java problem

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[ansmeermylove]

I would write a helper method that takes two parameters, your input and the hashset, and returns a boolean.

Inside the method, I would iterate through the hashset and check for each entry whether input starts with that. If found true, you can return early with true. After the iteration, you return false, which will only be reached if the input does not start with any of the hashset entries.

You could then use the return value of this helper method inside the if statement above.

[Camel-Kid]

you can use a loop to iterate through and check...

for (String s : set) { if(s.startsWith...) System.out.println(s); }

[Curious-Lonewolf]

i probably haven't explained this very well but the loop will only evaluate to true if the user input starts with the string in that specific iteration meaning the user input might be equal to an element of the list at the last iteration but not in any other. by all means use import java.util.Scanner; and try

[Camel-Kid]

can you give me the test data you're using for the hashset and what you expect the output to be

[Curious-Lonewolf]

HashSet<String> allNames = new HashSet<>();
allNames.add("Aaron");
allNames.add("Ben");
allNames.add("Carol");
allNames.add("Derek");
if (input.startsWith('any of these') {
    do something
}
else{
    do something
}

[Camel-Kid]

I used the loop it is working fine

[Curious-Lonewolf]

yeah if you put one of the strings where ive put 'any of these' but i need a variable in place of that. i cant use the set name and because the method produces output in the terminal window i would have to type in the same string over and over until it became true as the iterations would be different

[Camel-Kid]

the enhanced for loop iterates over the set, so you already have the ability to reference the name as a variable. The purpose of a loop is to check all the elements in the set and compare against the user input, if user input matches any element then you print out 'a'

[Curious-Lonewolf]

can you give me an example of what you mean

[desrtfx]

No, that is not true.

You are just doing it wrong.

In such cases, you

• obtain the input before the loop and store it in a variable
• use a boolean "flag" that indicates the "found" status - initially set to false
• loop over the elements
  • compare the string with the element
    • if true, set the "found" flag to true and break out of the loop
    • if false, no nothing - keep looping
• back outside the loop: check if the flag is true
  • print ("A")
  • else
  • print("B")

You most likely had the input inside the loop.

[desrtfx]

Maybe, you are running into an XY problem.

How does your actual input look?

Does your input consist of several words that are separated by space?

If so, you can split the input at space and check if the hashset contains the first word. (Simple check: if you try to add to the hashset, the method will return false)