I am modifying a delphi app.In it I'm getting a text from a combo box. The problem is that when I save the text in the table, it contains a carriage return. In debug mode it shows like this.
newStr := 'Projector Ex320u-st Short Throw '#$A'1024 X 768 2700lm'
Then I have put
newStr := StringReplace(newStr,'#$A','',[rfReplaceAll]);
to remove the '#$A' thing. But this doesn't remove it.
Is there any other way to do this..
Thanks
Remove the quotes around the #$A:
newStr := StringReplace(newStr,#$A,'',[rfReplaceAll]);
The # tells delphi that you are specifying a character by its numerical code.
The $ says you are specifying in Hexadecimal.
The A is the value.
With the quotes you are searching for the presence of the #$A characters in the string, which aren't found, so nothing is replaced.
Adapted from http://www.delphipages.com/forum/showthread.php?t=195756
The '#' denotes an ASCII character followed by a byte value (0..255).
The $A is hexadecimal which equals 10 and $D is hexadecimal which equals 13.
#$A and #$D (or #10 and #13) are ASCII line feed and carriage return characters respectively.
Line feed = ASCII character $A (hex) or 10 (dec): #$A or #10
Carriage return = ASCII character $D (hex) or 13 (dec): #$D or #13
So if you wanted to add 'Ok' and another line:
Memo.Lines.Add('Ok' + #13#10)
or
Memo.Lines.Add('Ok' + #$D#$A)
To remove the control characters (and white spaces) from the beginning
and end of a string:
MyString := Trim(MyString)
Why doesn't Pos() find them?
That is how Delphi displays control characters
to you, if you were to do Pos(#13, MyString) or Pos(#10, MyString) then it
would return the position.
Related
MyString = "CfegoAsZEM/sP\u{10}\u{10}}"
MyString.replacingOccurrences(of: "\"", with: "")
with print(MyString) I got this : "CfegoAsZEM/sP" (that's what I need)
with po MyString (on the debugger) : "CfegoAsZEM/sP\u{10}\u{10}}"
\u{10} is a linefeed character
Maybe a better way is to trim the string, it removes all whitespace and newline characters from the beginning and the end of the string
let myString = "CfegoAsZEM/sP\u{10}\u{10}"
let trimmedString = myString.trimmingCharacters(in: .whitespacesAndNewlines)
Your string doesn't contain literal backslash characters. Rather, the \u{} sequence is an escaped sequence that introduces a Unicode character. This is why you can't remove it using replacingOccurrences.
In this case, as Vadian pointed out it is the "new line" character (0x10). Since this is an invisible "white space" character you don't see it when you print the string, but you do see it when you use po. The debugger shows you escape sequences for non-printable characters. You will also see the sequence if you print(MyString.debugDescription)
Unfortunately the trimmingCharactersIn function doesn't appear to consider Unicode sequences.
We can use the filter function to examine each character in the string. If the character is ASCII and has a value greater than 31 ( 32 is the space character, the first "printable" character in the ASCII sequence) we can include it. We also need to ensure that values that aren't ASCII are included so as not to strip printable Unicode characters (e.g. emoji or non-Latin characters).
let MyString = "CfegoAsZEM/sP\u{10}\u{13}$}🔅\u{1F600}".filter { $0.asciiValue ?? 32 > 31 }
print(MyString.debugDescription)
print(MyString)
Output
"CfegoAsZEM/sP}🔅😀"
CfegoAsZEM/sP}🔅😀
asciiValue returns an optional, which is nil if the character isn't plain ASCII. I have used a nil-coalescing operator to return 32 in this case so that the character isn't filtered.
I modified the initial string to include some printable Unicode to demonstrate that it isn't stripped by the filter.
I need to build a string which contains Alt Codes, specifically Alt-16 (Arrow symbol). I have a line of text (aka a string). I append a Carriage return, then want an ARROW symbol and the new line of text. This line will then be passed to PPT. If I manually go into PPT, into a text box, I can hit Alt+16, and get the arrow symbol. This is what I programatically want to do.
Alt symbols found here.
Here is what I am trying, but it gives me a totally different symbol.
line := line + #13 + Chr(VK_MENU) + #16 + NewLine;
How do I build a string with ALT Codes as part of the string?
Not that those characters are not called Alt-codes or Alt-characters. The Alt-codes are just a way to type some special character, but they can't be used as such in a string.
You can just type in that character using the Alt-code, or copy it from the alt codes website. You can use the character as-is in the string. The code below would work fine:
// Show it
ShowMessage('►');
// Or use it in your string:
line := line + #13 + '►' + NewLine;
When I use AES128 encrypt string, if the encrypted string is too long then it will contain \r\n in it. like this
Now I have to use empty string to replace it. Why does the encrypt-string contain the \r\n and any better way to avoid it or fix it.
Thanks.
