I'm developing a game in Erlang, and now i need to read the standard input. I tried the following calls:
io:fread()
io:read()
The problem is that i can't read a whole string, when it contains white spaces. So i have the following questions:
How can i read the string typed from the user when he press the enter key? (remember that the string contains white spaces)
How can i convert a string like "56" in the number 56?
Read line
You can use io:get_line/1 to get string terminated by line feed from console.
3> io:get_line("Prompt> ").
Prompt> hello world how are you?
"hello world how are you?\n"
io:read will get you erlang term, so you can't read a string, unless you want to make your users wrap string in quotes.
Patterns in io:fread does not seem to let you read arbitrary length string containing spaces.
Parse integer
You can convert "56" to 56 using erlang:list_to_integer/1.
5> erlang:list_to_integer("56").
56
or using string:to_integer/1 which will also return you the rest of a string
10> string:to_integer("56hello").
{56,"hello"}
11> string:to_integer("56").
{56,[]}
The erlang documentation about io:fread/2 should help you out.
You can use field lengths in order to read an arbitrary length of characters (including whitespace):
io:fread("Prompt> ","~20c").
Prompt> This is a sentence!!
{ok,["This is a sentence!!"]}
As for converting a string (a list of characters) to an integer, erlang:list_to_integer/1 does the job:
7> erlang:list_to_integer("645").
645
Edit: try experimenting with io:fread/2, the format sequence can ease the parsing of data by applying some form of pattern matching:
9> io:fread("Prompt> ","~s ~s").
Prompt> John Doe
{ok,["John","Doe"]}
The console is not really a good place to do your stuff, because you need to know in advance the format of the answer. Considering that you allow spaces, you need to know how many words will be entered before getting the answer. Knowing that, you can use a string as entry, and then parse it later:
1> io:read("Enter a text > ").
Enter a text > "hello guy, this is my answer :o)".
{ok,"hello guy, this is my answer :o)"}
2>
The bad news is that the user must enter the quotes and a final dot, not user friendly...
Related
I'm new to GREP in BBEdit. I need to find a string inside an XML file. Such string is enclosed in quotes. I need to replace only what's inside the quotes.
The problem is that the replacement string starts with a number thus confuses BBEdit when I put together the replacement pattern. Example:
Original string in XML looks like this:
What I need to replace it with:
01 new file name.png
My grep search and replace patterns:
Using the replacement pattern above, BBEdit wrongly thinks that the first backreference is "\101" when what I really need it understand is that I mean "\01".
TIA for any help.
Your example is highly artificial because in fact there is no need for your \1 or \3 as you know their value: it is " and you can just type that directly to get the desired result.
"01 new file name.png"
However, just for the sake of completeness, the answer to your actual question (how to write a replacement group number followed by a number) is that you write this:
\0101 new file name.png\3
The reason that works is that there can only be 99 capture groups, so \0101 is parsed as \01 (the first capture group) followed by literal 01.
I have been looking around and have read a lot of different answers but none seems to answer my specific request.
I make watchfaces for Wear OS 2 with an app called ''WATCHMAKER'' witch uses LUA as language. I want to make a watch face with a special clock pointing to a number depending on a blood sugar value sent by an transmitter connected to the body.
The string values I want to parse follows this syntax:
<DECIMAL NUMBER> <ARROW> (<TIME>)
One example would be
5,6 -> (1m)
I want to extract the <DECIMAL NUMBER> part of the reading. In the above example, I want the value 5,6.
Every 5 minutes, the transmitter sends another reading, all of those informations change:
5,8 - (30 secondes)
Thank you so much
Say you have a string, in LUA, s="14,11 -> (something)" and you want this first number of the string to be parsed to a float so you can do maths on it.
s='9,6 -> (24m)'
-- Now we use so called regular expressions
-- to parse the string
new_s=string.match(s, '[0-9]+,[0-9]+')
-- news now has the number 9,6. Which is now parsed
-- however it's still a string and to be able to treat
-- it like a number, we have to do more:
-- But we have to switch the comma for a period
new_s=new_s:gsub(",",".")
