I'm a grade 11 student and I am coding in Basic language. I asked the user a question, and the user gets to choose yes or no. In the coding, a "type mismatch" error displayed. Also in the coding, after the question, I coded in some IFs and END IFs, and before I had this problem, it printed 0 when I ran the program.
I'm confused as to why I'm getting a type mismatch, even though my variables in the area of the IFs and END IFs. I have a string with $ at the end as the variable. This is what I have.
A part of the program is supposed to calculate the total insurance of someone renting a car, but the computer needs to know if the user even wants insurance. Even if I say I do want insurance, the program still displays 0 when it outputs the insurance at the end of the program.
So my question is, why am I getting a type mismatch?
Print "Do you want insurance?"
Print "1.Yes"
Print "2.No"
Input strInsurance$
IF strInsurance$=Yes THEN
strInsurance$=sngDailyInsurance*sngDays
END IF
IF strInsurance$=No THEN
strInsurance$=0
END IF
strInsurance$ is a string variable, so you'd have to surround Yes with quotes, like:
IF strInsurance$="Yes" THEN
And:
IF strInsurance$="No" THEN
Also, you cannot assign a number to a string variable like you do here:
strInsurance$=0
And here:
strInsurance$=sngDailyInsurance*sngDays
It's strange that you reuse the input variable for the result in any case.
The correct code would be:
Print "Do you want insurance?"
Print "1.Yes"
Print "2.No"
Input strInsurance$
IF strInsurance$="Yes" THEN
strInsurance$=str$(sngDailyInsurance*sngDays)
END IF
IF strInsurance$="No" THEN
strInsurance$="0"
END IF
Related
I'm comparing values of numbers from 2 data sheets, and I've dropped the relevant data from both into their own arrays. I need to find matching values to run other steps of analysis.
For i = Lbound(Array1) to UBound(Array1)
For j = LBound(Array2) to UBound(Array2)
If (criteria for Array2) then
variable = 11111
Else
variable = 22222
End if
If variable = Array1(i,1) Or variable = Array1(i,2) or variable = Array1(i,3) then
more steps
Else
more steps
End if
next j
next i
The first if statement sets the variable correctly, but the variable doesn't match any of the criteria. It doesn't go to the else like it should. Now I only know this because I walked through the code step by step. If I just F5 and run the thing, "Excel is not responding". I don't know what the hang up it. All of my variables are declared and assigned a type, I'm not missing any closing statements. And I have no idea what I'm doing wrong.
What do I need to check for in my code?
EDIT
Sorry, but in this instance I'm not allowed to upload any code here. It's work related, NDA kind of stuff. Hence the pseudo code. What I need to show wouldn't be a big deal(at least I don't think it would), but I'm not risking it.
My apologies.
The solution, as it turns out, has to do with a poorly named array(not me) and a simple typo(definitely me). I'm certain that would have been an easy solve for the good citizens of Stack Overflow if I would have been allowed to post actual code.
For what's it worth, I think it's dumb that I couldn't in this case. Thanks #ScottCraner and #SuperSymmertry for trying to be helpful even without much to go on.
Super, I'm still curious about Val. If you've got a minute, I would appreciate more knowledge on that. Anything from an actual person is better than Microsoft documentation.
I've started work on a Variable banner program, and I've hit a "phantom" syntax error.
name = input('Type here: ')
namelist = list(namelist)
print(namelist)
length_of_name=len(namelist)
asterisk=('*')
for length_of_name:
print (asterisk)
it throws up a syntax error, as I mentioned, but can anyone spot what I did wrong?
The SyntaxError is coming from your for statement. A for statement has to look like for variable in sequence_object:. Here sequence_object is any type that can be iterated, such as a list or tuple, but in your particular case for i in range(0,length_of_name): will make the code syntactically correct. But don't use this because there is a faster way which makes exactly the same output.
Instead of printing a single character in a loop, set asterisk to '*\n' and use print(asterisk*length_of_name, end=''). This prints the same output as the for loop but this multiplies the asterisk character by an integer to make a string that is repeated that many times. That way, you only print once. Setting end to an empty string ensures that a blank line is not printed.
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
Is there a way to prevent the program from executing characters in "Menu" type program where you choose options while running console and enter numbers only to do steps and browse data.
My program is a book catalog where you can review, change, add or delete information.
Things wouldn't be as bad but the the thing is, that when you enter a letter or non-number character the program gets messy and stuck.
I am not adding my code since I think that there should be a universal command to get rid of my problem + it would take some time to translate my code into EN.
The easiest way to do it is use the getline function I think. Instead of using cin >> num; for example, you would use getline(cin,value); where "value" is a string that you first store your input in. Then you can do num = atoi(value.c_str()); to save the string input as an integer value into "num". If "num" is a float, then just use atof instead of atoi. This way if the string is not a number, it just sets the value to 0. You could then use an if statement to display an error message if the value of num == 0. Hope that helps. Good luck!
If your application runs in a console and you are worried about wrong (non-number) input, perhaps you should the input not as a number, but as a text and then parse it to see it is really a number. If not, tell the user the input is incorrect and let them enter it again.
I'm having some trouble automating a matlab script which should prompt the user for the variable they are interested in, as well as the date range they want. I then want the script to concatenate their answers within a naming convention for the file they will ultimately load.
variable=input('please input variable of interest');
% temp
start=input('please state the start date in the form yymmdd: ');
%130418
enddate=input('please state the end date in the form yymmdd: ');
%140418
file=sprintf('%s_dailydata_%d_%d.csv',variable,start,enddate);
%so I thought 'file' would look like: temp_dailydata_130418_140418.csv
vardata=load(file);
the two numbers representing the dates are not causing any issues, but the fact that 'variable' is a string is. I know if I put apostrophes before and after 'temp' when I'm promted, it will work, but I have to assume that the end user won't know to do this. I tried putting curly braces around 'please input your variable..', but that didn't help either. Obviously this approach assumes that the date being requested exists in a filename.
Can anyone offer any advice? Perhaps the sprintf function is not the best option here?
Don't use 'end' as a variable name, it is a reserved name and using it could create conflicts with any function or logic block you're defining.
If you know your input is going to be a string: from the documentation for input()
str = input(prompt,'s')
Returns the entered text as a MATLAB string, without evaluating expressions.
As for knowing whether or not the file exists, that's something you'd have to incorporate some error logic for. Either a try/catch block with your load() call or you could use uigetfile() to get the filename.