How to solve the multiple name problem?

Since you set found = true in your loop if a name matches, but still loop through all names, your code will always print the last student if a student has been found. So you will need a means of identifying which student was found.

Instead of a boolean you could make found an array of found students. Then outside the loop check the length of the array to determine if there are students to print. You'd have to deal with the same name searched more than once. But this would allow you to print multiple positive search results as well as two or more students with the same name.

The suggestion of the boolean test was intended to go after the search (inner) loop, but still inside the main loop. And that test was just for showing the "not found" message.

The printing of the student record can be done in the same place as where the boolean is set to true, and that way it will run for each student found. And don't forget to set it back to false before beginning another search.

Kris's suggestion of building an array of found students is another viable strategy, but it would require another loop after the main one to display the array.

i tried to implement what u suggested and it doesn't working as expected this is the current snapshot: https://w.trhou.se/eg63ptww9i

i set the flag to false after the loop ends and when even if i serach an existing name i got the message - user not found.

Here is my solution, for what it's worth...

var message = ''
var student;
var search;
var correct = []
var searchHistory = []
function print(message) {
    let div = document.getElementById('output');
    div.innerHTML = message
}

function getStudentReport (student) {
    var report = '<h2>Student: ' + student.name + '</h2>'
    report += '<p>Track: ' + student.track + '</p>'
    report += '<p>Achievements: ' + student.achievements + '</p>'
    report += '<p>Points: ' + student.points + '</p>'
    return report
}


while (true) {
    search = prompt('Please enter the student name you require the records of, otherwise type \'quit\' to exit the program')
    if (search.toLowerCase() === 'quit' || search === null) {
        break
    }
    searchHistory.push(search)
    for (let i = 0; i < students.length; i++) {
        student = students[i]
        if (student.name === search) {
            message += getStudentReport(student)
            print(message)
            correct.unshift(student.name)
        } 
        if (correct[0] === correct[1]){
            correct.shift()
        }
    }
    if (correct.length < searchHistory.length) {
        alert('There are no students by that name, sorry!')
        searchHistory.pop()
    }
} 

The correct array was added primarily to later measure its length against the length of the searchHistory array.
So, for each time through the loop of students' names, if the search matched student.name, correct.length added 1. If there were 2 instances of the same name, correct length would still only be incremented by 1 because of the correct.shift(). The alert message would only be called after each prompt if , after iterating through the loop, there were no matching strings. The searchHistory.pop() removed the incorrect array item so that array.length wouldn't keep getting bigger casuing the alert message to continually pop up when it shouldn't.

Side note, I haven't tested this with 3 or more of the same student.name value. It probably wouldn't work... :)

If you want to see my modification to your code here it is. I'm sure it could use some work but this is what I was talking about.

Note: There are three idans. If you search for one all three are returned.

i need to put it before or after the while map? Steven Parker

In my code found is an array of student objects

if(student.name === searchName) 

matches the name then if true a check is done to see if the student object exists in the array

if (found.includes(student) === false) 

In this case the entire student object not just the name is verified. That way when the loop proceeds and finds the second idan by name, it is not a second item in the array as an object, so the second idan is added.

If you were to search for idan a second time the objects would be found so they would not be added. That way they wouldn't double up.

w3schools on includes()