I have a one line string that looks like this:
{"took":125,"timed_out":false,"_shards":{"total":10,"successful":10,"skipped":0,"failed":0}}{"took":365,"timed_out":false,"_shards":{"total":10,"successful":10,"skipped":0,"failed":0}}{"took":15,"timed_out":false,"_shards":{"total":10,"successful":10,"skipped":0,"failed":0}}
I would like to extract all the numbers after the "took" part, so in my case the output would look like this:
125
365
15
What I've tried so far is using took":(\d{1,6}),"(.*) as a regex. But since its a one line string, it only extracts the first occurence and ignores the others.
You can use
Find What: took":(\d+)|(?s)(?:(?!took":\d).)*
Replace With: (?{1}$1\n)
Details:
took": - literal text
(\d+) - one or more digits captured into Group 1
| - or
(?s) - set the DOTALL mode on (. matches line break chars now)
(?:(?!took":\d).)* - any single char, zero or more times, as many as possible, that does not start a took": + digit char sequence.
The (?{1}$1\n) conditional replacement pattern replaces this way:
(?{1} - if Group 1 is matched
$1\n - replace the match with Group 1 and a newline
) - else, replace with an empty string.
Related
Hi everyone i would like to extract the following Text. I have provided my regular expression below but its not going according to the output I want.
Output I want:
Extract Title
Extract 2nd line below title if there is, extract it. If not, move on.
Extract the Address (Only for address: regardless new line or not)
Regular expression:
/(.+?)\s*(\d+.*Singapore\s+\d{6}\b|\d+.*S\d{6})\b/g
/(^.+\n)(^.+\n)?(^\d+.*\sSingapore,?\s\d{6})/gm
(^.+\n) - capture title
() - defines capture group
^ - matches beginning of the line
.+ - matches 1 or more character
\n - matches new line
(^.+\n)? - capture 2nd line
? - matches the group 0 or 1 times (since this line is optional)
(^\d+.*\sSingapore,?\s\d{6}) - capture address
\d+ - matches 1 or more digit
.* - matches any character 0 or more times (maybe you need to modify it to be required)
\s - matches a whitespace
Singapore - matches the word Singapore
,? - matches a comma 0 or 1 times (remove ? if comma is required)
\s - matches a whitespace
\d{6} - matches 6 digits
gm
g - global flag, allows you to find multiple matches in the text. Only needed if your text contains more than one set of title/description/address.
m - multiline flag, looks for matches in the whole text, not in a single line.
I have a pipe delimited file which has a line
H||CUSTCHQH2H||PHPCCIPHP|1010032000|28092017|25001853||||
I want to substitute the date (28092017) with a regex "[0-9]{8}" if the first character is "H"
I tried the following example to test my understanding where Im trying to subtitute "a" with "i".
str = "|123||a|"
str.gsub /\|(.*?)\|(.*?)\|(.*?)\|/, "\|\\1\|\|\\1\|i\|"
But this is giving o/p as
"|123||123|i|"
Any clue how this can be achieved?
You may replace the first occurrence of 8 digits inside pipes if a string starts with H using
s = "H||CUSTCHQH2H||PHPCCIPHP|1010032000|28092017|25001853||||"
p s.gsub(/\A(H.*?\|)[0-9]{8}(?=\|)/, '\100000000')
# or
p s.gsub(/\AH.*?\|\K[0-9]{8}(?=\|)/, '00000000')
See the Ruby demo. Here, the value is replaced with 8 zeros.
Pattern details
\A - start of string (^ is the start of a line in Ruby)
(H.*?\|) - Capturing group 1 (you do not need it when using the variation with \K): H and then any 0+ chars as few as possible
\K - match reset operator that discards the text matched so far
[0-9]{8} - eight digits
(?=\|) - the next char must be |, but it is not added to the match value since it is a positive lookahead that does not consume text.
The \1 in the first gsub is a replacement backreference to the value in Group 1.
I have this two lines of text, that I want to manipulate using Regular Expression and substitute:
Obj.FieldNameA = Reader.GetEnumFromInt32<ClassName>(QueryGenerator,nameof(Obj.));
Obj.FieldNameB=Reader.GetTrimmedStringOrNull(QueryGenerator,nameof(Obj.));
Attached on the first Obj. there is a Field name, so in this case they are FieldNameA,FieldNameB
I want to attach these values to the second Obj. found on the same line, so the text should become:
Obj.FieldNameA = Reader.GetEnumFromInt32<ClassName>(QueryGenerator,nameof(Obj.FieldNameA));
Obj.FieldNameB=Reader.GetTrimmedStringOrNull(QueryGenerator,nameof(Obj.FieldNameB));
I have tested this very simple (and wrong) regex:
Obj\.(\w*).*\n
With substituition as $1
But I don't know how to use substitution...
Sample code here
Some Notes:
After FieldNameA there is always an equal sign that could be preceded or followed by a space.
Before the second Obj. there could be any character, including < ( etc...
Could this be achieved?
You may use
Find: (Obj\.(\w+).*\(Obj\.)\)
Replace: $1$2)
See the regex demo.
You may also add ^ to the start of the regex to match only at the start of a line/string.
