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I am trying to translate a huge mySQL database dump file from mySQL syntax into SQLite syntax.
At https://regex101.com/ I have successfully created a ECMAScript flavor regex to turn something like:
,'foo\'s bar!',
into:
,"foo\'s bar!"
with this regular expression:
/,'([^']+)\\'([^']+)',/"$1\\'$2"/g
testing against this short file:
(1058,'gpl5q0x51349lmdq3e0ijm4k9b6n','Henry\'s_1.csv','text/csv','{\"identified\":true,\"analyzed\":true}',33854,'mUVk0/XGX+afIpkrqBm7LQ==','2021-01-06 03:07:23'),
(1059,'xzj8mivsenkakkrurfjytxjsaj1h','Henry\'s_2.csv','text/csv','{\"identified\":true,\"analyzed\":true}',33555,'KfRYqfAWtSIYXZ6oQZyYbA==','2021-01-06 03:07:23'),
Resulting in:
(1058,'gpl5q0x51349lmdq3e0ijm4k9b6n'"Henry\'s_1.csv"'text/csv','{\"identified\":true,\"analyzed\":true}',33854,'mUVk0/XGX+afIpkrqBm7LQ==','2021-01-06 03:07:23'),
(1059,'xzj8mivsenkakkrurfjytxjsaj1h'"Henry\'s_2.csv"'text/csv','{\"identified\":true,\"analyzed\":true}',33555,'KfRYqfAWtSIYXZ6oQZyYbA==','2021-01-06 03:07:23'),
but for the life of me I cannot translate this into a GNU sed flavor regex.
For example, this command does not make any substitutions in the output:
sed -r s/,'([^']+)\\'([^']+)',/"$1\\'$2"/g <test.sql
...
sed -r s/,'([^']+)\\'([^']+)',/"\1\\'\2"/g <test.sql: doesn't work either.
I have looked for a regex tool online that translates between different flavors of regex but cannot find one that works on GNU sed (shipped with GIT: sed (GNU sed) 4.8). PCRE seems to be close to what sed has but that doesn't work. I tried perl as well, no luck.
Anyone know a regex expression that works or a translator tool that works?
I am just about ready to write a nodejs program to do this for me.
Also, for extra credit, how can I write a sed script to handle any number of escaped quotes within a quoted string? I have that issue to deal with as well in my DB dump file.
Examples:
'foo\'-bar' // on instance
'foo\'and\'bar' // two instances
'foo\'and\'bar\'s on the deck' // three instances
and so on...
Thanks!
You can use
sed -E "s/,'([^']+)\\\\'([^']+)',/"'"'"\\1\\\\'\\2"'"'/g test.sql
The "s/,'([^']+)\\\\'([^']+)',/"'"'"\\1\\\\'\\2"'"'/g consists of
"s/,'([^']+)\\\\'([^']+)',/" - a s/,'([^']+)\\'([^']+)',/ part (inside double quotes, so backslashes need doubling)
'"' - a " char (inside single quotes)
"\\1\\\\'\\2" - \1\\'\2 pattern (inside double quotes, so backslashes are doubled)
'"' - a " char (inside single quotes)
/g - the global flag (no need quoting here).
First look at your command
sed -r s/,'([^']+)\\'([^']+)',/"\1\\'\2"/g test.sql
I prefer writing the whole sed command in single quotes. When you need a single quote, you must close the string ('), use an escaped single quote (\') and open the next string with a ', all joined: '\''.
I also added two , characters.
sed -r 's/,'\''([^'\'']+)\\'\''([^'\'']+)'\'',/,"\1\\'\''\2",/g' test.sql
# Shorter
sed -r 's/,'\''([^'\'']+\\'\''[^'\'']+)'\'',/,"\1",/g' test.sql
# Using another way to write the single quotes, with the hex notation
sed -r 's/,\x27([^\x27]+\\\x27[^\x27]+)\x27,/,"\1",/g' test.sql
This works for simple cases, not for 'foo\'and\'bar\'s on the deck'.
I think you want to replace the quotes in the simple fields too.
