I use the following code in PHP to validate if a password includes invalid characters
function validate_password($str) {
if(strlen($str) < 5 || strlen($str) > 30) {
return false;
} else {
return preg_match('/^[-._A-Za-z0-9]*$/',$str);
}
}
I'm trying to do the same in Lua but not sure the best way to check if a string contains only certain characters. My valid characters are alphanumeric with the addition of "-" and "."
local function validate_password(str)
if string.len(str) < 5 or string.len(str) > 30 then
return false
else
if(str:match("%W")) then
-- this isn't right. need to check if alpha numeric or - or .
return false
else
return true
end
end
end
Appreciate any help!
return #str >= 5 and #str <= 30 and str:find('^[%-%.%w_]+$') ~= nil
Note 1: as mentioned in the comments, ~= nil part can be removed as the position returned by find will be evaluated as true (when it's not nil) in logical operators. However, since it's a number, it will fail when used in validate_password(str) == true comparisons (since 1 == true is not true). An alternative is to use not str:find('[^...]') as suggested by Egor.
Note 2: as mentioned in the comments, - and . don't need to be escaped in this particular case. However, since - is a range operator inside square brackets [_-.] is not the same as [-._], as it will match nothing as the range of symbols between _ and . is empty. It's safe to escape - and . even inside square brackets.
Related
I am having some issues with some code I wrote for this problem:
“Write a function namedd calc that will evaluate a simple arithmetic expression. The input to your program will be a string of the form:
operand1 operator operand2
where operand1 and operand2 are non-negative integers and operator is a single-character operator, which is either +, -, or *. You may assume that there is a space between each operand and the operator. You may further assume that the input is a valid mathemat- ical expression, i.e. your program is not responsible for the case where the user enters gibberish.
Your function will return an integer, such that the returned value is equal to the value produced by applying the given operation to the given operands.
Sample execution:
calc("5 + 10") # 15
“You may not use the split or eval functions in your solution.
Hint: the hard part here is breaking the input string into its three component. You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters. Be careful of off-by-one errors in using the slice operator.
Hint: it’s best to test your code as you work. The first step should be to break the input string into its three components. Write a program that does that, have it print out the operator and the two operands on separate lines, and test it until you are convinced that it works. Then, modifying it to perform the desired mathematical operation should be straightforward. Test your program with several different inputs to make sure it works as you expect.”
Here is my code:
def calc(exp):
operand1 = int(exp[:1])
operand2 = int(exp[4:6])
operator = exp[2:3]
if(operator == "+"):
addition = operand1+operand2
return addition
if(operator == "-"):
subtraction = operand1-operand2
return subtraction
if(operator == "*"):
multiplication = operand1*operand2
return multiplication
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
My code does not fully meet the criteria of this question. It only works for single digit numbers. How can I make my code work for any number?
Like:
“504 + 507”
”5678 + 76890”
and so on?
Thank you. Any help is appreciated.
As the hint says, get the position of the first and last space of the expression, use it to extract the operand and the operators, and then evaluate accordingly.
def calc(exp):
#Get the position for first space with find
low_idx = exp.find(' ')
#Get the position for last space with rfind
high_idx = exp.rfind(' ')
#Extract operators and operand with slice, converting operands to int
operand1 = int(exp[0:low_idx])
operator = exp[low_idx+1:high_idx]
operand2 = int(exp[high_idx:])
result = 0
#Evaluate based on operator
if operator == '+':
result = operand1 + operand2
elif operator == '-':
result = operand1 - operand2
elif operator == '*':
result = operand1 * operand2
return result
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
print(calc("504 + 507"))
print(calc("5678 + 76890"))
#15
#-4
#12
#1011
#82568
The answer is in the specification:
You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters.
find and rfind are methods of string objects.
You could split it into three components using this code: (note: this doesn't use split or eval)
def splitExpression(e):
numbers = ["1","2","3","4","5","6","7","8","9","0"] # list of all numbers
operations = ["+","-","*","/"] # list of all operations
output = [] # output components
currentlyParsing = "number" # the component we're currently parsing
buildstring = "" # temporary variable
for c in e:
if c == " ":
continue # ignore whitespace
if currentlyParsing == "number": # we are currently parsing a number
if c in numbers:
buildstring += c # this is a number, continue
elif c in operations:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "operation" # we are expecting an operation now
else:
pass # unknown symbol!
elif currentlyParsing == "operation": # we are currently parsing an operation
if c in operations:
buildstring += c # this is an operation, continue
elif c in numbers:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "number" # we are expecting a number now
else:
pass # unknown symbol!
if buildstring: # anything left in the buffer?
output.append(buildstring)
buildstring = ""
return output
Usage: splitExpression("281*14") returns ["281","*","14"]
This function also accepts spaces between numbers and operations
You can simply take the string and use the split method for the string object, which will return a list of strings based on some separator.
For example:
stringList = "504 + 507".split(" ")
stringList will now be a list such as ["504", "+", "507"] due to the separator " " which is a whitespace. Then just use stringList[1] with your conditionals to solve the problem. Additionally, you can use int(stringList[0]) and int(stringList[2]) to convert the strings to int objects.
