In the default letters that I see and use, the address of the from address is as following:
Max Mustermann, Beispielweg 23 a, 12345 Köln
I would like to change the separation symbol (in the above example ",") to "|".
So it will look like:
Max Mustermann | Beispielweg 23 a | 12345 Köln
I based my question on the text in example
Using
\setkomavar{backaddressseparator}{~|~}
solved my problem
Related
I need to find all the records which contains the given sub-string(art), and the condition is the given substring is either a complete word within the string or the start of any word within the string.
Sample data. Program
id | name
-----------
10 | Report of Quarter 1
11 | The Art Program
12 | The Artificial Program
From the above data, I must be able to get the record numbers 11 and 12 but not 10.
I am trying like this
Program.where("name ~* ?",'art\b')
but it's not working
I have also tried with
Program.where("regexp_match(name, ?)",'/art\b/i')
this too is not working
any help is really appreciable. Thanks!
EDITED
I guess you are using Postgres since you gave an example with regexp_match
Try Program.where("name ~* ?",'\mart') :)
You can see hidden in the Postgres docs ("Regular Expression Escapes" section) that \b means "backspace, as in C" whereas \m "matches only at the beginning of a word"
How do I turn Bay 13 | Brand PS/CPP into 13 using REGEXREPLACE in Google Sheets?
I can't figure out how to NOT include a pattern, as it seems whatever I type as the pattern it takes it out completely. I'm not used to this behavior in REGEX.
For instance:
REGEXREPLACE("(?i)bay ", )
returns 13 | Brand PS/CPP.
I want to keep the 13 (which could be any number between 1-35 including halves (12.5)), and get rid of everything after it including the "|" pipe.
try "everything after bay and before a pipe":
=REGEXEXTRACT(A1, "(?i)bay (.+) \|")
or if you want specifics:
=REGEXEXTRACT(A1, "(?i)bay (\d+(?:\.\d+)?) \|")
I think you want to use REGEXEXTRACT here rather than REGEXREPLACE:
REGEXEXTRACT("\d+(?:\.\d+)?", "Bay 13 | Brand PS/CPP")
Please, allow me ask a question about formula of counting word base on last number follow that word.
example:
| A | B
--------------------
1 | thumbnail20 | 20
2 | gallery13 | 13
3 | girl45 | 45
I'm so appreciate for all answer, sorry for duplicate question
thanks for #ziganotschka and #BHAWANI SINGH, it's all work, case close :)
There are several options depending on your data structure, e.g.
=VALUE(REGEXREPLACE(A1,"[^[:digit:]]", ""))
will extract all digits from the A column to the B column
Should you have several numbers within your string,
=SPLIT(lower(A4),"qwertyuiopasdfghjklzxcvbnm`-=[]\;',./!##$%^&*()")
will extract the first number into column B, the second into column C etc.
If you want to extract only the digits to the right, then
=arrayformula(RIGHT(A1,LEN(A1)+1-min(SEARCH({0,1,2,3,4,5,6,7,8,9},A1&"0123456789"))))
I have a non-delimited text file and want to parse it to add tabs at specific spots to delimit columns. The columns are sometimes empty or vary in length, which is why I need to add tabs to those specific spots. I had found the answer to this once a couple of years ago on the net using batch, but now can't find it or the code. I already have the following code to replace more than 2 spaces in the file, but this doesn't account for when the columns are empty.
gc $FileToOpen | % { $_ -replace ' +',"`t" } | set-content $FileToSave
So, I need to read each line, but be able to only read a portion (certain number of characters) of it and add the tabs after each portion to itself.
Here is a sample of the data file, the top row is the header and the data rows have no blank lines in between them:
MRUN Number Name X Exception Reason Data CDM# Quantity D.O.S
000000 00000000 Name W MODIFIER CANNOT BE FILED WITHOUT 08/13/2015 0000000 0 08/13/2015
000000 00000000 Name W MODIFIER CANNOT BE FILED WITHOUT 0000000 0 08/13/2015
The second data row is missing Data.
Using Ansgar's answer, my code that does find empty fields:
gc $FileToOpen |
? { $_ -match '^(.{8})(.{12})(.{20})(.{3})(.{34})(.{62})(.{10})(.{22})(.{10})$' } |
% { "{0}`t{1}`t{2}`t{3}`t{4}`t{5}`t{6}`t{7}`t{8}" -f $matches[1].Trim(), $matches[2].Trim(), $matches[3].Trim(), $matches[4].Trim(), $matches[5].Trim(), $matches[6].Trim(), $matches[7].Trim(), $matches[8].Trim(), $matches[9].Trim() } |
Set-Content $FileToSave
Thanks for your patience Ansgar, I know I tried it! I really do appreciate the help!
