gsub(/^/, "\t" * num)
What character is being substituted?
No character is being substituted, it is just inserting num tabs at the beginning so you could say that it is substituting the zero width "beginning of line" marker. Whoever wrote that would have been better off with something more like this:
tabbed = "\t" * num + original
A regular expression really isn't the right tool for simple string concatenation.
Clarification: If you're expecting your string to contain multiple lines then using:
gsub(/^/, "\t" * num)
to prefix all the lines with tabs is a reasonably thing to do and less noisy than splitting, prefixing, and re-joining. If you're only expecting to deal with a single line in your string then simple string concatenation would be the better choice.
^ means "start of line" in regex syntax, so this will insert num tab characters at the beginning of every line. Technically you could say that it replaces the empty string at the start of every line.
Related
Is there a way to convert a quoted string to a multiline string?
Something like "This string \66 here" to [[This string \66 here]] since I would like to ignore the interpretation of escaped characters.
Lua 5.3 Reference Manual 3.1: Lexical Conventions
Literal strings can also be defined using a long format enclosed by
long brackets. We define an opening long bracket of level n as an
opening square bracket followed by n equal signs followed by another
opening square bracket. So, an opening long bracket of level 0 is
written as [[, an opening long bracket of level 1 is written as [=[,
and so on. A closing long bracket is defined similarly; for instance,
a closing long bracket of level 4 is written as ]====]. A long literal
starts with an opening long bracket of any level and ends at the first
closing long bracket of the same level. It can contain any text except
a closing bracket of the same level. Literals in this bracketed form
can run for several lines, do not interpret any escape sequences, and
ignore long brackets of any other level. Any kind of end-of-line
sequence (carriage return, newline, carriage return followed by
newline, or newline followed by carriage return) is converted to a
simple newline.
For convenience, when the opening long bracket is immediately followed
by a newline, the newline is not included in the string.
That's all you need to know about long strings.
It does not make much sense to convert a string that has been defined using quotes "some string" to a string like [[some string]] as both quotes and square brackets are not actually part of that string and the string itself is the same.
The only difference would be a leading newline which is ignored in square brackets or escape sequences which are not interpreted.
Quotes and square brackets are only part of the string if you have nested strings. In this case conversion also doesn't make much sense because you cannot nest strings with quotes like strings with brackets.
Maybe your whole approach is a bit off?
Do you look for something like this?
local db = "google"
local tbl = "accounts"
local where = "field = 'VALUE' AND TRUE"
local order = "id DESC"
local query = string.format([[
SELECT *
FROM `%s`.`%s`
WHERE %s
ORDER BY %s
]], db, tbl, where, order)
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.
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.
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