I need to find a pattern that starts with a $ is followed by two numbers, a single character that is not a number, and anything else.
I know how to find a pattern starting in a dollar sign and followed by two numbers but I can't figure out how to check for one character that is not a number.
I also need to count how many lines have this pattern.
I have this so far:
grep -Ec '\$[0-9][0-9].....
I don't know what to do. Can someone please help? Any help would be much appreciated.
The caret character inverts a selection group, so if [0-9] is "match any digit" then [^0-9] is "match any non-digit".
You can possibly try this regex \$[0-9][0-9][^0-9].*
\$[0-9][0-9][^0-9].*
\$ matches the character $ literally
[0-9] match a single character present in the list below.
0-9 a single character in the range between 0 and 9
[0-9] match a single character present in the list below.
0-9 a single character in the range between 0 and 9
[^0-9] match a single character not present in the list below.
0-9 a single character in the range between 0 and 9
.* matches any character (except newline)
Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
I would second #realspirituals answer, and if you need to count how many lines have this pattern, you can count how many lines grep ouputs by piping to wc -l. In order to both show the lines and count them in one fell swoop, pipe the output like so
grep "\$[0-9]{2}[^0-9].*" | tee >(wl -l)
where tee will split the output between wl and STDOUT. {2} will cause the prior [0-9] to match twice.
Related
I have to accept the strings that only have a single number, it doesn't matter the content of the string, it just needs to be a single number.
I was trying something like this:
echo "exaaaamplee1" | grep '[0-9]\{1\}'
This string is accepted, but this string also is accepted:
echo "exaaaamplee11" | grep '[0-9]\{1\}'
You probably want to use something like [^0-9]. This represents any character except a digit 0-9, and you can use [0-9] (or \d) for the one digit that is allowed.
Something like ^[^0-9]*[0-9][^0-9]*$ should match any string with exactly one digit. (^ being the start and $ the end of the string)
If you want to match a string with only one digit character using grep, it's
echo whatever1 | grep '^[^[:digit:]]*[[:digit:]][^[:digit:]]*$'
Start of line followed by any number of non-digits, one digit, and then any number of non-digits until the end of the line.
I've got a text-file with the following line:
201174480 11-01-1911 J Student 25-07 11585 2 0 SPOED BEZORGEN 1ST 25,00
320819019 11-01-1911 T. Student 28-07 13561 1 15786986 DESLORATADINE TABL OMH 5MG 60ST 3,60
706059901 11-01-1911 ST Student-Student 30-06 14956 1 15356221 METOPROLOLSUCC RET T 100MG 180ST 12,90-
I want to change this line with SED into:
201174480 11-01-1911 J Student 25-07 11585 2 0 SPOED BEZORGEN 1ST 25,00
320819019 11-01-1911 T. Student 28-07 13561 1 15786986 DESLORATADINE TABL OMH 5MG 60ST 3,60
706059901 11-01-1911 ST Student-Student 30-06 14956 1 15356221 METOPROLOLSUCC RET T 100MG 180ST -12,90
So I want to swap the minus sign so that I get-12,90 in stead of 12,90- with SED. I tried:
try 1:
sed 's/\([0-9.]\+\)-/-\1/g' file.txt > file1.txt
try 2:
sed 's/\([0-9].\+\)-$/-\1/g' file.txt > file1.txt
So there must be something wrong with the REGEX but I donot really understand it. Please help.
You may use
sed 's/\([0-9][0-9,.]\+\)-\($\|[^0-9]\)/-\1\2/g'
See the online demo
The point is that after matching a number and a - (see \([0-9][0-9,.]\+\)-), there should come either end of string or non-digit (\($\|[^0-9]\)). Thus, we have 2 capturing groups now, and that is why we need a second backreference in the replacement pattern (\2).
I added a dot . to the bracket expression just in case you have mixed number formats, you may remove it if you always have a comma as the decimal separator.
Pattern details:
\([0-9][0-9,.]\+\) - Group 1 capturing
[0-9] - a digit
[0-9,.]\+ - one or more digits, commas or dots
- - a literal hyphen
\($\|[^0-9]\) - Group 2 capturing the end of string $ or a non-digit ([^0-9])
In your example, both files are identical, but I think I know what you mean.
