Context
In PeopleCode, the following declaration with MessageBox is valid:
MessageBox(0, "", 0, 0, "Something = %1, Something else = %2.", &something, &somethingElse);
This allows me to use bind variables for the MessageBox. The same is also true with SQL:
SQLExec("SELECT * FROM %Table(:1) WHERE VALUE = :2", Record.Y_SOMETHING, &value);
Question
Is there a way to do that with normal strings? I've never liked having to "pipe" strings together like this &string = Something = " | &something | ", Something else = " | &somethingElse | ".".
Is there a way to use this format for regular strings? I've looked through various of Oracle's PeopleBooks, but I haven't found anything.
Maybe this is what you are looking for:
Local number &message_set, &message_num;
Local string &default_msg_txt = "%1 %2 %3";
Local string &l_result= MsgGetText(&message_set, &message_num, &default_msg_txt, "hallo", "Frank", "!");
result:
"hallo Frank !"
You can use MsgGetText function to determine a message by message catalogue. In case the message is not found, the default text is used.
Related
I would like to check in KQL (Kusto Query Language) if a string starts with any prefix that is contained in a list.
Something like:
let MaxAge = ago(30d);
let prefix_list = pack_array(
'Mr',
'Ms',
'Mister',
'Miss'
);
| where Name startswith(prefix_list)
I know this example can be done with startswith("Mr","Ms","Mister","Miss") but this is not escalable.
an inefficient but functional option would be using matches regex - this can work well if the input data set is not too large:
let T = datatable(Name:string)
[
"hello" ,'world', "Mra", "Ms 2", "Miz", 'Missed'
]
;
let prefix_list = pack_array(
'Mr',
'Ms',
'Mister',
'Miss'
);
let prefix_regex = strcat("^(", strcat_array(prefix_list, ")|("), ")");
T
| where Name matches regex prefix_regex
Name
Mra
Ms 2
Missed
This function is not available in the Kusto query language, you are welcome to open a suggestion for it in the user feedback form
I have a multidimensional string vector, gameState. The contents of gameState can be seen below. I would like to "move" A up one space by making [2][1] = " " and [1][1] = "A". But I'm getting an error (error: invalid conversion from 'const char*' to '__gnu_cxx::__alloc_traitsstd::allocator<char, char>::value_type' {aka 'char'} [-fpermissive]). Not sure if there is a way to use vector.at() in this case. Here is a snippet. The full code reads in the initial game state from a text file and assigns it to the string vector gameState. I get the row and col indexes for each letter in the grid below and assign them to variables (eg. Arow and Acol). I'm very new to C++, so any help would be appreciated. If you have any questions about the rest of the code, I'd be happy to elaborate. Thanks in advance.
int Arow;
int Acol;
vector<string> gameState;
string tempU = gameState[Arow-1][Acol];
if (tempU != "#") {
// if no wall is up, move up
gameState[Arow][Acol] = " ";
gameState[Arow-1][Acol] = "A";
Arow = Arow-1;
Amovesequence.push_back("U");
}
###########
#...#P...B#
#A#.$.###*#
#...#D...R#
###########
The type of gameState[Arow-1][Acol] is a char, not a string. Modify the code accordingly, e.g.: char tempU = ... and replace " " with ' 'etc.
I have a string that contains some text followed by a blank line. What's the best way to keep the part with text, but remove the whitespace newline from the end?
Use String.trim() method to get rid of whitespaces (spaces, new lines etc.) from the beginning and end of the string.
String trimmedString = myString.trim();
String.replaceAll("[\n\r]", "");
This Java code does exactly what is asked in the title of the question, that is "remove newlines from beginning and end of a string-java":
String.replaceAll("^[\n\r]", "").replaceAll("[\n\r]$", "")
Remove newlines only from the end of the line:
String.replaceAll("[\n\r]$", "")
Remove newlines only from the beginning of the line:
String.replaceAll("^[\n\r]", "")
tl;dr
String cleanString = dirtyString.strip() ; // Call new `String::string` method.
String::strip…
The old String::trim method has a strange definition of whitespace.
As discussed here, Java 11 adds new strip… methods to the String class. These use a more Unicode-savvy definition of whitespace. See the rules of this definition in the class JavaDoc for Character::isWhitespace.
Example code.
String input = " some Thing ";
System.out.println("before->>"+input+"<<-");
input = input.strip();
System.out.println("after->>"+input+"<<-");
Or you can strip just the leading or just the trailing whitespace.
You do not mention exactly what code point(s) make up your newlines. I imagine your newline is likely included in this list of code points targeted by strip:
It is a Unicode space character (SPACE_SEPARATOR, LINE_SEPARATOR, or PARAGRAPH_SEPARATOR) but is not also a non-breaking space ('\u00A0', '\u2007', '\u202F').
