My delay codes are always 3 digits. Two letters a dash (-) and a number. I am trying to use a single line of code to detect either MT or DA, the actual classification number is irrelevant, so I want the message box to fire on the two letters only.
The code looks right, but it doesn't fire as it should. If I take out the wild card it works. I think I have a problem with the concatenation, but I'm not sure. I tried putting () brackets around it but that was not help.
Additionally I tried using an or statement to capture the MT code on the other side but got nothing but an error code for type mismatch. Any ideas?
If Range("L24").Value = "DA" & "*" Then
MsgBox "The flight had a Maintenance delay"
Else
End If
A simple solution to this kind of problem would be to ignore the wildcard altogether and check the first two digits:
If Left(Range("L24").Value, 2) = "DA" Then
Related
i am a complete noob to programming and i got an assignment from the online course where I am learning, mine also gives the output, but it is not same as the instructor's method. This works for me and this method was easy for me.
but i am not sure is this the right method or had i done something wrong? could someone help?
I was not allowed to use choice()
**
import random
names_string = input("Give me everybody's names, separated by a comma. ")
names = names_string.split(", ")
a=len(names)
random_name=random.randint(0,a)
print(f"{names[random_name]} is going to pay the bill")
**
Welcome!
First of all you need to describe what exactly your problem and what is the problem you are facing.
It looks like you even have not tried to run that code. I will recommend an online interpreter to you to test your code on. You may use that
https://www.online-python.com/
Secondly "import" statement must be in lower case not "Import"
Finally the code works only incase of the string (names) is separated by a comma followed by space ", " same as the split string used. for example "a,b,c" is not going to work, but "a, b, c" does
random.randint(0,a) (a is included; may be returned) so it must be a-1 to avoid IndexError: list index out of range
Fixed code
import random
names_string = input("Give me everybody's names, separated by a comma. ")
names = names_string.split(", ")
a=len(names)
random_name=random.randint(0,a-1)
print(f"{names[random_name]} is going to pay the bill")
no, the code is not right.The error can be solved by replacing "Import" by "import".
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
I have a calculation which will remove a blank space and replace with a full stop. This is correct for 90% of my cases. However, sometimes two blanks will appear in my value. For the second space I want to delete it. Is this possible?
I think it may be possible using a code stage, but I am not sure what the code would be.
My current calculation is Replace([Item Data.Name], " ", ".")
Example data John B Smith I want the result to be John.BSmith
For anything that'd like to do with the strings, there is a really powerful tool called Regular Expressions (regex). I encourage you to play with it, because it's a really powerful tool in the hands of RPA developer.
To replace the second space in any string with a "." you can use the following action.
Object: Utility - Strings
Action: Regex - Find and Replace
Input:
Regex Pattern: "(?<= .*) "
Text: "John B Smith"
Replacement: "."
The above action is not a standard Blueprism one, so it has to be added to your VBO. The action looks as follows:
The VB.net code for that action is as follows:
Dim R as New Regex(Regex_Pattern, RegexOptions.SingleLine)
Dim M as Match = R.Match(Text)
replacement_result = R.Replace(Text,Regex_Pattern,replacement_string)
There might be a need for some additional assemblies, so please see below a printscreen of references and namespaces used in my object:
I resolved this issue by using the Utility - Strings object and the split text action. I split my name by space. This outputted a collection which I was then able to loop through and add a full stop after the fist instance but then trim the other instances.
Please see screenshot
I think the simplest solution would be
Replace(Replace(Text," "," ")" ","."))
if you know that it will give one or two spaces
First replace the two white spaces to single and then again single white space to dot(.)
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
I just wrote a code with the criteria above, but it doesn't seem to work properly because I either miss a letter at the end or in the middle.
Could anyone please check out my code an tell me what I'm doing wrong. By the way I already checked other threads on this similar problem, but I'm not allowed to use regex or print function.
phrase=('my room is cold')
allSpaces=findstr(' ',phrase);
k=length(allSpaces)
acr=phrase(1:allSpaces(1):allSpaces(k)-1)
Output:
acr= mrms
Change last line to
acr = phrase([1 allSpaces+1])
That way you get the first letter, and then the first after each space.