Ay caramba!
grep -v is used to list files which do not match. Remove the -v option.
I have a string variable x=tmp/variable/custom-sqr-sample/test/example
in the script, what I want to do is to replace all the “-” with the /,
after that,I should get the following string
x=tmp/variable/custom/sqr/sample/test/example
Can anyone help me?
I tried the following syntax
it didnot work
exa=tmp/variable/custom-sqr-sample/test/example
exa=$(echo $exa|sed 's/-///g')
sed basically supports any delimiter, which comes in handy when one tries to match a /, most common are |, # and #, pick one that's not in the string you need to work on.
$ echo $x
tmp/variable/custom-sqr-sample/test/example
$ sed 's#-#/#g' <<< $x
tmp/variable/custom/sqr/sample/test/example
In the commend you tried above, all you need is to escape the slash, i.e.
echo $exa | sed 's/-/\//g'
but choosing a different delimiter is nicer.
The tr tool may be a better choice than sed in this case:
x=tmp/variable/custom-sqr-sample/test/example
echo "$x" | tr -- - /
(The -- isn't strictly necessary, but keeps tr (and humans) from mistaking - for an option.)
In bash, you can use parameter substitution:
$ exa=tmp/variable/custom-sqr-sample/test/example
$ exa=${exa//-/\/}
$ echo $exa
tmp/variable/custom/sqr/sample/test/example