EDIT:
Now I realized that your problem said to use find() instead of split(). Simply use the logic above but instead find(" ") the first whitespace. You will then need to find the second whitespace by slicing past the first whitespace using the two additional arguments available for find().
You need to split the string out instead of hard coding the positions of the indexes.
When coding you want to try to make your code as dynamic as possible, that generally means not hard coding stuff that could be a variable or in this case could be grabbed from the spaces.
Also in the if statements I modified them to elif as it is all one contained statement and thus should be grouped.
def calc(exp):
vals = exp.split(' ')
operand1 = int(vals[0])
operand2 = int(vals[2])
operator = vals[1]
if operator == '+':
return operand1+operand2
elif operator == '-':
return operand1-operand2
else:
return operand1*operand2
I have a string-array, myString = ' 23.654 ' and a character, Char = '.'. Is there some way to use the find function to determine the index of Char in myString?
You can use strfind which will identify the locations where one string exists within another.
index = strfind(myString, '.')
Or you could use == combined with find. The == performs an element-wise equality check between each character in the string and your character of interest. It will yield a logical array that is true where the character occurs, and false otherwise. If you need the actual index, find will return the location of all of the true values.
index = find(myString == '.')
I'm trying to check whether or not all chars in my string are either one of these numbers: 1,2,3,4,5,6,7,8,9,0 OR a decimal point and whether they contain spaces or not. I'm using this piece of code but it's not working and I can't figure out why:
print("Confirming added item")
local validAmountString = true
print(addScreen.amountTxtField.text)
for i = 1, #addScreen.amountTxtField.text do
local c = addScreen.amountTxtField.text:sub(i,i)
print("Char " .. i .. " in the amount textflied is " .. c)
if c == "0" or c == "1" or c == "2" or c == "3" or c == "4" or c == "5" or c == "6" or c == "7" or c == "8" or c == "9" or c == "." then
validAmountString = true
else
validAmountString = false
end
print(validAmountString)
end
if string.gsub(addScreen.nameTxtField.text, "%s+", "") ~= "" and string.gsub(addScreen.amountTxtField.text, "%s+", "") ~= "" and validAmountString == true then
--Proceed...
end
The spaces part is working, but for some reason, when I try to check whether validAmountString == true in my if statement, it stops working. The validAmountString returns the correct boolean value according to my print statement though.
Any help? :)
An easier way to go about this would be string.find
local validAmountString = not addScreen.amountTxtField.text:find( "[^%d%.]" )
I'm afraid I can't tell why validAmountString is not true, but you don't have to compare boolean values.
if (validAmountString) then
--is equivilent to
if (validAmountString == true) then
For checking your strings refer to KingOfGamesYami's answer. He's using something called string patterns btw. Check the Lua reference http://www.lua.org/manual/5.3/manual.html#6.4.1 for details.
"[^%d%.]" matches every character that not a digit or a dot. So if find returns something but nil, your string is invalid
For why your program isn't working as is (in case you want to stick to your complicated way:
You check every character of your string. If it is a number or a dot you asign true, if it's not you asign false.
So let's say you have "12A3" as a string.
Then the final value of validAmountString is true because the last character is a number. So you assign true 2 times, then false for the "A" and after that you asign true again. You mess up your total result.
What you should do here is either assign validAmountString = validAmountString and true, else validAmountString = false so you can never overwrite a false, once set.
The even better and more efficient way would be to simply stop your check once you've hit an invalid character. There is no need to check further characters if you find one invalid.
So simply add a break after validAmountString = false
You do not need the loop to check whether it does or does not have valid characters.
You can simply do tonumber(string). It will return nil if there is any character other than a digit in the string.
I'm trying to do a library in Lua with some function that manipulate strings.
I want to do a function that changes the letter case to upper only on odd characters of the word.
This is an example:
Input: This LIBRARY should work with any string!
Result: ThIs LiBrArY ShOuLd WoRk WiTh AnY StRiNg!
I tried with the "gsub" function but i found it really difficult to use.
This almost works:
original = "This LIBRARY should work with any string!"
print(original:gsub("(.)(.)",function (x,y) return x:upper()..y end))
It fails when the string has odd length and the last char is a letter, as in
original = "This LIBRARY should work with any strings"
I'll leave that case as an exercise.
First, split the string into an array of words:
local original = "This LIBRARY should work with any string!"
local words = {}
for v in original:gmatch("%w+") do
words[#words + 1] = v
end
Then, make a function to turn words like expected, odd characters to upper, even characters to lower:
function changeCase(str)
local u = ""
for i = 1, #str do
if i % 2 == 1 then
u = u .. string.upper(str:sub(i, i))
else
u = u .. string.lower(str:sub(i, i))
end
end
return u
end
Using the function to modify every words:
for i,v in ipairs(words) do
words[i] = changeCase(v)
end
Finally, using table.concat to concatenate to one string:
local result = table.concat(words, " ")
print(result)
-- Output: ThIs LiBrArY ShOuLd WoRk WiTh AnY StRiNg
Since I am coding mostly in Haskell lately, functional-ish solution comes to mind:
local function head(str) return str[1] end
local function tail(str) return substr(str, 2) end
local function helper(str, c)
if #str == 0 then
return ""
end
if c % 2 == 1 then
return toupper(head(str)) .. helper(tail(str),c+1)
else
return head(str) .. helper(tail(str), c+1)
end
end
function foo(str)
return helper(str, 1)
end
Disclaimer: Not tested, just showing the idea.
