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使用Java获取客户端的每个IP地址的问题

[英]Issue With Getting Every IP Address of the Client in Java

I am having an issue with getting every IP address in Java. 我在使用Java获取每个IP地址时遇到问题。 When I open the GUI to select which IP you want to use, I call: 当我打开GUI以选择您要使用的IP时,我打电话给:

private List<String> getIP() {
    List<String> outputList = new ArrayList<String>();
    try {
        InetAddress localIP = InetAddress.getLocalHost();
        InetAddress[] everyIPAddress = InetAddress.getAllByName(localIP
                .getCanonicalHostName());
        if (everyIPAddress != null && everyIPAddress.length > 1) {
            for (int i = 0; i < everyIPAddress.length; i++) {
                if (!everyIPAddress[i].toString().contains(":")) {
                    outputList.add(everyIPAddress[i].toString());
                }
            }
        }
    } catch (UnknownHostException e) {
        System.out.println("Error finding IP Address");
    }
    return outputList;
}

This method gets all of the IPv4 addresses that the client has. 此方法获取客户端具有的所有IPv4地址。 I know IPv6 addresses contain colons, so I don't add any with a colon to the list. 我知道IPv6地址包含冒号,因此我不会在列表中添加冒号。

Then, pressing the button changes the IP address. 然后,按下按钮可更改IP地址。 However, I have noticed that when there is only one IPv4 address that the machine has (You get two from having a service like Hamachi) it will return a null exception. 但是,我注意到当机器只有一个IPv4地址时(你得到两个像Hamachi这样的服务),它将返回一个空例外。 How would I go about getting every IP address of the client without returning a null exception if there is only one address? 如果只有一个地址,如何在不返回空例外的情况下获取客户端的每个IP地址?

if (everyIPAddress != null && everyIPAddress.length > 1) {

should be 应该

if (everyIPAddress != null && everyIPAddress.length >= 1) {

or 要么

if (everyIPAddress != null && everyIPAddress.length > 0) {

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