What's the difference between "grep -e" and "grep -E" [closed] - linux

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I have a file test.txt, in which there are some formatted phone numbers. I'm trying to use grep to find the lines containing a phone number.
It seems that grep -e "[0-9]{3}-[0-9]{3}-[0-9]{4}" test.txt doesn't work and gives no results. But grep -E "[0-9]{3}-[0-9]{3}-[0-9]{4}" test.txtworks. So I wonder what's the difference between these 2 options.
According to man grep:
-E, --extended-regexp
Interpret pattern as an extended regular expression (i.e. force
grep to behave as egrep).
-e pattern, --regexp=pattern
Specify a pattern used during the search of the input: an input
line is selected if it matches any of the specified patterns.
This option is most useful when multiple -e options are used to
specify multiple patterns, or when a pattern begins with a dash
(`-').
But I don't quite understand it. What is an extended regex?

As you mentioned, grep -E is for extended regular expressions whereas -e is for basic regular expressions. From the man page:
EDIT: As Jonathan pointed out below, grep -e "specifies that the following argument is (one of) the regular expression(s) to be matched."
Basic vs Extended Regular Expressions
In basic regular expressions the meta-characters ?, +, {, |, (, and ) lose
their special meaning; instead use the backslashed versions \?, \+, \{,
\|, \(, and \).
Traditional egrep did not support the { meta-character, and some egrep
implementations support \{ instead, so portable scripts should avoid { in
grep -E patterns and should use [{] to match a literal {.
GNU grep -E attempts to support traditional usage by assuming that { is
not special if it would be the start of an invalid interval specification.
For example, the command grep -E '{1' searches for the two-character
string {1 instead of reporting a syntax error in the regular expression.
POSIX.2 allows this behavior as an extension, but portable scripts should
avoid it.
But man pages are pretty terse, so for further info, check out this link:
http://www.regular-expressions.info/posix.html
The part of the manpage regarding the { meta character though specifically talks about what you are seeing with respect to the difference.
grep -e "[0-9]{3}-[0-9]{3}-[0-9]{4}"
won't work because it is not treating the { character as you expect. Whereas
grep -E "[0-9]{3}-[0-9]{3}-[0-9]{4}"
does because that is the extended grep version — or the egrep version for example.

Here is a simple test:
$ cat file
apple is a fruit
so is orange
but onion is not
$ grep -e 'but' -e 'fruit' file #Allows you to pass multiple patterns explicitly
apple is a fruit
but onion is not
$ grep -E 'is (a|not)' file #Allows you to use extended regular expressions like ?, +, | etc
apple is a fruit
but onion is not

The -e option to grep simply says that the following argument is the regular expression. Thus:
grep -e 'some.*thing' -r -l .
looks for some followed by thing on a line in all the files in the current directory and all its sub-directories. The same could be achieved by:
grep -r -l 'some.*thing' .
(On Linux, the situation is confused by the behaviour of GNU getopt() which, unless you set POSIXLY_CORRECT in the environment, permutes options, so you could also run:
grep 'some.*thing' -r -l .
and get the same result. Under POSIX and other systems not using GNU getopt(), options need to precede arguments, and the grep would look for a file called -r and another called -l.)
The -E option changes the regular expressions from 'basic' to 'extended'. It can be used with -e:
grep -e "[0-9]{3}-[0-9]{3}-[0-9]{4}" test.txt
grep -E -e "[0-9]{3}-[0-9]{3}-[0-9]{4}" test.txt
The ERE option means the same regular expressions, more or less, as used to be recognized by the egrep command, which is no longer a part of POSIX (having been replaced by grep -E, and fgrep by grep -F).

