I have a list of Employee, and I want to retrieve only one Employee information with the specific name:
public static Employee getAllEmployeeDetails(String employeeName) {
List<Employee> empList = getAllEmployeeDetails();
Employee employee = empList.stream().filter(x -> x.equals(employeeName));
return employee;
}
Please let me know how to filter the data and return a single element.
This will locate the first Employee matching the given name, or null if not found:
Employee employee = empList.stream()
.filter(x -> x.getName().equals(employeeName))
.findFirst()
.orElse(null);
You're looking for findFirst or findAny:
empList.stream()
.filter(x -> x.getName().equals(employeeName))
.findFirst()
or
empList.stream()
.filter(x -> x.getName().equals(employeeName))
.findAny();
However, I'd suggest changing the method return type with the use of Optional which is intended for use as a method return type where
there is a need to represent "no result" . in other words we let the "client" decide what to do in the "no value" case. i.e.
public static Optional<Employee> getAllEmployeeDetails(String employeeName) {
return empList.stream()
.filter(x -> x.getName().equals(employeeName))
.findFirst()
}
you could have done .findFirst().orElse(null) but dealing with nullity is sometimes dangerous and this is a perfect opportunity to leverage the Optional<T> API.
orElseThrow if the get is critical based on the employeeName, thoughts?
Hi You can use multiple ways to get single object from list here i just modified the method names and also some variable names so that the code should be readable and also try to use Optional as it handles null pointer
public static Employee findEmployeeByName(String employeeName) {
List<Employee> empList = getAllEmployeeDetails();
Optional<Employee> employee = empList.stream()
.filter(employee -> employee.getName().equalsIgnoreCase(employeeName))
.findFirst();
return employee.isPresent() ? employee.get() : null; // Instead of null you can also return empty Employee Object
}