It is '\t', U+0009 HORIZONTAL TABULATION.
It is '\n', U+000A LINE FEED.
It is '\u000B', U+000B VERTICAL TABULATION.
It is '\f', U+000C FORM FEED.
It is '\r', U+000D CARRIAGE RETURN.
It is '\u001C', U+001C FILE SEPARATOR.
It is '\u001D', U+001D GROUP SEPARATOR.
It is '\u001E', U+001E RECORD SEPARATOR.
It is '\u001F', U+0
If your string is potentially null, consider using StringUtils.trim() - the null-safe version of String.trim().
If you only want to remove line breaks (not spaces, tabs) at the beginning and end of a String (not inbetween), then you can use this approach:
Use a regular expressions to remove carriage returns (\\r) and line feeds (\\n) from the beginning (^) and ending ($) of a string:
s = s.replaceAll("(^[\\r\\n]+|[\\r\\n]+$)", "")
Complete Example:
public class RemoveLineBreaks {
public static void main(String[] args) {
var s = "\nHello world\nHello everyone\n";
System.out.println("before: >"+s+"<");
s = s.replaceAll("(^[\\r\\n]+|[\\r\\n]+$)", "");
System.out.println("after: >"+s+"<");
}
}
It outputs:
before: >
Hello world
Hello everyone
<
after: >Hello world
Hello everyone<
I'm going to add an answer to this as well because, while I had the same question, the provided answer did not suffice. Given some thought, I realized that this can be done very easily with a regular expression.
To remove newlines from the beginning:
// Trim left
String[] a = "\n\nfrom the beginning\n\n".split("^\\n+", 2);
System.out.println("-" + (a.length > 1 ? a[1] : a[0]) + "-");
and end of a string:
// Trim right
String z = "\n\nfrom the end\n\n";
System.out.println("-" + z.split("\\n+$", 2)[0] + "-");
I'm certain that this is not the most performance efficient way of trimming a string. But it does appear to be the cleanest and simplest way to inline such an operation.
Note that the same method can be done to trim any variation and combination of characters from either end as it's a simple regex.
Try this
function replaceNewLine(str) {
return str.replace(/[\n\r]/g, "");
}
String trimStartEnd = "\n TestString1 linebreak1\nlinebreak2\nlinebreak3\n TestString2 \n";
System.out.println("Original String : [" + trimStartEnd + "]");
System.out.println("-----------------------------");
System.out.println("Result String : [" + trimStartEnd.replaceAll("^(\\r\\n|[\\n\\x0B\\x0C\\r\\u0085\\u2028\\u2029])|(\\r\\n|[\\n\\x0B\\x0C\\r\\u0085\\u2028\\u2029])$", "") + "]");
Start of a string = ^ ,
End of a string = $ ,
regex combination = | ,
Linebreak = \r\n|[\n\x0B\x0C\r\u0085\u2028\u2029]
Another elegant solution.
String myString = "\nLogbasex\n";
myString = org.apache.commons.lang3.StringUtils.strip(myString, "\n");
For anyone else looking for answer to the question when dealing with different linebreaks:
string.replaceAll("(\n|\r|\r\n)$", ""); // Java 7
string.replaceAll("\\R$", ""); // Java 8
This should remove exactly the last line break and preserve all other whitespace from string and work with Unix (\n), Windows (\r\n) and old Mac (\r) line breaks: https://stackoverflow.com/a/20056634, https://stackoverflow.com/a/49791415. "\\R" is matcher introduced in Java 8 in Pattern class: https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html
This passes these tests:
// Windows:
value = "\r\n test \r\n value \r\n";
assertEquals("\r\n test \r\n value ", value.replaceAll("\\R$", ""));
// Unix:
value = "\n test \n value \n";
assertEquals("\n test \n value ", value.replaceAll("\\R$", ""));
// Old Mac:
value = "\r test \r value \r";
assertEquals("\r test \r value ", value.replaceAll("\\R$", ""));
String text = readFileAsString("textfile.txt");
text = text.replace("\n", "").replace("\r", "");
I'm trying to concat 2 variables Address and Payload. After that I want to send them with http to a server but I have 2 problems. When i try to concat the 2 variables with a delimiter ';' it doesn't work. Also sending the data of Payload or Address doesn't work. This is my code:
handle_rx(Gateway, #link{devaddr=DevAddr}=Link, #rxdata{port=Port, data= RxData }, RxQ)->
Data = base64:encode(RxData),
Devaddr = base64:encode(DevAddr),
TextAddr="Device address: ",
TextPayload="Payload: ",
Address = string:concat(TextAddr, Devaddr),
Payload = string:concat(TextPayload, Data),
Json=string:join([Address,Payload], "; "),
file:write_file("/tmp/foo.txt", io_lib:fwrite("~s.\n", [Json] )),
inets:start(),
ssl:start(),
httpc:request(post, {"http://192.168.0.121/apiv1/lorapacket/rx", [], "application/x-www-form-urlencoded", Address },[],[]),
ok;
handle_rx(_Gateway, _Link, RxData, _RxQ) ->
{error, {unexpected_data, RxData}}.
