可以用string.match如果你知道的格式來解決:
str = "1791 (AR6K Async) S 2 "
s1 = str:match("(%d%d%d%d)%s%(.*%)%s.+%s.+")
s2 = str:match("%d%d%d%d%s(%(.*%))%s.+%s.+")
s3 = str:match("%d%d%d%d%s%(.*%)%s(.+)%s.+")
s4 = str:match("%d%d%d%d%s%(.*%)%s.+%s(.+)")
print(s1)
print(s2)
print(s3)
print(s4)
另一個通用的解決方案,允許可變數量的條目(嘗試它:簡單地過去在lua解釋器中):
function get_separate_words(str)
local i = 1
local words = {}
function get_parentheses_content(str,is_recursively_called)
local i = 1
local function split(s, sep)
local fields = {}
local sep = sep or ":"
local pattern = string.format("([^%s]+)", sep)
string.gsub(s, pattern, function(c) fields[#fields + 1] = c end)
return fields
end
for j = 1,#str do
local c = string.sub(str,j,j)
local d = string.sub(str,j+1,j+1)
if j <= i then
elseif c == "(" then
i = j + #get_parentheses_content(string.sub(str,j+1,#str),true) + 2
elseif c == ")" and (is_recursively_called or (d == " ") or (not d)) then
print('c')
local parentheses_content = string.sub(str,1,j-1)
return {parentheses_content}
end
end
local parentheses_content = string.match(str,"^(.*)%)%s+[^)]*$")
if parentheses_content then print('a') end
parentheses_content = parentheses_content or string.match(str,"^(.*)%)$")
if parentheses_content then
print("A")
return {parentheses_content}
else
print("B")
return split("("..str," ")
end
end
local function merge(table_a, table_b)
table_a = table_a or {}
table_b = table_b or {}
for k_b, v_b in pairs(table_b) do
if type(v_b) == "table" and type(table_a[k_b] or false) == "table" then
merge(table_a[k_b], table_b[k_b])
else
table_a[k_b] = v_b
end
end
return table_a
end
for j = 1,#str do
local c = string.sub(str,j,j)
if j < i then
elseif c == " " or j == #str then
local word = string.gsub(string.sub(str,i,j)," ","")
if #word > 0 then
table.insert(words, word)
print(word)
end
i = j+1
elseif c == "(" then
local all_characters_after_opening_parentheses = string.sub(str,j+1,#str)
local parentheses_content = get_parentheses_content(all_characters_after_opening_parentheses)[1]
table.insert(words, parentheses_content)
j= j+#parentheses_content+2
i = j
end
end
return words
end
separate_words = get_separate_words("1791 (AR6(K As)ync) S 2)")
for k,v in ipairs(separate_words) do print(k,v) end
可能重複[分割一個項目在括號中的字符串](http://stackoverflow.com/questions/39755445/splitting-a-string-where-one-item-is-in-parentheses) –
嘗試改善你現有的問題,而不是打開一個關於同一問題的新問題。 – Piglet
你已經問過這個... – warspyking