Related

File Glob Patterns in Linux terminal

I want to search a filename which may contain kavi or kabhi.
I wrote command in the terminal:
ls -l *ka[vbh]i*
Between ka and i there may be v or bh .
The code I wrote isn't correct. What would be the correct command?
A nice way to do this is to use extended globs. With them, you can perform regular expressions on Bash.
To start you have to enable the extglob feature, since it is disabled by default:
shopt -s extglob
Then, write a regex with the required condition: stuff + ka + either v or bh + i + stuff. All together:
ls -l *ka#(v|bh)i*
The syntax is a bit different from the normal regular expressions, so you need to read in Extended Globs that...
#(list): Matches one of the given patterns.
Test
$ ls
a.php AABB AAkabhiBB AAkabiBB AAkaviBB s.sh
$ ls *ka#(v|bh)i*
AAkabhiBB AAkaviBB
a slightly longer cmd line could be using find, grep and xargs. it has the advantage of being easily extended to different search terms (by either extending the grep statement or by using additional options of find), a bit more readability (imho) and flexibility in being able to execute specific commands on the files which are found
find . | grep -e "kabhi" -e "kavi" | xargs ls -l
You can get what you want by using curly braces in bash:
ls -l *ka{v,bh}i*
Note: this is not a regular expression question so much as a "shell globbing" question. Shell "glob patterns" are different from regular expressions, though they are similar in many ways.

Some help needed on grep

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.

Specify multiple possible patterns for a single command

Basically there a few lines which contain a common format, but different wording at the end. The command will work for all of them, but I want to match all possible pattern, thereby needing only 1 line in the script. As an example, I know how to make the script work like so:
/pattern1/ s/asdf/ghjk/g
/pattern2/ s/asdf/ghjk/g
/pattern3/ s/asdf/ghjk/g
Any ideas?
If your patterns are really as similar as in your example, you can use
sed -e '/pattern[1-3]/ s/asdf/ghjk/g'
If the patterns aren't so similar and your sed command supports extended regular expressions, you can use
sed -E -e '/(pattern1|pattern2|pattern3)/ s/asdf/ghjk/g'
# ^^ use extended regular expressions
# for GNU sed, use -r or escape (, |, and ) with \
If your sed command doesn't support extended regular expressions, you might have to turn to awk or perl:
perl -ple '/(pattern1|pattern2|pattern3)/ && s/asdf/ghjk/g'

Sed:Replace a series of dots with one underscore

I want to do some simple string replace in Bash with sed. I am Ubuntu 10.10.
Just see the following code, it is self-explanatory:
name="A%20Google.."
echo $name|sed 's/\%20/_/'|sed 's/\.+/_/'
I want to get A_Google_ but I get A_Google..
The sed 's/\.+/_/' part is obviously wrong.
BTW, sed 's/\%20/_/' and sed 's/%20/_/' both work. Which is better?
sed speaks POSIX basic regular expressions, which don't include + as a metacharacter. Portably, rewrite to use *:
sed 's/\.\.*/_/'
or if all you will ever care about is Linux, you can use various GNU-isms:
sed -r 's/\.\.*/_/' # turn on POSIX EREs (use -E instead of -r on OS X)
sed 's/\.\+/_/' # GNU regexes invert behavior when backslash added/removed
That last example answers your other question: a character which is literal when used as is may take on a special meaning when backslashed, and even though at the moment % doesn't have a special meaning when backslashed, future-proofing means not assuming that \% is safe.
Additional note: you don't need two separate sed commands in the pipeline there.
echo $name | sed -e 's/\%20/_/' -e 's/\.+/_/'
(Also, do you only need to do that once per line, or for all occurrences? You may want the /g modifier.)
The sed command doesn't understand + so you'll have to expand it by hand:
sed 's/\.\.*/_/'
Or tell sed that you want to use extended regexes:
sed -r 's/\.+/_/' # GNU
sed -E 's/\.+/_/' # OSX
Which switch, -r or -E, depends on your sed and it might not even support extended regexes so the portable solution is to use \.\.* in place of \.+. But, since you're on Linux, you should have GNU sed so sed -r should do the trick.

