Im trying to perform a search in DXL of a string that ends with specific characteres Im not able to find the way to perform this.
Example, I'm looking for
" A: 23.1.23.2.4"
But if this contains at the end the character "~" the find function does not work
Example Where the skip list contains "A: 12.2.1.4.5~ text text text text"
I just need to know in the object.text contains A: 12.2.1.4.5
string string_text = "A: 12.2.1.4.5"
if(find(skip[i],string_text,string_text)){
modify_attributes(req_text)
}else{
output << "stgring not found : "
}
use a regular expression, like this
void modify_attributes (string fulltext) {print "modifying.."}
string fulltext = "A: 12.2.1.4.5~ text text text text"
Regexp searchme = regexp2 "A: 12.2.1.4.5"
if(searchme (fulltext)){
modify_attributes(fulltext)
}else{
print "string not found "
}
The "find"-method for Skip lists is O(1), if I am not mistaken. But for that to work properly, the key, you are asking for, has to match exactly.
So, to benefit from the speed of value-retrieval by the find method, I suggest, that you have a look at your code part, where you put stuff into your Skip, (only put "clean" information in the Skip, which you know, you want to ask for later on).
That of course only works, if you have the possibility to do so, i.e. you don't get the Skip from somewhere you don't have control over..
Related
Often I'm facing lines like
result = 'Some text'
result += some_text_variable if some_text_variable.present?
And every time I want to replace that with something more accurate but I don't know how
Any ideas plz?
result += some_text_variable.to_s
It will work if some_text_variable is nil or empty string for example
But it always will concat empty string to original string
You can also use
result += some_text_variable.presence.to_s
It will work for all presence cases (for example for " " string)
You could "compact" and join an array, e.g.
['Some text', some_text_variable].select(&:present?).join
I realise this is a longhand form, just offering as an alternative to mutating strings.
This can look a bit nicer, if you have a large number of variables to munge together, or you want to join them in some other way e.g.
[
var_1,
var_2,
var_3,
var_4
].select(&:present?).join("\n")
Again, nothing gets mutated - which may or may not suit your coding style.
I am working on a project, in which you type your input sentence, and I need to be able to use " and ' in the sentence, such as Input = "I said, "Hi what's up?" print(Input) in which I get an error. If anyone knows how to fix this that would be great.
See https://www.lua.org/pil/2.4.html. Lua has very interesting feature to declare string with square brackets:
input = [[I said, "Hi what's up?"]]
input = "I said, \"Hi what's up?\""
input = 'I said, "Hi what\'s up?"'
I will tell some things in addition to what #Darius told above
When you tried to add a quatation mark inside a string, the lua interpreter get confused and break your string after the next quation mark without reaching the end of the line. That's the reason for the error.
Try to understand it by the following code
str = "Hello I"m somebody" -- here the interpreter will think str equals to "Hello I" at first, and then it will find some random characters after which may make it confused (as m somebody is neither a variable nor a keyword)"
-- you can also see the way it got confused by looking at the highlighted code
--What you can do to avoid this is escaping the quotes
str = "Hello I\"m somebody" -- here the interpreter will treat \" as a raw character (") and parse the rest.
You can also use the escape character () with others such as \', \", \[, \n (newline character), \t (tab) and so on.
I'm fairly new to this forum. I am having trouble with manipulating the correct string to achieve this.
Basically, what I'm trying to do is receive an input string like this example:
str = "Say hello to=Stack overflow, Say goodbye to=other resources"
for question, answer in pairs(string.gmatch(s, "(%w+)=(%w+)"))
print(question, answer)
end
I want it to return: question = "Say hello to" and answer = "Stack overflow, question = "Say goodbye to" and so on and so forth. but instead, it picks up the word just before the equal sign and the word just after. I've even tried the * quantifier, and it does the same exact thing.
I've also tried this pattern
[%w%s]*=[%w%s]
I just want to be able to sort this string into a key-value table where the key is all words before each = and the value is all words after that equal but before the comma.
Does anyone have a suggestion?
You can use something like this:
local str = "Say hello to=Stack overflow, Say goodbye to=other resources"
for question, answer in string.gmatch(str..",", "([^=]+)=([^,]+),%s*") do
print(question, answer)
end
"([^=]+)=([^,]+),%s*" means the following: anything except = ([^=]) repeated 1 or more times (+) followed by = and then anything except ',', followed by comma and optional whitespaces (to avoid including them in the next question). I also added comma to the string, so it parses the last pair as well.
To elaborate a bit further per request in the comments: in the expression [^=]+, [=] designates a set with one allowed character (=) and [^=] negates that, so it's a set with any character allowed except = and + allows the set to be repeated 1 or more times.
As #lhf suggested you can use a simpler expression: (.-)=(.-),%s*, which means: take all characters until the first = (- makes matching non-greedy) and then take all characters until the first ,.
Given a string like "Whatup <b>whatever<b> \n", i need to turn that into "Whatup whatever".
I'm pretty close with my below method, but I can't find a good way to remove dynamic & and type codes. I don't want to gsub each out (like i'm doing with the comma) -- There are hundreds of thousands of rows and many different codes in them...blah
Any pointers are welcome.
def self.clean_string(st)
return strip_tags(st).force_encoding("UTF-8").gsub(",","").squish if st and st != ""
end
For the HTML entities, add this regex replacement:
.gsub(/&[^;]+;/, '')
It will remove any &-style entity from the text.
Frustratingly, any my previous Lua tries went in extensive Google searching of more/less same Lua resources, and then resulted in some multi-line code to get basic things, which i.e. I get from Python with simple command.
Same again, I want to replace substring from string, and use i.e.:
string.gsub("My string", "str", "th")
which results in:
My thing 1
I imagine replacement count can be useful, but who would expect it by default, and without option to suppress it, but maybe I miss something?
How to print just string result, without counter?
Enclose in parentheses: (string.gsub("My string", "str", "th")).
The results are only a problem because you are using print, which takes multiple parameters. Lua allows multiple assignments, so normally the code would look like
newstr, n = string.gsub("My string", "str", "th")
but the count is only provided if there is a place to put it, so
newstr = string.gsub("My string", "str", "th")
is also fine, and causes the count to be discarded. If you are using print directly (the same applies to return) then you should enclose the call in parentheses to discard all but the first result.