I have a string which is looking like this for example Lübeck Nürnberg Österreich with "Umlaute".
Now I want to split it and add the new "words" on the string like this Lübeck Nürnberg Österreich Luebeck Nuernberg Oesterreich.
So ä should be ae, ö should be oe and ü should be ue.
But I have no clue how to achieve this, thank you!
DATA: lt_split TYPE TABLE OF char40.
SPLIT string AT space INTO TABLE lt_split.
You can use REPLACE to replace a string with another string inside a string:
REPLACE ALL OCCURENCES OF: 'ä'
IN string
WITH 'ae',
'ü'
IN string
WITH 'ue'.
There is another variation of this statement, it can be used with an internal table:
REPLACE 'ä'
IN TABLE lt_split
WITH 'ae'.
https://help.sap.com/doc/abapdocu_751_index_htm/7.51/de-DE/abapreplace_in_pattern.htm
Please note, the new string will be longer (if at least one replacement happened), this can be a problem, if the string was character type with fixed length and the new string would be longer, than allowed.
You can use the function module SCP_REPLACE_STRANGE_CHARS, but you have less control than replace on what is changed.
Example:
DATA name TYPE string.
name = 'Lübeck Österreich Haßberge Eichstätt'.
CALL FUNCTION 'SCP_REPLACE_STRANGE_CHARS'
EXPORTING
intext = name
IMPORTING
outtext = name
EXCEPTIONS
invalid_codepage = 1
codepage_mismatch = 2
internal_error = 3
cannot_convert = 4
fields_not_type_c = 5
OTHERS = 6.
ASSERT name = 'Luebeck Oesterreich Hassberge Eichstaett'.
Be careful, the function module affects lots of characters with accent and lots of special characters, for instance, depending on the values of parameters, we may have these replacements:
á ==> AE
ā ==> A (Acircumflex)
Ă ==> Ae (Adieresis)
£ ==> L (sterling)
ß ==> ss (eszett)
¼ ==> 1/4
You may use the program RSCP0007 to test the function module.
Related
I am searching for a way to use a formatter to put a space between two characters. i thought it would be easy with a string formatter.
here is what i am trying to accomplish:
given: "AB" it will produce "A B"
Here is what i have tried so far:
"AB".format("%#s")
but this keep returning "AB" i want "A B". i thought the number sign could be used for space.
i also tried this:
"26".format("%#d") but its still prints "26"
is there anyway to do this with string.formatter.
It is kind of possible with the string formatter although not directly with a pattern.
jshell> String.format("%1$c %2$c", "AB".chars().boxed().toArray())
$10 ==> "A B"
We need to turn the string into an object array so it can be passed in as varargs and the formatter pattern can extract characters based on index (1$ and 2$) and format them as characters (c).
A much simpler regex solution is the following which scales to any number of characters:
jshell> "ABC^&*123".replaceAll(".", "$0 ").trim()
$3 ==> "A B C ^ & * 1 2 3"
All single characters are replaced with them-self ($0) followed by a space. Then the last extra space is removed with the trim() call.
I could not find way to do this using String#format. But here is a way to accomplish this using regex replacement:
String input = "AB";
String output = input.replaceAll("(?<=[A-Z])(?=[A-Z])", " ");
System.out.println(output);
The regex pattern (?<=[A-Z])(?=[A-Z]) will match every position in between two capital letters, and interpolate a space at that point. The above script prints:
A B
In Apex, I want to remove all the special characters in a string except for "+". This string is actually a phone number. I have done the following.
String sampleText = '+44 597/58-31-30';
sampleText = sampleText.replaceAll('\\D','');
System.debug(sampleText);
So, what it prints is 44597583130.
But I want to keep the sign + as it is represents 00.
Can someone help me with this ?
Possible solutions
String sampleText = '+44 597/58-31-30';
// exclude all characters which you want to keep
System.debug(sampleText.replaceAll('[^\\+|\\d]',''));
// list explicitly each char which must be replaced
System.debug(sampleText.replaceAll('/|-| ',''));
Output in both case will be the same
|DEBUG| +44597583130
|DEBUG| +44597583130
Edit
String sampleText = '+0032 +497/+59-31-40';
System.debug(sampleText.replaceAll('(?!^\\+)[^\\d]',''));
|DEBUG|+0032497593140
In Swift I can create a String variable such as this:
let s = "Hello\nMy name is Jack!"