Suppose you want to transform
(1058,'gpl5q0x51349lmdq3e0ijm4k9b6n','Henry\'s_1.csv','text/csv','{\"identified\":true,\"analyzed\":true}',33854,'mUVk0/XGX+afIpkrqBm7LQ==','2021-01-06 03:07:23'),
(1059,'xzj8mivsenkakkrurfjytxjsaj1h','Henry\'s_2.csv','text/csv','{\"identified\":true,\"analyzed\":true}',33555,'KfRYqfAWtSIYXZ6oQZyYbA==','2021-01-06 03:07:23'),
(2000,'extra credit from question','foo\'and\'bar\'s on the deck','text/csv','{\"identified\":true,\"analyzed\":true}',33999,'KgSBFstbdthdsssssstvbA==','2022-01-02 13:07:23'),
into
(1058,"gpl5q0x51349lmdq3e0ijm4k9b6n","Henry\'s_1.csv","text/csv","{\"identified\":true,\"analyzed\":true}",33854,"mUVk0/XGX+afIpkrqBm7LQ==","2021-01-06 03:07:23"),
(1059,"xzj8mivsenkakkrurfjytxjsaj1h","Henry\'s_2.csv","text/csv","{\"identified\":true,\"analyzed\":true}",33555,"KfRYqfAWtSIYXZ6oQZyYbA==","2021-01-06 03:07:23"),
(2000,"extra credit from question","foo\'and\'bar\'s on the deck","text/csv","{\"identified\":true,\"analyzed\":true}",33999,"KgSBFstbdthdsssssstvbA==","2022-01-02 13:07:23"),
In this answer I don't use the '\'' but the hexadecimal notation \x27.
First "backup" the \' combinations (replace them by an unused character like \r), replace all normal quotes by double quotes and "restore the backup" (change back the \r).
sed 's/\\\x27/\r/g; s/\x27/"/g; s/\r/\\\x27/g' test.sql
# or hex value for double quote "
sed 's/\\\x27/\r/g; s/\x27/\x22/g; s/\r/\\\x27/g' test.sql
I know this type of search has been address in a few other questions here, but for some reason I can not get it to work in my scenario.
I have a text file that contains something similar to the following patter:
some text here done
12345678_123456 226-
more text
some more text here done
12345678_234567 226-
I'm trying to find all cases where done is followed by 226- on the next line, with the 16 characters proceeding. I tried grep -Pzo and pcregrep -M but all return nothing.
I attempted multiple combinations of regex to take in account the 2 lines and the 16 chars in between. This is one of the examples I tried with grep:
grep -Pzo '(?s)done\n.\{16\}226-' filename
Related posts:
How to find patterns across multiple lines using grep?
Regex (grep) for multi-line search needed [duplicate]
How can I search for a multiline pattern in a file?
Generalize it to this (?m)done$\s+.*226-$
Because requiring a \n after 226- at end of string is a bad thing.
And not requiring a \n after 226- is also a bad thing.
Thus, the paradox is solved with (\n|$) but why the \n at all?
Both problems solved with multiline and $.
https://regex101.com/r/A33cj5/1
You must not escape { and } while using -P (PCRE) option in grep. That escaping is only for BRE.
You can use:
grep -ozP 'done\R.{16}226-\R' file
done
12345678_123456 226-
done
12345678_234567 226-
\R will match any unicode newline character. If you are only dealing with \n then you may just use:
grep -ozP 'done\n.{16}226-\n' file
I am attempting to parse (with sed) just First Last from the following DN(s) returned by the DSCL command in OSX terminal bash environment...
CN=First Last,OU=PCS,OU=guests,DC=domain,DC=edu
I have tried multiple regexs from this site and others with questions very close to what I wanted... mainly this question... I have tried following the advice to the best of my ability (I don't necessarily consider myself a newbie...but definitely a newbie to regex..)
DSCL returns a list of DNs, and I would like to only have First Last printed to a text file. I have attempted using sed, but I can't seem to get the correct function. I am open to other commands to parse the output. Every line begins with CN= and then there is a comma between Last and OU=.
Thank you very much for your help!
I think all of the regular expression answers provided so far are buggy, insofar as they do not properly handle quoted ',' characters in the common name. For example, consider a distinguishedName like:
CN=Doe\, John,CN=Users,DC=example,DC=local
Better to use a real library able to parse the components of a distinguishedName. If you're looking for something quick on the command line, try piping your DN to a command like this:
echo "CN=Doe\, John,CN=Users,DC=activedir,DC=local" | python -c 'import ldap; import sys; print ldap.dn.explode_dn(sys.stdin.read().strip(), notypes=1)[0]'
(depends on having the python-ldap library installed). You could cook up something similar with PHP's built-in ldap_explode_dn() function.