Answers: it's caused by the Base64 encoding process, every 64 characters will insert a \r\n .
That is a Base64 encoded string.
Actual encryption output is an array of 8-bit bytes, not characters. The code is Base64 encoding the encrypted data with an option to insert line breaks every 64 characters, this is sometimes to allow better printing of the output. When it is decoded use the NSDataBase64DecodingIgnoreUnknownCharacters option to remove line breaks .
In particular for Objective-C the to create a Base64 string from NSData is:
- (NSString *)base64EncodedStringWithOptions:(NSDataBase64EncodingOptions)options
The options include:
NSDataBase64Encoding64CharacterLineLength
Set the maximum line length to 64 characters, after which a line ending is inserted.
Which inserts "\r\n" (carriage return, new line) characters each 64 characters.
If that is not what you want pass 0 as the option value.
To decode Base64 use the Objective-C method:
- (instancetype)initWithBase64EncodedString:(NSString *)base64String options:(NSDataBase64DecodingOptions)options
With the option: NSDataBase64DecodingIgnoreUnknownCharacters.
Apple code:
The default implementation of this method will reject non-alphabet characters, including line break characters. To support different encodings and ignore non-alphabet characters, specify an options value of NSDataBase64DecodingIgnoreUnknownCharacters.
The thing that gives it away as a Base64 string is a length that is a multiple of 4, the characters used "a-zA-Z0-9+/" and the trailing "=" characters.
Historic note: These days on OSX and iOS line breaks are a single "\n" (0x0a) line feed character. Back when we used teletypes as terminals "\r" (0x0d) carriage return moved the carriage or print head back but did not move the paper up to the next line. "\n" newline moved the paper up one line but did move the carriage or print head back. They were two distinct mechanical operations. Later some systems used either "\r\n", "\n\r", "\r" or "\n". Unix choose "\n" and thus OSX and iOS.
How to extract the values from a csv like string dropping the new lines characters (\r\n or \n) with a pattern.
A line looks like:
1.1;2.2;Example, 3
Notice there are only 3 values and the separator is ;. The problem I'm having is to come up with a pattern that reads the values while dropping the new line characters (the file comes from a windows machine so it has \r\n, reading it from a linux and would like to be independent from the new line character used).
My simple example right now is:
s = "1.1;2.2;Example, 3\r\n";
p = "(.-);(.-);(.-)";
a, b, c = string.match(s, p);
print(c:byte(1, -1));
The two last characters printed by the code above are the \r\n.
The problem is that both, \r and \n are detected by the %c and %s classes (control characters and space characters), as show by this code:
s = "a\r";
print(s:match("%c"));
print(s:match("%s"));
print(s:match("%d"));
So, is it possible to left out from the match the new lines characters? (It should not be assumed that the last two characters will be new lines characters)
The 3º value may contain spaces, punctuation and alphanumeric characters and since \r\n are detected as space characters a pattern like `"(.-);(.-);([%w%s%c]-).*" does not work.
Your pattern
p = "(.-);(.-);(.-)";
does not work: the third field is always empty because .- matches a little as possible. You need to anchor it at the end of the string, but then the third field will contain trailing newline chars:
p = "(.-);(.-);(.-)$";
So, just stop at the first trailing newline char. This also anchors the last match. Try this pattern instead:
p = "(.-);(.-);(.-)[\r\n]";
If trailing newline chars are optional, try this pattern:
p = "(.-);(.-);(.-)[\r\n]*$";
Without any lua experience I found a naive solution:
clean_CR = s:gsub("\r","");
clean_NL = clean_CR:gsub("\n","");
With POSIX regex syntax I'd use
^([^;]*);([^;]*);([^\n\r]*).*$
.. with "\n" and "\r" possibly included as "^M", "^#" (control/unicode characters) .. depending on your editor.
I have a string
String :='this is my string | yes';
I need delphi to get whatever the text is after the |
So something like:
getTextAftertoCharacter('|',String);
This should return "yes" in the example above.
Has Delphi get a function for this?
I'm not aware of a single function that does this. I think the easiest way would be to use Pos and then Copy:
answer := Copy(str, Pos('|', str) + 1, Length(str));
Start character is at Pos('|', str) + 1, amount of characters to copy is actually Length(str)-Pos('|', str), but passing a greater value to Copy also works.
Note: this will return the whole contents of str if there is no '|'. Check for Pos('|', str) being non-zero if you need different behavior.
You can use Controls.GetLongHint():
GetLongHint(String);
That's because the separator of a two part hint is a |. In your example it would return ' yes', with the leading space. If | is not found, the function returns the String.
I'd use then following code:
Trim(RightStr(Text, Length(Text)-LastDelimiter('|', Text)));
It'll locate the last delimiter in the string and I used Trim to get rid of the whitespace. LastDelimiter can take more than one delimiter which may be usefull. If no delimiter is found it will return the whole string.