For this particular file, you want to match a space, followed by zero or more digits, followed by a comma, followed by at least one digit, followed by a dash,
followed by zero or more spaces to the end of the line.
Then you want to replace the space in front of the matched digits and the comma with a dash. This will do the trick:
sed -e 's/ \([0-9]*,[0-9][0-9]*\)- *$/-\1/' <file.txt >file1.txt
Your first regular expression attempts to match against a string of numbers and .s, but the text contains a comma, not a .. It does the substitution you want if you replace [0-9.] with [0-9,], giving:
sed 's/\([0-9,]\+\)-/-\1/g' file.txt > file1.txt
However, it also replaces 25-07 in that case with -2507. I suggest you explicitly match against the end of the line:
sed 's/\([0-9,]\+\)-$/-\1/g'
or alternatively, you can demand that the match contains exactly one comma:
sed 's/\([0-9]\+,[0-9]\+\)-$/-\1/g'
I also find these things easier to read if you use the -r option to sed, which enables "extended regular expressions":
sed -r 's/([0-9]+,[0-9]+)-$/-\1/g'
Fewer special characters need to be escaped (on the other hand, more literal characters need to be escaped, but I find that tends to be a rarer occurrence).
(Aside: note that . usually means "any character", but inside a character class [.] it means "literally a .", since after all having it mean "any character" in there would be pretty useless.)
I am attempting to replace a date in the format 08/09/2014 but at the same time also the format 8/9/14 using sed. I know the + sign is supposed to match one or more occurrences, and ? 0 or more. I've tried both but none of the dates are being replaced with "testing". I was expecting this would find 1 or more digits followed by a slash, 1 or more digits followed by a slash, 4 digits.
Do I need to escape the special character, or what is wrong here?
sed -f mySed.sed dates.csv
# mySed.sed file
s#[0-9]+/[0-9]+/[0-9][0-9][0-9][0-9]#testing#g
# sample line in dates.csv
...,20/01/2001,2/1/2009,...
You have made several mistakes. Here is a working example:
echo '20/01/2001,2/1/2009' | sed 's~[0-9]\{1,2\}/[0-9]\{1,2\}/\([0-9]\{2\}\)\{1,2\}~toto~g'
Note that the ? means "optional" (in other words 0 or 1 time) and must be escaped.
To be more precise, I have choosen to use this quantifier {m,n} instead of +. But if you use + don't forget to escape it \+ otherwise it will be seen as a literal character.
You need to escape the + quantifiers in your regular expression, and you can use a range for the last set.
s#[0-9]\+/[0-9]\+/[0-9]\{2,4\}#testing#g
Or you can use the range quantifier throughout your pattern.
s#[0-9]\{1,2\}/[0-9]\{1,2\}/[0-9]\{2,4\}#testing#g
I have a regular expression that I'm going to be using to verify that an inputted number is in standard U.S. telephone format (i.e (###) ###-####). I am new to regex and still having some trouble figuring out the exact function of each character. If someone would go through this piece by piece/verify that I am understanding I would really appreciate it. Also if the regex is wrong I would obviously like to know that.
\D*?(\d\D*?){10}
What I think is happening:
\D*?( indicates an escape sequence for the parenthesis metacharacter... not sure why the \D*? is necessary
\d indicating digits
\D*? indicating there is a non-digit character (-) followed by the closing parenthesis.
{10} for the 10 digits
I feel very unsure explaining this, like my understanding is very vague in terms of why the regex is in the order that it is etc. Thanks in advance for help/explanations.
EDIT
It seems like this is not the best regex for what I want. Another possibility was [(][0-9]{3}[)] [0-9]{3}-[0-9]{4}, but I was told this would fail. I suppose I'll have to do a little more work with regular expressions to figure this out.
\D matches any non-digit character.
* means that the previous character is repeated 0 or more times.
*? means that the previous character is repeated 0 or more times, but until the match of the following character in the regex. It is a bit difficult perhaps at the start, but in your regex, the next character is \d, meaning \D*? will match the least amount of characters until the next \d character.
( ... ) is a capture group, and is also used to group things. For instance {10} means that the previous character or group is repeated 10 times exactly.