Two cut commands is probably the simplest (although not necessarily the best):
DSCL | cut -d, -f1 | cut -d= -f2
First, split the output from DSCL on commas and print the first field ("CN=First Last"); then split that on equal signs and print the second field.
Using sed:
sed 's/^CN=\([^,]*\).*/\1/' input_file
^ matches start of line
CN= literal string match
\([^,]*\) everything until a comma
.* rest
http://www.gnu.org/software/gawk/manual/gawk.html#Field-Separators
awk -v RS=',' -v FS='=' '$1=="CN"{print $2}' foo.txt
I like awk too, so I print the substring from the fourth char:
DSCL | awk '{FS=","}; {print substr($1,4)}' > filterednames.txt
This regex will parse a distinguished name, giving name and val a capture groups for each match.
When DN strings contain commas, they are meant to be quoted - this regex correctly handles both quoted and unquotes strings, and also handles escaped quotes in quoted strings:
(?:^|,\s?)(?:(?<name>[A-Z]+)=(?<val>"(?:[^"]|"")+"|[^,]+))+
Here is is nicely formatted:
(?:^|,\s?)
(?:
(?<name>[A-Z]+)=
(?<val>"(?:[^"]|"")+"|[^,]+)
)+
Here's a link so you can see it in action:
https://regex101.com/r/zfZX3f/2
If you want a regex to get only the CN, then this adapted version will do it:
(?:^|,\s?)(?:CN=(?<val>"(?:[^"]|"")+"|[^,]+))
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input
This question already has answers here:
How do I match any character across multiple lines in a regular expression?
(26 answers)
Closed 3 years ago.
Really basic question here. So I'm told that a dot . matches any character EXCEPT a line break. I'm looking for something that matches any character, including line breaks.
All I want to do is to capture all the text in a website page between two specific strings, stripping the header and the footer. Something like HEADER TEXT(.+)FOOTER TEXT and then extract what's in the parentheses, but I can't find a way to include all text AND line breaks between header and footer, does this make sense? Thanks in advance!
When I need to match several characters, including line breaks, I do:
[\s\S]*?
Note I'm using a non-greedy pattern
You could do it with Perl:
$ perl -ne 'print if /HEADER TEXT/ .. /FOOTER TEXT/' file.html
To print only the text between the delimiters, use
$ perl -000 -lne 'print $1 while /HEADER TEXT(.+?)FOOTER TEXT/sg' file.html
The /s switch makes the regular expression matcher treat the entire string as a single line, which means dot matches newlines, and /g means match as many times as possible.
The examples above assume you're cranking on HTML files on the local disk. If you need to fetch them first, use get from LWP::Simple:
$ perl -MLWP::Simple -le '$_ = get "http://stackoverflow.com";
print $1 while m!<head>(.+?)</head>!sg'
Please note that parsing HTML with regular expressions as above does not work in the general case! If you're working on a quick-and-dirty scanner, fine, but for an application that needs to be more robust, use a real parser.
By definition, grep looks for lines which match; it reads a line, sees whether it matches, and prints the line.
One possible way to do what you want is with sed:
sed -n '/HEADER TEXT/,/FOOTER TEXT/p' "$#"
This prints from the first line that matches 'HEADER TEXT' to the first line that matches 'FOOTER TEXT', and then iterates; the '-n' stops the default 'print each line' operation. This won't work well if the header and footer text appear on the same line.
To do what you want, I'd probably use perl (but you could use Python if you prefer). I'd consider slurping the whole file, and then use a suitably qualified regex to find the matching portions of the file. However, the Perl one-liner given by '#gbacon' is an almost exact transliteration into Perl of the 'sed' script above and is neater than slurping.
The man page of grep says:
grep, egrep, fgrep, rgrep - print lines matching a pattern
grep is not made for matching more than a single line. You should try to solve this task with perl or awk.
As this is tagged with 'bbedit' and BBedit supports Perl-Style Pattern Modifiers you can allow the dot to match linebreaks with the switch (?s)
(?s).
will match ANY character. And yes,
(?s).+
will match the whole text.
As pointed elsewhere, grep will work for single line stuff.
For multiple-lines (in ruby with Regexp::MULTILINE, or in python, awk, sed, whatever), "\s" should also capture line breaks, so
HEADER TEXT(.*\s*)FOOTER TEXT
might work ...
here's one way to do it with gawk, if you have it
awk -vRS="FOOTER" '/HEADER/{gsub(/.*HEADER/,"");print}' file