And now for real, you can treat a string like a list of characters with random-access with reference semantics on []. Simple for loop with index should do the trick just fine.
I'm looking for a Visual FoxPro function which is similar to the PHP function is_numeric().
I have found this, but I could not use VARTYPE or TYPE because the variable is always a character string which contains digits only.
I found ISDIGIT() function, but the manual says that it only checks the first character.
Determines whether the leftmost character of the specified character
expression is a digit (0 through 9).
ISDIGIT(cExpression)
Parameters
cExpression
Specifies the character expression that ISDIGIT( ) tests. Any
characters after the first character in cExpression are ignored.
I would create my own function using the regular expression object VBScript.RegExp
FUNCTION isNumeric( tcValue )
LOCAL oRE
oRE = CreateObject("VBScript.RegExp")
oRE.Pattern = '^[0-9]+$'
RETURN oRE.test( tcValue )
ENDFUNC
? isNumeric( '123' )
But, is there any function provided by FoxPro for this purpose?
Am I just overlooking?
Also same for ISALHPA() which determines whether the leftmost character in a character expression is alphabetic. I want to check if the variable contain only alphabets.
You can create your own function like this.
FUNCTION IsAllDigits
LPARAMETERS tcSearched, tcOptionalSearch
* tcSearched = the string of characters to test.
* tcOptionalSearch = optional, additional characters to allow.
LOCAL lcSearch
m.lcSearch = "01234567989" + IIF(VARTYPE(m.tcOptionalSearch) = "C", m.tcOptionalSearch, "")
LOCAL lcRemaining
m.lcRemaining = CHRTRAN(m.tcSearched, m.lcSearch, "")
RETURN ( LEN(m.lcRemaining) = 0 )
ENDFUNC
FUNCTION ISNUMERIC
LPARAMETERS cVal
LOCAL llNumeric, lnLen, lcChr, lnDecs, lnVal
llNumeric = VARTYPE(cVal) = "N" && Donkey has sent a numeric value
lnDecs = 0
DO CASE
CASE llNumeric
CASE VARTYPE(cVal)<>"C" && Not a character
OTHERWISE
cVal = ALLTRIM(cVal) && Trim spaces
lnLen = LEN(cVal) && How many characters
llNumeric = .T. && Assume
i = 0
DO WHILE llNumeric AND i<lnLen
i = i+1
lcChr = SUBSTR(cVal,i,1) && Get next char
lnVal = VAL(lcChr)
DO CASE
CASE lcChr = "0" && Allowed
CASE lnVal>0 && 1 - 9 OK
CASE INLIST(lcChr, "-", "+") && Allowed but ONLY at the start
llNumeric = i = 1
CASE lcChr = "." && Decimal point but ONLY one
lnDecs = lnDecs+1
llNumeric = lnDecs = 1
OTHERWISE
llNumeric = .F.
ENDCASE
ENDDO
ENDCASE
RETURN llNumeric
ENDFUNC
This could work for ISDIGIT() or ISALPHA().
Function IsAllDigits(myValue)
lReturn = .t.
FOR i = 1 TO LEN(myvalue)
IF !ISDIGIT( SUBSTR(myValue, i, 1) )
lReturn = .f.
EXIT
ENDIF
ENDFOR
RETURN lReturn
ENDFUNC
How about a one liner?
Function IsNumeric
Lparameters pString
Return m.pString == Chrtran(m.pString, Chrtran(m.pString, "0123456789", ""), "")
EndFunc
You can any other valid characters to "0123456789" like "." or ","
There is more simple to test if a string is numeric or not :
If String="123" => val('String')>0
If String="AB123" => val('String')=0
That's all...
Using only Visual Fox Pro, you can do something like this:
FUNCTION is_numeric(var_name)
error_trigger=.f. &&** first initialization
&&** evaluate("any_char") will generate an error so I'm using TRY-CATCH-ENDTRY
TRY
EVALUATE(var_name)
CATCH &&** evaluate() generates an error then the string is character
WAIT WINDOW "character"
error_trigger=.t. &&** there was an error returned by evaluate()
EXIT && stop and jump after ENDTRY
ENDTRY
IF error_trigger=.f. &&** there was no error returned by evaluate() then the string was numeric
WAIT WINDOW "numeric"
ENDIF
ENDFUNC
Then call to the function:
is_numeric("333") will show: numeric
is_numeric("aaa") will show: character
is_numeric("333a333") will show: character
I hope it will help you