I want replace a Letter with a literal $. I tried:
var s = string.replaceAll("Register","$10")
I want that this text Register saved to be changed to: $10 saved
Illegal group reference is the error I get.
If you look at the scaladoc for replaceAll, you'll see that it takes a regular expression string as the parameter. Escape the $ with a \, or use replaceAllLiterally
replaceAll uses a regular expressions to find the match. In the replacement string $ is a special character that refers to a specific capture group in the matching string. You have no capture groups so this is an error. It's not what you want anyway since you want the literal text "$10".
Usereplaceinstead ofreplaceAll`. It just does a direct string replacement.
Related
I have the following line:
>XXX-220_5004_COVID-A6
TTTATTTGACATGAGTAAATTTCCCCTTAAATTAAGGGGTACTGCTGTTATGTCTTTAAA
AGAAGGTCAAATCAATGATATGATTTTATCTCTTCTTAGTAAAGGTAGACTTATAATTAG
AGAAAACAAC
I would like to convert the first line as follows:
>INITWORD/XXX-220_5004_COVID-A6/FINALWORD
TTTATTTGACATGAGTAAATTTCCCCTTAAATTAAGGGGTACTGCTGTTATGTCTTTAAA
AGAAGGT...
So far I have managed to add the first word as follows:
sed 's/>/>INITTWORD\//I'
That returns:
>INITWORD/XXX-220_5004_COVID-A6
TTTATTTGACATGAGTAAATTTCCCCTTAAATTAAGGGGTACTGCTGTTATGTCTTTAAA
AGAAGGT
How can i add the FINALWORD at the end of the first line?
Just substitute more. sed conveniently allows you to recall the text you matched with a back reference, so just embed that between the things you want to add.
sed 's%^>\(.*\)%>INITWORD/\1/FINALWORD%I' file.fasta
I also added a ^ beginning-of-line anchor, and switched to % delimiters so the slashes don't need to be escaped.
In some more detail, the s command's syntax is s/regex/replacement/flags where regex is a regular expression to match the text you want to replace, and replacement is the text to replace it with. In the regex, you can use grouping parentheses \(...\) to extract some of the matched text into the replacement; so \1 refers to whatever matched the first set of grouping parentheses, \2 to the second, etc. The /flags are optional single-character specifiers which modify the behavior of the command; so for example, a /g flag says to replace every match on a line, instead of just the first one (but we only expect one match per line so it's not necessary or useful here).
The I flag is non-standard but since you are using that, I assume it does something useful for you.
This appears to be a bug in fail2ban, with different behaviour between the fail2ban-regex tool and a failregex filter
I am attempting to develop a new regex rule for fail2ban, to match:
\"%20and%20\"x\"%3D\"x
When using fail2ban-regex, this appears to produce the desired result:
^<HOST>.*GET.*\\"%20and%20\\"x\\"%3D\\"x.* 200.*$
As does this:
^<HOST>.*GET.*\\\"%20and%20\\\"x\\\"%3D\\\"x.* 200.*$
However, when I put either of these into a filter, I get the following error:
Failed during configuration: '%' must be followed by '%' or '(', found:…
To have this work in a filter you have to double-up the ‘%’, ie ‘%%’:
^<HOST>.*GET.*\\\"%%20and%%20\\\"x\\\"%%3D\\\"x.* 200.*$
While this gets the required hits running as a filter, it gets none running through fail2ban-regex.
I tried the \\\\ as Andre suggested below, but this gets no results in fail2ban-regex.
So, as this appears to be differential behaviour, I am going to file it as a bug.
According to Python's own site a singe backslash "\" has to be written as "\\\\" and there's no mention of %.
Regular expressions use the backslash character ('') to indicate
special forms or to allow special characters to be used without
invoking their special meaning. This collides with Python’s usage of
the same character for the same purpose in string literals; for
example, to match a literal backslash, one might have to write '\\'
as the pattern string, because the regular expression must be \, and
each backslash must be expressed as \ inside a regular Python string
literal
I would just go with:
failregex = (?i)^<HOST> -.*"(GET|POST|HEAD|PUT).*20and.*3d.*$
the .* wil match anything inbetween anyways and (?i) makes the entire regex case-insensitive
I have a database entry that has entries that look like this:
id | name | code_set_id
I have this particular entry that I need to find:
674272310 | raphodo/qrc_resources.py | 782732
In my rails app (2.3.8), I have a statement that evaluates to this:
SELECT * from fyles WHERE code_set_id = 782732 AND name LIKE 'raphodo/qrc\\_resources.py%';
From reading up on escaping, the above query is correct. This is supposed to correctly double escape the underscore. However this query does not find the record in the database. These queries will:
SELECT * from fyles WHERE code_set_id = 782732 AND name LIKE 'raphodo/qrc\_resources.py%';
SELECT * from fyles WHERE code_set_id = 782732 AND name LIKE 'raphodo/qrc_resources.py%';
Am I missing something here? Why is the first SQL statement not finding the correct entry?
A single backslash in the RHS of a LIKE escapes the following character:
9.7.1. LIKE
[...]
To match a literal underscore or percent sign without matching other characters, the respective character in pattern must be preceded by the escape character. The default escape character is the backslash but a different one can be selected by using the ESCAPE clause. To match the escape character itself, write two escape characters.
So this is a literal underscore in a LIKE pattern:
\_
and this is a single backslash followed by an "any character" pattern:
\\_
You want LIKE to see this:
raphodo/qrc\_resources.py%
PostgreSQL used to interpret C-stye backslash escapes in strings by default but no longer, now you have to use E'...' to use backslash escapes in string literals (unless you've changed the configuration options). The String Constants with C-style Escapes section of the manual covers this but the simple version is that these two:
name LIKE E'raphodo/qrc\\_resources.py%'
name LIKE 'raphodo/qrc\_resources.py%'
do the same thing as of PostgreSQL 9.1.
Presumably your Rails 2.3.8 app (or whatever is preparing your LIKE patterns) is assuming an older version of PostgreSQL than the one you're actually using. You'll need to adjust things to not double your backslashes (or prefix the pattern string literals with Es).
I have tried [0-9] and checked the use wildcard box but it replaces the individual numbers with the literal [0-9] string. How do I replace with the number it found plus a character?
Backreferences. Your unspecified environment may or may not support them, but if it does, you would:
replace \([0-9]*\)
with \1 <then, whatever the character you want is>
please provide me a sed oneliner which provides this output:
sdc3 sdc2
for Input :
sdc3[1] sdc2[0]
I mean remove all subscript value from the string ..
sed 's/\[[^]]*\]//g'
reads: substitute any string with literal "[" followed by zero or more characters that aren't a "]", and then the closing "]", with an empty string.
You need the [^]] bit to prevent greedy matching treating "[1] sdc2[0]" as a single match in your sample string.
As for your comment:
sed 's#\([^[ ]*\)\[[^]]*\]#/dev/\1#g'
I switch the seperator from the usual '/' to '#', just to avoid escaping the /dev/ bit you asked for (I won't say "for clarity")
the \(...\) bit matches a subgroup, here sdc2 or whatever, so we can refer to it in the replacement
the subgroup uses a similar character class to the one we used discarding the index: [^[ ] means any character except an "[" (again, to avoid greedily matching the index) or a space (assuming your values are space-delimited as per your post)
the replacement is now the literal "/dev/" followed by the first (and only) subgroup match
the g flag at the end tells it to perform multiple matches per line, instead of stopping at the first one