And if I use s, the output will be:
Hello
My name is Jack!
(because the \n is a linefeed)
But what if I want to programmatically obtain the raw characters in the s variable? As in if I want to actually do something like:
let sRaw = s.raw
I made the .raw up, but something like this. So that the literal value of sRaw would be:
Hello\nMy name is Jack!
and it would literally print the string, complete with literal "\n"
Thank you!
The newline is the "raw character" contained in the string.
How exactly you formed the string (in this case from a string literal with an escape sequence in source code) is not retained (it is only available in the source code, but not preserved in the resulting program). It would look exactly the same if you read it from a file, a database, the concatenation of multiple literals, a multi-line literal, a numeric escape sequence, etc.
If you want to print newline as \n you have to convert it back (by doing text replacement) -- but again, you don't know if the string was really created from such a literal.
You can do this with escaped characters such as \n:
let secondaryString = "really"
let s = "Hello\nMy name is \(secondaryString) Jack!"
let find = Character("\n")
let r = String(s.characters.split(find).joinWithSeparator(["\\","n"]))
print(r) // -> "Hello\nMy name is really Jack!"
However, once the string s is generated the \(secondaryString) has already been interpolated to "really" and there is no trace of it other than the replaced word. I suppose if you already know the interpolated string you could search for it and replace it with "\\(secondaryString)" to get the result you want. Otherwise it's gone.
I am looking for elegant ways to "shorten" the (user provided) names of object. More precisely:
my users can enter free text (used as "name" of some object), they can use up to 64 chars (including whitespaces, punctuation marks, ...)
in addition to that "long" name; we also have a "reduced" name (exactly 8 characters); required for some legacy interface
Now I am looking for thoughts on how to generate these "reduced" names, based on the 64-char name.
With "elegant" I am wondering about any useful ideas that "might" allow the user to recognize something with value within the shortened string.
Like, if the name is "Production Test Item A5"; then maybe "PTIA5" might (or might not) tell the user something useful.
Apply a substring method to the long version, trim it, in case there are any whitespace characters at the end, optionally remove any special characters from the very end (such as dashes) and finally add a dot, in case you want to indicate your abbreviation that way.
Just a quick hack to get you started:
String longVersion = "Aswaghtde-5d";
// Get substring 0..8 characters
String shortVersion = longVersion.substring(0, (longVersion.length() < 8 ? longVersion.length() : 8));
// Remove whitespace characters from end of String
shortVersion = shortVersion.trim();
// Remove any non-characters from end of String
shortVersion = shortVersion.replaceAll("[^a-zA-Z0-9\\s]+$", "");
// Add dot to end
shortVersion = shortVersion.substring(0, (shortVersion.length() < 8 ? shortVersion.length() : shortVersion.length() - 1)) + ".";
System.out.println(shortVersion);
I needed to shorten names to function as column names in a database. Ideally, the names should be recognizable for users. I set up a dictionary of patterns for commonly occuring words with corresponding "abbreviations". This was applied ONLY to those names which were over the limit of 30 characters.
The string I am given is as follows:
scrap1 =
a le h
ke fd
zyq b
ner i
You'll notice there are 2 blank spaces indicating a space (ASCII 32) in each row. I need to find the mean ASCII value in each column without taking into account the spaces (32). So first I would convert to with double(scrap1) but then how do I find the mean without taking into account the spaces?
If it's only the ASCII 32 you want to omit:
d = double(scrap1);
result = mean(d(d~=32)); %// logical indexing to remove unwanted value, then mean
You can remove the intermediate spaces in the string with scrap1(scrap1 == ' ') = ''; This replaces any space in the input with an empty string. Then you can do the conversion to double and average the result. See here for other methods.
Probably, you can use regex to find the space and ignore it. "\s"
findSpace = regexp(scrap1, '\s', 'ignore')
% I am not sure about the ignore case, this what comes to my mind. but u can read more about regexp by typying doc regexp.