I have customer input that may include letters, digits or spaces. For instance:
local customer_input = 'I need 2 tomatoes';
or
local customer_input = 'I need two tomatoes';
However, due to the nature of my application, I may get #, *, #, etc, in the customer_input string. I want to remove any non alphanumeric characters but the space.
I tried with these:
customer_input , _ = customer_input:gsub("%W%S+", "");
This one drops everything but the first word in the phrase.
or
customer_input , _ = customer_input:gsub("%W%S", "");
This one actually drops the space and the first letter of each word.
So, I know I am doing it wrong but I am not really sure how to match alphanumeric + space. I am sure this must be simple but I have not been able to figure it out.
Thanks very much for any help!
You may use
customer_input , _ = customer_input:gsub("[^%w%s]+", "");
See the Lua demo online
Pattern details
[^ - start of a negated character class that matches any char but:
%w - an alphanumeric
%s - a whitespace
]+ - 1 or more times.
Related
how can I extract a few words separated by symbols in a string so that nothing is extracted if the symbols change?
for example I wrote this code:
function split(str)
result = {};
for match in string.gmatch(str, "[^%<%|:%,%FS:%>,%s]+" ) do
table.insert(result, match);
end
return result
end
--------------------------Example--------------------------------------------
str = "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
my_status={}
status=split(str)
for key, value in pairs(status) do
table.insert(my_status,value)
end
print(my_status[1]) --
print(my_status[2]) --
print(my_status[3]) --
print(my_status[4]) --
print(my_status[5]) --
print(my_status[6]) --
print(my_status[7]) --
output :
busy
MPos
-750.222
900.853
1450.808
2
10
This code works fine, but if the characters and text in the str string change, the extraction is still done, which I do not want to be.
If the string change to
str = "Hello stack overFlow"
Output:
Hello
stack
over
low
nil
nil
nil
In other words, I only want to extract if the string is in this format: "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
In lua patterns, you can use captures, which are perfect for things like this. I use something like the following:
--------------------------Example--------------------------------------------
str = "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
local status, mpos1, mpos2, mpos3, fs1, fs2 = string.match(str, "%<(%w+)%|MPos:(%--%d+%.%d+),(%--%d+%.%d+),(%--%d+%.%d+)%|FS:(%d+),(%d+)%>")
print(status, mpos1, mpos2, mpos3, fs1, fs2)
I use string.match, not string.gmatch here, because we don't have an arbitrary number of entries (if that is the case, you have to have a different approach). Let's break down the pattern: All captures are surrounded by parantheses () and get returned, so there are as many return values as captures. The individual captures are:
the status flag (or whatever that is): busy is a simple word, so we can use the %w character class (alphanumeric characters, maybe %a, only letters would also do). Then apply the + operator (you already know that one). The + is within the capture
the three numbers for the MPos entry each get (%--%d+%.%d+), which looks weird at first. I use % in front of any non-alphanumeric character, since it turns all magic characters (such as + into normal ones). - is a magic character, so it is required here to match a literal -, but lua allows to put that in front of any non-alphanumerical character, which I do. So the minus is optional, so the capture starts with %-- which is one or zero repetitions (- operator) of a literal - (%-). Then I just match two integers separated by a dot (%d is a digit, %. matches a literal dot). We do this three times, separated by a comma (which I don't escape since I'm sure it is not a magical character).
the last entry (FS) works practically the same as the MPos entry
all entries are separated by |, which I simply match with %|
So putting it together:
start of string: %<
status field: (%w+)
separator: %|
MPos (three numbers): MPos:(%--%d+%.%d+),(%--%d+%.%d+),(%--%d+%.%d+)
separator: %|
FS entry (two integers): FS:(%d+),(%d+)
end of string: %>
With this approach you have the data in local variables with sensible names, which you can then put into a table (for example).
If the match failes (for instance, when you use "Hello stack overFlow"), nil` is returned, which can simply be checked for (you could check any of the local variables, but it is common to check the first one.
So for I game I want the user to be able to do commands.
For simplicity all parameters are put into a table.
Example: "message all Hello" -> {"message","all","Hello"}
For that I've used the alphanumeric pattern (%w).
Problem is that characters like: _ ; : . Simply can not be used, since they're not alphanumeric.
Is there anyway to use the all characters pattern(.), but ignore spaces.
Or is there any better way to do it?
Thank you for your help
According to Lua docs:
Making the letter after the % uppercase inverts the class, so %D will match all non-digit characters.
So the pattern you're looking for is %S+.