Now, \D*?(\d\D*?){10} will match exactly 10 numbers, starting with non-digit characters or not, with non-digit characters in between the digits if they are present.
[(][0-9]{3}[)] [0-9]{3}-[0-9]{4}
This regex is a bit better since it doesn't just accept anything (like the first regex does) and will match the format (###) ###-#### (notice the space is a character in regex!).
The new things introduced here are the square brackets. These represent character classes. [0-9] means any character between 0 to 9 inclusive, which means it will match 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. Adding {3} after it makes it match 3 similar character class, and since this character class contains only digits, it will match exactly 3 digits.
A character class can be used to escape certain characters, such as ( or ) (note I mentioned earlier they are for capturing groups, or grouping) and thus, [(] and [)] are literal ( and ) instead of being used for capturing/grouping.
You can also use backslashes (\) to escape characters. Thus:
\([0-9]{3}\) [0-9]{3}-[0-9]{4}
Will also work. I would also recommend the use of line anchors ^ and $ if you're only trying to see if a phone number matches the above format. This ensures that the string has only the phone number, and nothing else. ^ matches the beginning of a line and $ matches the end of a line. Thus, the regex will become:
^\([0-9]{3}\) [0-9]{3}-[0-9]{4}$
However, I don't know all the combinations of the different formats of phone numbers in the US, so this regex might need some tweaking if you have different phone number formats.
\D is "not a digit"; \d is "digit". With that in mind:
This matches zero or more non-digits, then it matches a digit and any number of non-digit characters 10 times. This won't actually verify that the number if formatted properly, just that it contains 10 digits. I suspect that the regex isn't what you want in the first place.
For example, the following will match your regex:
this is some bad text 1 and some more 2 and more 34567890
\D matches a character that is not a digit
* repeats the previous item 0 or more times
? find the first occurrence
\d matches a digit
so your group is matches 10 digits or non digits
I need to find the text of all the one-digit number.
My code:
$string = 'text 4 78 text 558 my.name#gmail.com 5 text 78998 text';
$pattern = '/ [\d]{1} /';
(result: 4 and 5)
Everything works perfectly, just wanted to ask it is correct to use spaces?
Maybe there is some other way to distinguish one-digit number.
Thanks
First of all, [\d]{1} is equivalent to \d.
As for your question, it would be better to use a zero width assertion like a lookbehind/lookahead or word boundary (\b). Otherwise you will not match consecutive single digits because the leading space of the second digit will be matched as the trailing space of the first digit (and overlapping matches won't be found).
Here is how I would write this:
(?<!\S)\d(?!\S)
This means "match a digit only if there is not a non-whitespace character before it, and there is not a non-whitespace character after it".
I used the double negative like (?!\S) instead of (?=\s) so that you will also match single digits that are at the beginning or end of the string.
I prefer this over \b\d\b for your example because it looks like you really only want to match when the digit is surrounded by spaces, and \b\d\b would match the 4 and the 5 in a string like 192.168.4.5
To allow punctuation at the end, you could use the following:
(?<!\S)\d(?![^\s.,?!])
Add any additional punctuation characters that you want to allow after the digit to the character class (inside of the square brackets, but make sure it is after the ^).
Use word boundaries. Note that the range quantifier {1} (a single \d will only match one digit) and the character class [] is redundant because it only consists of one character.
\b\d\b
Search around word boundaries:
\b\d\b
As explained by the others, this will extract single digits meaning that some special characters might not be respected like "." in an ip address. To address that, see F.J and Mike Brant's answer(s).
It really depends on where the numbers can appear and whether you care if they are adjacent to other characters (like . at the end of a sentence). At the very least, I would use word boundaries so that you can get numbers at the beginning and end of the input string:
$pattern = '/\b\d\b/';
But you might consider punctuation at the end like:
$pattern = '/\b\d(\b|\.|\?|\!)/';
If one-digit numbers can be preceded or followed by characters other than digits (e.g., "a1 cat" or "Call agent 7, pronto!") use
(?<!\d)\d(?!\d)
Demo
The regular expression reads, match a digit (\d) that is neither preceded nor followed by digit, (?<!\d) being a negative lookbehind and (?!\d) being a negative lookahead.