Details
^ - start of string
(Obj\.(\w+).*\(Obj\.) - Group 1 ($1 in the replacement):
Obj\. - Obj. text
(\w+) - Group 2 ($2): 1 or more word chars
.* - any 0+ chars other than line break chars as many as possible (you may use .*? to only match the second Obj. on a line, your current input only has two with the second one closer to the end of a line, so .* will work better)
\(Obj\. - (Obj. text
\) - a ) char.
<.*>|\n.*\s.*\sid="(\w*)".*\n+|.*>\n|\n.+
and replace $1
This regex can take all id out from file
<a href="java" class="total" id="maker" placeholder="getTheResult('local6')">master6<a>
Result is maker
How can I extract getTheResult key name?
so my result will be local6
Tried <.*>|\n.*\s.*\sgetTheResult('(\w*)').*\n+|.*>\n|\n.+ but didn't helped
I assume that:
you have files with text like getTheResult('local6')
you may have several values like that on a line
you'd like to keep those text only, one value per line.
I suggest
getTheResult\('([^']*)'\)|(?:(?!getTheResult\(')[\s\S])*
and replace with $1\n. The \n will insert a newline between the values. You can then use ^\n regex (to replace with empty string) to remove empty lines.
Pattern details:
getTheResult\(' - matches getTheResult(' as a literal string (note the ( is escaped)
([^']*) - Group 1 capturing 0+ chars other than '
'\) - a literal ')
| - or
(?:(?!getTheResult\(')[\s\S])* - 0+ chars that are not starting chars of the getTheResult(' character sequence (this is a tempered greedy token).
I have the some data and i'd like to convert it into a table format.
Here's the input data
1- This is the 1st line with a
newline character
2- This is the 2nd line
Each line may contain multiple newline characters.
Output
<td>1- This the 1st line with
a new line character</td>
<td>2- This is the 2nd line</td>
I've tried the following
^(\d{1,3}-)[^\d]*
but it seems to match only till the digit 1 in 1st.
I'd like to be able to stop matching after i find another \d{1,3}\- in my string.
Any suggestions?
EDIT:
I'm using EditPad Lite.
This is for vim, and uses zerowidth positive-lookahead:
/^\d\{1,3\}-\_.*[\r\n]\(\d\{1,3\}-\)\#=
Steps:
/^\d\{1,3\}- 1 to 3 digits followed by -
\_.* any number of characters including newlines/linefeeds
[\r\n]\(\d\{1,3\}-\)\#= followed by a newline/linefeed ONLY if it is followed
by 1 to 3 digits followed by - (the first condition)
EDIT: This is how it would be in pcre/ruby:
/(\d{1,3}-.*?[\r\n])(?=(?:\d{1,3}-)|\Z)/m
Note you need a string ending with a newline to match the last entry.
SEARCH: ^\d+-.*(?:[\r\n]++(?!\d+-).*)*
REPLACE: <td>$0</td>
[\r\n]++ matches one or more carriage-returns or linefeeds, so you don't have to worry about whether the file use Unix (\n), DOS (\r\n), or older Mac (\r) line separators.
(?!\d+-) asserts that the first thing after the line separator is not another line number.
I used the possessive + in [\r\n]++ to make sure it matches the whole separator. Otherwise, if the separator is \r\n, [\r\n]+ could match the \r and (?!\d+-) could match the \n.
Tested in EditPad Pro, but it should work in Lite as well.
You did not specify a language (there are many regexp implementations), but in general, what you are looking for is called "positive lookahead", which lets you add patterns that will influence the match, but will not become part of it.
Search for lookahead in the documentation of whatever language you are using.
Edit: the following sample seems to work in vim.
:%s#\v(^\d+-\_.{-})\ze(\n\d+-|%$)#<td>\1</td>
Annotation below:
% - for all lines
s# - substitute the following (you can use any delimiter, and slash is most
common, but as that will require that we escape slashes in the command
I chose to use the number sign)
\v - very magic mode, let's us use less backslashes
( - start group for back referencing
^ - start of line
\d+ - one or more digits (as many as possible)
- - a literal dash!
\_. - any character, including a newline
{-} - zero or more of these (as few as possible)
) - end group
\ze - end match (anything beyond this point will not be included in the match)
( - start a new group
[\n\r] - newline (in any format - thanks Alan)
\d+ - one or more digits
- - a dash
| - or
%$ - end of file
) - end group
# - start substitute string
<td>\1</td> - a TD tag around the first matched group
(\d+-.+(\r|$)((?!^\d-).+(\r|$))?)
You can match only the separators and split on them. In C#, for example, it could be done like this:
string s = "1- This is the 1st line with a \r\nnewline character\r\n2- This is the 2nd line";
string ss = "<td>" + string.Join("</td>\r\n<td>", Regex.Split(s.Substring(3), "\r\n\\d{1,3}- ")) + "</td>";
MessageBox.Show(ss);
Would it be good for you to do it in 3 steps?
(these are perl regex):
Replace the first:
$input =~ s/^(\d{1,3})/<td>\1/;
Replace the rest
$input =~ s/\n(\d{1,3})/<\/td>\n<td>\1/gm;
Add the last:
$input .= '</td>';