Currently I have code that looks like this:
somestring = "param=valueZ&456"
local stringToPrint = (somestring):gsub("(param=)[^&]+", "%1hello", 1)
StringToPrint will look like this:
param=hello&456
I have replaced all of the characters before the & with the string "hello". This is where my question becomes a little strange and specific.
I want my string to appear as: param=helloZ&456. In other words, I want to preserve the character right before the & when replacing the string valueZ with hello to make it helloZ instead. How can this be done?
I suggest:
somestring:gsub("param=[^&]*([^&])", "param=hello%1", 1)
See the Lua demo
Here, the pattern matches:
param= - literal substring param=
[^&]* - 0 or more chars other than & as many as possible
([^&]) - Group 1 capturing a symbol other than & (here, backtracking will occur, as the previous pattern grabs all such chars other than & and then the engine will take a step back and place the last char from that chunk into Group 1).
There are probably other ways to do this, but here is one:
somestring = "param=valueZ&456"
local stringToPrint = (somestring):gsub("(param=).-([^&]&)", "%1hello%2", 1)
print(stringToPrint)
The thing here is that I match the shortest string that ends with a character that is not & and a character that is &. Then I add the two ending characters to the replaced part.
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've been given a large file with a funny CSV format to parse into a database.
The separator character is a semicolon (;). If one of the fields contains a semicolon it is "escaped" by wrapping it in doublequotes, like this ";".
I have been assured that there will never be two adjacent fields with trailing/ leading doublequotes, so this format should technically be ok.
Now, for parsing it in VBScript I was thinking of
Replacing each instance of ";" with a GUID,
Splitting the line into an array by semicolon,
Running back through the array, replacing the GUIDs with ";"
It seems to be the quickest way. Is there a better way? I guess I could use substrings but this method seems to be acceptable...
Your method sounds fine with the caveat that there's absolutely no possibility that your GUID will occur in the text itself.
On approach I've used for this type of data before is to just split on the semi-colons regardless then, if two adjacent fields end and start with a quote, combine them.
For example:
Pax;is;a;good;guy";" so;says;his;wife.
becomes:
0 Pax
1 is
2 a
3 good
4 guy"
5 " so
6 says
7 his
8 wife.
Then, when you discover that fields 4 and 5 end and start (respectively) with a quote, you combine them by replacing the field 4 closing quote with a semicolon and removing the field 5 opening quote (and joining them of course).
0 Pax
1 is
2 a
3 good
4 guy; so
5 says
6 his
7 wife.
In pseudo-code, given:
input: A string, first character is input[0]; last
character is input[length]. Further, assume one dummy
character, input[length+1]. It can be anything except
; and ". This string is one line of the "CSV" file.
length: positive integer, number of characters in input
Do this:
set start = 0
if input[0] = ';':
you have a blank field in the beginning; do whatever with it
set start = 2
endif
for each c between 1 and length:
next iteration unless string[c] = ';'
if input[c-1] ≠ '"' or input[c+1] ≠ '"': // test for escape sequence ";"
found field consting of half-open range [start,c); do whatever
with it. Note that in the case of empty fields, start≥c, leaving
an empty range
set start = c+1
endif
end foreach
Untested, of course. Debugging code like this is always fun….
The special case of input[0] is to make sure we don't ever look at input[-1]. If you can make input[-1] safe, then you can get rid of that special case. You can also put a dummy character in input[0] and then start your data—and your parsing—from input[1].
One option would be to find instances of the regex:
[^"];[^"]
and then break the string apart with substring:
List<string> ret = new List<string>();
Regex r = new Regex(#"[^""];[^""]");
Match m;
while((m = r.Match(line)).Success)
{
ret.Add(line.Substring(0,m.Index + 1);
line = line.Substring(m.Index + 2);
}
(Sorry about the C#, I don't known VBScript)
Using quotes is normal for .csv files. If you have quotes in the field then you may see opening and closing and the embedded quote all strung together two or three in a row.
If you're using SQL Server you could try using T-SQL to handle everything for you.
SELECT * INTO MyTable FROM OPENDATASOURCE('Microsoft.JET.OLEDB.4.0',
'Data Source=F:\MyDirectory;Extended Properties="text;HDR=No"')...
[MyCsvFile#csv]
That will create and populate "MyTable". Read more on this subject here on SO.
I would recommend using RegEx to break up the strings.
Find every ';' that is not a part of
";" and change it to something else
that does not appear in your fields.
Then go through and replace ";" with ;
Now you have your fields with the correct data.
Most importers can swap out separator characters pretty easily.
This is basically your GUID idea. Just make sure the GUID is unique to your file before you start and you will be fine. I tend to start using 'Z'. After enough 'Z's, you will be unique (sometimes as few as 1-3 will do).
Jacob