Since you seem to have an input file with fixed-width columns, you should probably use a regular expression for transforming the input into a tab-delimited format.
Assume the following input file:
A B C
foo 13 22
bar 4 17
baz 142 23
The file has 3 columns. The first column is 6 characters wide, the other two columns 4 characters each.
The transformation could be done with a regular expression like this:
Get-Content 'C:\path\to\input.txt' |
? { $_ -match '^(.{6})(.{4})(.{4})$' } |
% { "{0}`t{1}`t{2}" -f $matches[1].Trim(), $matches[2].Trim(), $matches[3].Trim() } |
Set-Content 'C:\path\to\output.txt'
The regular expression defines the columns by character count and captures them in groups (parentheses). The groups can then be accessed as the indexes 1 and above of the resulting $matches collection. Trimming removes the leading/trailing whitespace. The format operator (-f) then inserts the trimmed values into the tab-separated format string.
If the last column has a variable width (because its values are aligned to the left and don't have trailing spaces) you may need to change the regular expression to ^(.{6})(.{4})(.{,4})$ to take care of that. The quantifier {,4} (or {0,4}) means up to four times the preceding expression.
This question already has answers here:
URL encoding the space character: + or %20?
(5 answers)
Closed 6 years ago.
In a URL, should I encode the spaces using %20 or +? For example, in the following example, which one is correct?
www.mydomain.com?type=xbox%20360
www.mydomain.com?type=xbox+360
Our company is leaning to the former, but using the Java method URLEncoder.encode(String, String) with "xbox 360" (and "UTF-8") returns the latter.
So, what's the difference?
Form data (for GET or POST) is usually encoded as application/x-www-form-urlencoded: this specifies + for spaces.
URLs are encoded as RFC 1738 which specifies %20.
In theory I think you should have %20 before the ? and + after:
example.com/foo%20bar?foo+bar
According to the W3C (and they are the official source on these things), a space character in the query string (and in the query string only) may be encoded as either "%20" or "+". From the section "Query strings" under "Recommendations":
Within the query string, the plus sign is reserved as shorthand notation for a space. Therefore, real plus signs must be encoded. This method was used to make query URIs easier to pass in systems which did not allow spaces.
According to section 3.4 of RFC2396 which is the official specification on URIs in general, the "query" component is URL-dependent:
3.4. Query Component
The query component is a string of information to be interpreted by
the resource.
query = *uric
Within a query component, the characters ";", "/", "?", ":", "#",
"&", "=", "+", ",", and "$" are reserved.
It is therefore a bug in the other software if it does not accept URLs with spaces in the query string encoded as "+" characters.
As for the third part of your question, one way (though slightly ugly) to fix the output from URLEncoder.encode() is to then call replaceAll("\\+","%20") on the return value.
This confusion is because URL is still 'broken' to this day
Take "http://www.google.com" for instance. This is a URL. A URL
is a Uniform Resource Locator and is really a pointer to a web page
(in most cases). URLs actually have a very well-defined structure
since the first specification in 1994.
We can extract detailed information about the "http://www.google.com"
URL:
+---------------+-------------------+
| Part | Data |
+---------------+-------------------+
| Scheme | http |
| Host address | www.google.com |
+---------------+-------------------+
If we look at a more
complex URL such as
"https://bob:bobby#www.lunatech.com:8080/file;p=1?q=2#third" we can
extract the following information:
+-------------------+---------------------+
| Part | Data |
+-------------------+---------------------+
| Scheme | https |
| User | bob |
| Password | bobby |
| Host address | www.lunatech.com |
| Port | 8080 |
| Path | /file |
| Path parameters | p=1 |
| Query parameters | q=2 |
| Fragment | third |
+-------------------+---------------------+
The reserved characters are different for each part
For HTTP URLs, a space in a path fragment part has to be encoded to
"%20" (not, absolutely not "+"), while the "+" character in the path
fragment part can be left unencoded.
Now in the query part, spaces may be encoded to either "+" (for
backwards compatibility: do not try to search for it in the URI
standard) or "%20" while the "+" character (as a result of this
ambiguity) has to be escaped to "%2B".
This means that the "blue+light blue" string has to be encoded
differently in the path and query parts:
"http://example.com/blue+light%20blue?blue%2Blight+blue". From there
you can deduce that encoding a fully constructed URL is impossible
without a syntactical awareness of the URL structure.
What this boils down to is
you should have %20 before the ? and + after
Source
It shouldn't matter, any more than if you encoded the letter A as %41.
However, if you're dealing with a system that doesn't recognize one form, it seems like you're just going to have to give it what it expects regardless of what the "spec" says.
You can use either - which means most people opt for "+" as it's more human readable.
When encoding query values, either form, plus or percent-20, is valid; however, since the bandwidth of the internet isn't infinite, you should use plus, since it's two fewer bytes.