I have no errors that I can show you. When I write Address or Payload individually to the file it works but sending doesn't work...
Thank you for your help!
When i try to concat the 2 variables with a delimiter ';' it doesn't work.
5> string:join(["hello", <<"world">>], ";").
[104,101,108,108,111,59|<<"world">>]
6> string:join(["hello", "world"], ";").
"hello;world"
base64:encode() returns a binary, yet string:join() requires string arguments. You can do this:
7> string:join(["hello", binary_to_list(<<"world">>)], ";").
"hello;world"
Response to comment:
In erlang the string "abc" is equivalent to the list [97,98,99]. However, the binary syntax <<"abc">> is not equivalent to <<[97,98,99]>>, rather the binary syntax <<"abc">> is special short hand notation for the binary <<97, 98, 99>>.
Therefore, if you write:
Address = [97,98,99].
then the code:
Bin = <<Address>>.
after variable substitution becomes:
Bin = <<[97,98,99]>>.
and that isn't legal binary syntax.
If you need to convert a string/list contained in a variable, like Address, to a binary, you use list_to_binary(Address)--not <<Address>>.
In your code here:
Json = string:join([binary_to_list(<<Address>>),
binary_to_list(<<Payload>>)],
";").
Address and Payload were previously assigned the return value of string:concat(), which returns a string, so there is no reason to (attempt) to convert Address to a binary with <<Address>>, then immediately convert the binary back to a string with binary_to_list(). Instead, you would just write:
Json = string:join(Address, Payload, ";")
The problem with your original code is that you called string:concat() with a string as the first argument and a binary as the second argument--yet string:concat() takes two string arguments. You can use binary_to_list() to convert a binary to the string that you need for the second argument.
Sorry I'm new to Erlang
As with any language, you have to study the basics and write numerous toy examples before you can start writing code that actually does something.
You don't have to concatenate strings. It is called iolist and is one of best things in Erlang:
1> RxData = "Hello World!", DevAddr = "Earth",
1> Data = base64:encode(RxData), Devaddr = base64:encode(DevAddr),
1> TextAddr="Device address", TextPayload="Payload",
1> Json=["{'", TextAddr, "': '", Devaddr, "', '", TextPayload, "': '", Data, "'}"].
["{'","Device address","': '",<<"RWFydGg=">>,"', '",
"Payload","': '",<<"SGVsbG8gV29ybGQh">>,"'}"]
2> file:write_file("/tmp/foo.txt", Json).
ok
3> file:read_file("/tmp/foo.txt").
{ok,<<"{'Device address': 'RWFydGg=', 'Payload': 'SGVsbG8gV29ybGQh'}">>}
This might be a really easy one but I couldn't seem to find an answer anywhere
I'm trying to comment my code as follows
Session("test") = "JAMIE" _
'TEST INFO
& "TEST" _
'ADDRESS INFO
& "ADDRESS = TEST"
With the code above i'm getting the error
Syntax error
But when I remove the comments like so
Session("test") = "JAMIE" _
& "TEST" _
& "ADDRESS = TEST"
It works fine so my guess is that I cannot comment my code between the _ character.
Is there some way I can get around this as I'd like to comment my code ideally
The _ character is the line continuation. It means that the next line is interpreted as if it was on the same line.
So, putting a comment in the middle of the line is a syntax error.
Since you want a solution:
Either put a comment before the continued line or after it
As Tim Schmelter points out in his answer, you can construct the value that will go into the Session object before you put it into the Session object - you can do that is separate statements and comment those to your hearts content.
As Oded has mentioned, The _ character continues the line so you cannot comment between.
You could write:
Dim value = "JAMIE"
'TEST INFO
value &= "TEST"
'ADDRESS INFO
value &= "ADDRESS = TEST"
Session("test") = value
Because that may create separate strings internally just to comment them, you could use a StringBuilder here. You could show us what you're really tring to do, so that we can suggest a different approach(if you need to comment each "line" of a single variable, you should consider to redesign the way you assign the value to the variable).
System.Text.StringBuilder str = new System.Text.StringBuilder();
str.Append("JAMIE");
str.Append("TEST");//TEST INFO
str.Append("ADDRESS");//ADDRESS INFO
public string Test
{
get
{
return Convert.ToString(Session["TEST"]);
}
set
{
Session["Test"] = value;
}
}
Test = st.ToString();