String.split（）导致创建一个空字符串，如何防止这种情况？

Question

I have a string that is a concatenation of smaller strings of the form refNum orderNum , as in a string that contains a reference number and the amount of items on an order eg BRICK1 10 . The larger concatenated string is a number of these smaller string pieced together and using a pound sign to divide them eg #BRICK1 10#BRICK2 20#BRICK3 30 .

When I use String.split() I'd like the result to be an array that has divided the larger string into its smaller elements so BRICK1 10 , BRICK2 20 and BRICK3 30 , however when I actually try to split the string using "#" as a delimiter I get a fourth string at the start of the array that is empty.

I'm not sure why this is happening, how do I prevent it?

My long string is produced using this method

public String getAllOrders()
    {
        String orderDetails = "";

        for(BrickOrder o : orderList)
        {
            String order = "#" + o.getReferenceNumber() + " " + o.getNumberOfBricks();
            System.out.println("ORDER: " + order);
            orderDetails = orderDetails.concat(order);
        }

        return orderDetails;
    }

Whilst the split method is used within a test class, here.

@Test
    public void testFullOrderListReturnedCorrectly()
    {
        sys.createOrder(10);
        sys.createOrder(100);
        sys.createOrder(1000); //Create three orders

        String orderList = sys.getOrders(); //Return full order list
        System.out.println(orderList);

        String[] orders = orderList.split("#");
        System.out.println(orders.length);
    }

Answer 1

You get an empty string at the start of the array because the string you are splitting starts with you chosen delimiter ( # ). You can prevent it by removing it from the beginning before splitting (using method String.substring(int) for example).

Example solution:

String[] orders = orderList.substring(1).split("#");

Answer 2

That's what split does. It splits your string by using the delimiter and keeps the string, even if its empty, if it's not the last string.(trailing empty strings are excluded) You could use a regex instead and a Pattern match. But in this case I'd recommend to just check your orderList before you split it and remove the # if it's the first character:

String[] orders;
if (orderList.charAt(0) == '#') {
    orders = orderList.substring(1).split("#");
} else {
    orders = orderList.split("#");
}

Answer 3

I believe the issue is because of the "#" that you are appending to the start of the order string.

The way split works is, it will break the order string at every "#", and the first "#" is encountered right at the beginning of the order string, and hence that is an empty.

Depending on your requirement, either you can avoid appending this "#" at the start, or you remove the "#" before spiting, or you ignore the element of the spitted array.

Hope this clarifies.

Answer 4

Try this (Java 8+ is required):

import java.util.Arrays;
    ...
String s = "#BRICK1 10#BRICK2 20#BRICK3 30";
String[] tokens = Arrays.stream(s.split("#")).filter(t -> !t.isEmpty()).toArray(String[]::new);

Answer 5

how do I prevent it?

By placing delimiter between the parts you want to get with split.

Imagine, you want to split a stick in two.

How do you do it? Where would be the point of application of force?

Option one:  ---V---

Option two: V---V--- (looks familiar?)

The former gives you two sticks, obviously, and the latter would give you an empty piece of stick in the beginning. It just doesn't exist in a real world, while an empty string happen to exist in Java.

Check String#split documentation, it says exactly that:

Splits this string around matches of the given regular expression.

Answer 6

Use regex to safely (ie if any exist) remove all leading # chars, then split:

String[] ordersPairs = orderList.replaceAll("^#*", "").split("#+");

By splitting on "#+" , empty orders are also ignored/discarded.