-- Now s has "9.6" as string
-- now we convert it to a number
number = string.format('%.10g', tonumber(new_s))
print(number)
Now number contains the number 9.6.
Tuple={<<"jid">>,Member},
Tuple_in_string=lists:flatten(io_lib:format("~p", [Tuple])),
it gives output as:
"{<<\"jid\">>,\"sdfs\"}"
But i want this output without these slashes like
"{<<"jid">>,Member}"
Any pointers?
I have tried all the answers but at the end with io:format("\"~s\"~n", [Tuple_in_string]). what am geeting is "{<<"jid">>,Member}" but it is not a string.it is a atom.I need string on which i can apply concat operation.Any pointers?
You can print it like this:
io:format("\"~s\"~n", [Tuple_in_string]).
It prints:
"{<<"jid">>,"sdfs"}"
The \ are here to denote that the following " is part of the string and not a string delimiter. they do not exist in the string itself. They appear because you use the pretty print format ~p. If you use the string format ~s they wont appear in the display.
1> io:format("~p~n",["a \"string\""]).
"a \"string\""
ok
2> io:format("~s~n",["a \"string\""]).
a "string"
ok
3> length("a \"string\""). % is 10 and not 12
10
Firstly, you don't need to flatten the list here:
Tuple_in_string=lists:flatten(io_lib:format("~p", [Tuple])),
Erlang has the concept of iodata(), which means that printable things can be in nested lists and most functions can handle them, so you should leave only:
Tuple_in_string = io_lib:format("~p", [Tuple]),
Secondly, when you use ~p, you tell Erlang to print the term in such way, that it can be copied and pasted into console. That is why all double quotes are escaped \". Use ~s, which means "treat as string".
1> 38> Tuple = {<<"jid">>,"asdf"}.
{<<"jid">>,"asdf"}
2> IODATA = io_lib:format("~p", [Tuple]).
[[123,[[60,60,"\"jid\"",62,62],44,"\"asdf\""],125]]
3> io:format("~s~n", [IODATA]).
{<<"jid">>,"asdf"}
ok
L = Packet_in_tuple_form={xmlel,<<"message">>,[{<<"id">>,<<"rkX6Q-8">>},{<<"to">>,<<"multicast.devlab">>}],[{xmlel,<<"body">>,[],[{xmlcdata,"Hello"}]},{xmlel,<<"addresses">>,[{<<"xmlns">>,<<"http://jabber.org/protocol/address">>}],[{xmlel,<<"address">>,[{<<"type">>,<<"to">>},"{<<\"jid\">>,\"sds\"}",{<<"desc">>,"Description"}],[]}]}]}.
Gives me:
{xmlel,<<"message">>,
[{<<"id">>,<<"rkX6Q-8">>},{<<"to">>,<<"multicast.devlab">>}],
[{xmlel,<<"body">>,[],[{xmlcdata,"Hello"}]},
{xmlel,<<"addresses">>,
[{<<"xmlns">>,<<"http://jabber.org/protocol/address">>}],
[{xmlel,<<"address">>,
[{<<"type">>,<<"to">>},
"{<<\"jid\">>,\"sds\"}",
{<<"desc">>,"Description"}],
[]}]}]}
The \ in the address field are escape characters.
You can verify the same by checking the length of string.
Trying to generate a list through comprehension and at some point I start seeing strange character strings. Unable to explain their presence at this point (guessing the escape chars to be ASCII codes - but why?):
45> [[round(math:pow(X,2))] ++ [Y]|| X <- lists:seq(5,10), Y <- lists:seq(5,10)].
[[25,5],
[25,6],
[25,7],
[25,8],
[25,9],
[25,10],
[36,5],
[36,6],
[36,7],
"$\b","$\t","$\n",
[49,5],
[49,6],
[49,7],
"1\b","1\t","1\n",
[64,5],
[64,6],
[64,7],
"#\b","#\t","#\n",
[81,5],
[81,6],
[81,7],
"Q\b",
[...]|...]
In Erlang all strings are just list of small integers (like chars in C). And shell to help you out a little tries to interpret any list as printable string. So what you get are numbers, they are just printed in a way you would not expect.
If you would like to change this behaviour you can look at this answer.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.