I'm new to the Lua .. I want to get the username , using string.match
That's like the way I use them
local text = 'hi my name #saleh and my friend #chris ..'
print(string.match(text, "(#[%a%d]+)"));
Result
#saleh
I want all the usernames
You want string.gmatch. Use it in a loop:
for name in text:gmatch "(#[%a%d]+)" do
print( name )
end
above answer not work
This code is exact
for name in text:gmatch "(#[%a%d]+)" do
print(name:sub(2, -1))
end
Related
I have email format like
iba#test.com
Expected output
#test.com
I need to remove the name after the mail id how can I do that I have tried str_replace but that does not give me a solution.
Please try like this :
let string = "iba#test.com"
const result = `#${string.split('#')[1]}`
using explode
$email= "iba#test.com"
$domain= explode("#",$email);
echo $domain[0]; // iba
echo $domain[1]; //test.com
I want to find a specific string within a string.
For example, let's say I have the string
string = "username:quantopia;password:blabla
How can I then find quantopia?
I am using python 3.
Update: I am sorry I did not mention what I try before..
string.split('username:',1)[1].split(';',1)[0]
But this look very bad and not efficient, I was hoping for something better.
Just use regex as such:
import re
username = re.search("username:(.*);password", "username:quantopia;password:blabla").group(1)
print("username:", username)
This will output quantopia.
In this expression "username:(.*);password" you are saying "give me everything from username: to ;password" So this is why you're getting quantopia. This might as well be ":(.*);" as it will output the same thing in this case.
The simple solution is:
string = "username:quantopia;password:blabla"
username = "username"
if username in string:
# do work.
You might be better to just use split to create a dictionary so you dont need to use multiple regex to extract different parts of data sets. The below will split stirng into key value pairs then split key value pairs then pass the list of lists to dict to create a dictionary.
string = "username:quantopia;password:blabla"
data = dict([pairs.split(':') for pairs in string.split(';')])
print(f'username is "{data["username"]}" and password is "{data["password"]}"')
OUTPUT
username is "quantopia" and password is "blabla"
Something like that is coming in:
str="Hello;this;is;a;text"
What I do want as result is this:
result="72:101:108:108:111;116:104:105:115;..."
which should be the Text in ASCII.
You could use string matching to get each word separated by ; and then convert, concat:
local str = "Hello;this;is;a;text"
for word in str:gmatch("[^;]+") do
ascii = table.pack(word:byte(1, -1))
local converted = table.concat(ascii, ":")
print(converted)
end
The output of the above code is:
72:101:108:108:111
116:104:105:115
105:115
97
116:101:120:116
I'll leave the rest of work to you. Hint: use table.concat.
Here is another approach, which exploits that fact that gsub accepts a table where it reads replacements:
T={}
for c=0,255 do
T[string.char(c)]=c..":"
end
T[";"]=";"
str="Hello;this;is;a;text"
result=str:gsub(".",T):gsub(":;",";")
print(result)
Another possibility:
function convert(s)
return (s:gsub('.',function (s)
if s == ';' then return s end
return s:byte()..':'
end)
:gsub(':;',';')
:gsub(':$',''))
end
print(convert 'Hello;this;is;a;text')
Finding certain character or string (such as ";") can be done by using string.find - https://www.lua.org/pil/20.1.html
Converting character to its ASCII code can be done by string.byte - https://www.lua.org/pil/20.html
What you need to do is build a new string using two functions mentioned above. If you need more string-based functions please visit official Lua site: https://www.lua.org/pil/contents.html
Okay...I got way further, but I can't find how to return a string made up of two seperate strings like
str=str1&" "&str2
I have created a structure containing a few different fields. The fields contain data from a number of different subjects/participants.
At the beginning of the script I prompt the user to enter the "Subject number" like so:
prompt='Enter the subject number in the format SUB_n: ';
SUB=input(prompt,'s');
Example SUB_34 for the 34th subject.
I want to then name my structure such that it contains this string... i.e. I want the name of my structure to be SUB_34, e.g. SUB_34.field1. But I don't know how to do this.
I know that you can assign strings to a specific field name for example for structure S if I want field1 to be called z then
S=struct;
field1='z';
S.(field1);
works but it does not work for the structure name.
Can anyone help?
Thanks
Rather than creating structures named SUB_34 I would strongly recommend just using an array of structures instead and having the user simply input the subject number.
number = input('Subject Number')
S(number) = data_struct
Then you could simply find it again using:
subject = S(number);
If you really insist on it, you could use the method proposed in the comment by #Sembei using eval to get the struct. You really should not do this though
S = eval([SUB, ';']);
Or to set the structure
eval([SUB, ' = mydata;']);
One (of many) reasons not to do this is that I could enter the following at your prompt:
>> prompt = 'Enter the subject number in the format SUB_n: ';
>> SUB = input(prompt, 's');
>> eval([SUB, ' = mydata;']);
And I enter:
clear all; SUB_34
This would have the unforeseen consequence that it would remove all of your data since eval evaluates the input string as a command. Using eval on user input assumes that the user is never going to ever write something malformed or malicious, accidentally or otherwise.
I have quite simple question, but my google research did not help.. I am pretty new to Lua, so..
I have string "XXXX_YYYYYY_zzzzzz" stored in local variable and I want to parse it and get 3 new local variables. Should I use string.find?
local str_ = "XXXX_YYYYY_zzzzzz"
local first_, second_, third_ = strind.find(str_, "^(%w+)_(%w+)_(%w+)$")
Use string.match instead:
local str_ = "XXXX_YYYYY_zzzzzz"
local first_, second_, third_ = str_:match "^([^_]+)_([^_]+)_([^_]+)$"
Have a look at the string library on lua-users wiki.
string.find would additionally return the indexes where the matched substring was located/found. These two (start and end) indices are not useful to your case, which is why string.match would be a better tool.