What is the escape character for # in execute command activity? Well i was trying to replace one string in file with "#",but datastage treating # as job parameter and expecting the value to be assigned in datastage parameters.For that we need escape character for "#". I tried using \ and / as escape characters but none of them solved my problem. Thank You.
Just wanted to add that the actual answer to this question is:
\043
The expression mentioned previously (!/^\043/) is related to the first link #Damienknight posted.
Use awk to replace the # sign, and you can use the octal character code for # in the regular expression:
!/^\043/
There is a thread in DSXchange that discusses this.
Here is a guide on escape sequences in AWK expressions.
A table with ascii codes, including a column for octal values.
Related
I'm trying to replace below specific lines in a file
/ACCOUNT/passwd=
/BMC/CONFIRMATION/PASSWORD=
I need help in preparing the sed command
The required output would look something like this
/ACCOUNT/passwd=-2$-$A88CA7BD3DADDDFFC
/TMC/CONFIRMATION/PASSWORD=-2$-$A88CA7BD3DADDDFFC
Any help is appreciated.
There is nothing special about the forward slash, except if you choose to use it as the delimiter in your sed command, so don’t:
sed 's,ACCOUNT/passwd=,ACCOUNT/passwd=-2$-$A88CA7BD3DADDDFFC,g'
And similar for other target strings.
Here I’ve used a comma as the delimiter. You can choose another character as you prefer.
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.
Setup: Postgresql Server 9.3 - OS: CentOS 6.6
Attempting to bulk insert 250 million records into a Postgresql 9.3 server using the COPY command. The data is in delimited format using a pipe '|' as the delimiter.
Almost all columns in the table that I'm copying to are TEXT datatypes. Unfortunately, out of the 250 million records, there's about 2 million that have legitimate textual values with a "\0" in the text.
Example entry:
245150963|DATASOURCE|736778|XYZNR-1B5.1|10-DEC-1984 00:00:00|||XYZNR-1B5.1\1984-12-10\0.5\1\ASDF1|pH|Physical|Water|XYZNR|Estuary
As you can see, the 8th column has a legitimate \0 in its value.
XYZNR-1B5.1\1984-12-10\0.5\1\ASDF1
No matter how I escape this, the COPY command will either convert this \0 into an actual "\x0" or the COPY command fails with "ERROR: invalid byte sequence for encoding "UTF8": 0x00".
I have tried replacing the \0 with "sed -i" with:
\\0
\\\0
'\0'
\'\'0
\\\\\0
... and many others I can't remember and none of them work.
What would be the correct escaping of these types of strings?
Thanks!
Per Postgres doc on COPY:
Backslash characters () can be used in the COPY data to quote data
characters that might otherwise be taken as row or column delimiters.
In particular, the following characters must be preceded by a
backslash if they appear as part of a column value: backslash itself,
newline, carriage return, and the current delimiter character.
Try to convert all your backslash characters in that path in the field to \\, not just the \0.
FYI \b is shorthand for backslash as well.
So either of these should work:
XYZNR-1B5.1\b1984-12-10\b0.5\b1\bASDF1
XYZNR-1B5.1\\1984-12-10\\0.5\\1\\ASDF1
The one you needed was the one example you didn't give:
sed -e 's/\\/\\\\/g'
You want this for all occurrences of \, not just for \0.
From the perspective of the file and postgres, were trying to convert \ to \\.
In sed, \ is a special character which we need to self escape, so \ becomes \\, and \\ becomes \\\\, hence the above expression.
Have you confirmed that your sed command is actually giving you \\0?
I don't understand why the following sed command contains an # symbol:
sed 's#session\s*required\s*pam_loginuid.so#session optional pam_loginuid.so#g' -i /etc/pam.d/sshd
I've looked at /etc/pam.d/sshd for the before/after effects of this command:
BEFORE:
...
# Set the loginuid process attribute.
session required pam_loginuid.so
...
AFTER:
...
# Set the loginuid process attribute.
session optional pam_loginuid.so
....
Is the # symbol possibly part of regex or sed syntax?
Could not find any doco on this.
Note: The above sed command is actually part of a Dockerfile RUN command in tutorial:
https://docs.docker.com/examples/running_ssh_service/
These are alternate delimiters for the regular expressions and replacement string. Handy when your regex or replacement string includes '/'.
From the sed manual
The syntax of the s (as in substitute) command is ‘s/regexp/replacement/flags’. The / characters may be uniformly replaced by any other single character within any given s command. The / character (or whatever other character is used in its stead) can appear in the regexp or replacement only if it is preceded by a \ character.
From the POSIX specification:
[2addr]s/BRE/replacement/flags
Substitute the replacement string for instances of the BRE in the pattern space. Any character other than <backslash> or <newline> can be used instead of a to delimit the BRE and the replacement. Within the BRE and the replacement, the BRE delimiter itself can be used as a literal character if it is preceded by a <backslash>.
as other says, it is another delimiter than traditionnal / in the s///action. This is usually used when / is found/part of the pattern like searching (or replacing by) a unix path that need to escape the /
s/\/my\/path/\/Your\/path/
# same as
s#my/path#/Your/path#
You often use a character that is not alpha numeric (but you can). The only (logical) constraint is to avoid a special character (aka special meaning like ^$[]{}()+\*.) for regex that make it difficult to read (but functionnal) and without the feature of this character in the pattern
echo "b(a)l" | sed 's(.)()('
I am trying to write a sed script to convert LaTeX coded tables into tab delimited tables.
To do this I need to convert & into \t and strip out anything that is preceded by \.
This is what I have so far:
s/&/\t/g
s/\*/" "/g
The first line works as intended. In the second line I try to replace \ followed by anything with a space but it doesn't alter the lines with \ in them.
Any suggestions are appreciated. Also, can you briefly explain what suggested scripts "say"? I am new to sed and that really helps with the learning process!
Thanks
Assuming you're running this as a sed script, and not directly on the command line:
s/\\.*/ /g
Explanation:
\\ - double backslash to match a literal backslash (a single \ is interpreted as "escape the following character", followed by a .* (. - match any single character, * - arbitrarily many times).
You need to escape the backslash as it is a special character.
If you want to denote "any character" you need to use . (a period)
the second expression should be:
s/\\.//g
I hope I understood your intention and you want to strip the character after the backslash,
if you want to delete all the characters in the line after the backslash add a star (*)
after the period.