Using ? with sed

I just want to get the number of a file that may or may not be gzip'd. However, it appears that a regular expression in sed does not support a ?. Here's what I tried:
echo 'file_1.gz'|sed -n 's/.*_\(.*\)\(\.gz\)?/\1/p'
and nothing was returned. Then I added a ? to the string being analyzed:
echo 'file_1.gz?'|sed -n 's/.*_\(.*\)\(\.gz\)?/\1/p'
and got:
1
So, it looks like the ? used in most regex's is not supported in sed, right? Well then, I would just like sed to give a 1 for file_1 and file_1.gz. What's the best way to do that in a bash script if execution time is critical?
The equivalent to x? is \(x\|\).
However, many versions of sed support an option to enable "extended regular expressions" which includes ?. In GNU sed the flag is -r. Note that this also changes unescaped parens to do grouping. eg:
echo 'file_1.gz'|sed -n -r 's/.*_(.*)(\.gz)?/\1/p'
Actually, there's another bug in your regex which is that the greedy .* in the parens is going to swallow up the ".gz" if there is one. sed doesn't have a non-greedy equivalent to * as far as I know, but you can use | to work around this. | in sed (and many other regex implementations) will use the leftmost match that works, so you can do something like this:
echo 'file_1.gz'|sed -r 's/(.*_(.*)\.gz)|(.*_(.*))/\2\4/'
This tries to match with .gz, and only tries without it if that doesn't work. Only one of group 2 or 4 will actually exist (since they are on opposite sides of the same |) so we just concatenate them to get the value we want.
If you're looking for an answer to the specific example given in the question, or why it uses the ? incorrectly (regardless of syntax), see the answer by Laurence Gonsalves.
If you're looking instead for the answer to the general question of why ? doesn't exhibit its special meaning in sed as you might expect:
By default, sed uses the " POSIX basic regular expressions syntax", so the question mark must be escaped as \? to apply its special meaning, otherwise it matches a literal question mark. As an alternative, you can use the -r or --regexp-extended option to use the "extended regular expression syntax", which reverses the meaning of escaped and non-escaped special characters, including ?.
In the words of the GNU sed documentation (view by running 'info sed' on Linux):
The only difference between basic and extended regular expressions is in
the behavior of a few characters: '?', '+', parentheses, and braces
('{}'). While basic regular expressions require these to be escaped if
you want them to behave as special characters, when using extended
regular expressions you must escape them if you want them to match a
literal character.
and the option is explained:
-r
--regexp-extended
Use extended regular expressions rather than basic regular
expressions. Extended regexps are those that `egrep' accepts;
they can be clearer because they usually have less backslashes,
but are a GNU extension and hence scripts that use them are not
portable.
Update
Newer versions of GNU sed now say this:
-E
-r
--regexp-extended
Use extended regular expressions rather than basic regular
expressions. Extended regexps are those that 'egrep' accepts; they
can be clearer because they usually have fewer backslashes.
Historically this was a GNU extension, but the '-E' extension has
since been added to the POSIX standard
(http://austingroupbugs.net/view.php?id=528), so use '-E' for
portability. GNU sed has accepted '-E' as an undocumented option
for years, and *BSD seds have accepted '-E' for years as well, but
scripts that use '-E' might not port to other older systems.
So, if you need to preserve compatibility with ancient GNU sed, stick with -r. But if you prefer better cross-platform portability on more modern systems (e.g. Linux+Mac support), go with -E (but note that there are still some quirks and differences between GNU sed and BSD sed, so you'll have to make sure your scripts are portable in any case).
echo 'file_1.gz'|sed -n 's/.*_\(.*\)\?\(\.gz\)/\1/p'
Works. You have to put the return in the right spot, and you have to escape it.
A function that should return a number that follows the '_' in a filename, regardless of file extension:
realname () {
local n=${$1##*/}
local rn="${n%.*}"
sed 's/^.*\_//g' ${$rn:-$n}
}
You should use awk which is superior to sed when it comes to field grabbing/parsing:
$ awk -F'[._]' '{print $2}' <<<"file_1"
1
$ awk -F'[._]' '{print $2}' <<<"file_1.gz"
1
Alternatively you can just use Bash's parameter expansion like so:
var=file_1.gz;
temp=${var#*_};
file=${temp%.*}
echo $file
Note: works when var=file_1 as well
Part of the solution lies in escaping the question mark or using the -r option.
sed 's/.*_\([^.]*\)\(\.\?[^.]\+\)\?$/\1/'
or
sed -r 's/.*_([^.]*)(\.?[^.]+)?$/\1/'
will work for:
file_1.gz
file_12.txt
file_123
resulting in:
1
12
123
I just realized that could do something very easy:
echo 'file_1.gz'|sed -n 's/.*_\([0-9]*\).*/\1/p'
Notice the [0-9]* instead of a .*. #Laurence Gonsalves's answer made me realize the greediness of my previous post.

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