Java：字符串索引超出范围：-1

Java: String index out of range: -1

提问by gossfunkel

New code:

新代码：

I have no idea what's going on here, I'm getting this error:

我不知道这里发生了什么，我收到此错误：

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -1
    at java.lang.String.substring(String.java:1949)
    at client0_0_2.loginToClient(client0_0_2.java:98)
    at client0_0_2.general(client0_0_2.java:196)
    at client0_0_2.<init>(client0_0_2.java:208)
    at client0_0_2.main(client0_0_2.java:215)

and this is the contents of the method that's causing the error:

这是导致错误的方法的内容：

public String loginToClient() throws FileNotFoundException, IOException {
    //decryptUsers();
    int tries;
    tries = 5;
    while (tries > 0) {
        System.out.println("LOGIN");
        String usnm = c.readLine("Username: ");
        char [] passwd = c.readPassword("Password: ");
        users = new FileInputStream("users.fra");
        DataInputStream dis = new DataInputStream(users);
        BufferedReader br = new BufferedReader(new InputStreamReader(dis));
        String logindat = br.readLine();
        System.out.println(logindat);
        int startUsnm = logindat.indexOf(usnm);
        String logdat = logindat.substring(startUsnm, logindat.indexOf("."));
        if (startUsnm == -1) {
            System.err.println("Username not recognised, please try another or create user.");
            usnm = "INV";
            return usnm;
        }
        else {
            int tendUsnm = logdat.indexOf(':'); 
            int startPass = endUsnm + 1;
            int tendPass = logdat.indexOf('.');
            String Usnm = logdat.substring("0", tendUsnm);
            String Pass = logdat.substring(startPass, endPass);
            char [] Passwd = Pass.toCharArray();
            if (usnm.equals(Usnm)) {
                if (Arrays.equals(passwd,Passwd)) {
                    System.out.println ("Logged in. Welcome, " + usnm + ".");
                    String data = "LOGIN: " + usnm;
                    printLog(data);
                    //encryptUsers();
                    return usnm;
                }
                else {
                    System.out.println ("Incorrect password, please try again.");
                    String data = "PASWFAIL: " + usnm;
                    printLog(data);
                    tries -= 1;
                }
            }
            else {
                System.out.println ("Username not recognised.");
                printLog("USNAMFAIL");
                usnm = "INV";
                return usnm;
                //encrytUsers();
            }
        }
    }
    //encryptUsers();
    System.exit(2);
    return usnm;
}

It looks kinda like it's trying to access the last character of something and that's upsetting it, but I have no idea why. Any help for a total newbie?

看起来有点像它试图访问某个东西的最后一个字符，这让它感到不安，但我不知道为什么。对新手有什么帮助吗？

edits:

编辑：

users.fracontains username:passwordat the time of execution.

users.fra包含username:password在执行时。

note: line 98 is the

注意：第 98 行是

char [] Passwd = Pass.toCharArray();

n.b. not homework, a personal project.

nb 不是作业，是个人项目。

Will add in a method to deal with the event that username is not in users.fra.

将添加一个方法来处理用户名不在的事件users.fra。

回答by unholysampler

indexOf()returns -1 when the argument can't be found. I bet that the string you are searching does not have one of the characters you are searching for and it is causing an invalid index when you try to get a substring.

indexOf()当找不到参数时返回 -1。我敢打赌，您正在搜索的字符串没有您正在搜索的字符之一，并且在您尝试获取子字符串时会导致索引无效。

回答by Alex Stybaev

try to use contains()before indexOf()for yor string to avoid the situations when indexOf()returns -1 and you try to get substring with the -1 position.

尝试对 yor 字符串使用contains()beforeindexOf()以避免indexOf()返回 -1 并且您尝试获取具有 -1 位置的子字符串的情况。

回答by Dancrumb

String().indexOfwill return -1 if there is no match

String().indexOf如果没有匹配，将返回 -1

String().substringwill throw an StringIndexOutOfBoundsExceptionif you give it a negative beginvalue.

String().substringStringIndexOutOfBoundsException如果你给它一个负值会抛出一个begin。

This is all documented at http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html

这都记录在http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html

Thus, I'd look at this line of code:

因此，我会看这行代码：

String logdat = logindat.substring(startUsnm, logindat.indexOf("."));

This will fail if usnmis not in logindat

如果usnm不在，这将失败logindat

回答by Xeon

You should really look at those lines:

你真的应该看看这些行：

        int startUsnm = logindat.indexOf(usnm);
        int tendUsnm = logdat.indexOf(':');
        int endUsnm = startUsnm + tendUsnm;
        int startPass = endUsnm + 1;
        int tendPass = logdat.indexOf('.');
        int endPass = startPass + tendPass;
        String Usnm = logindat.substring(startUsnm, endUsnm);
        String Pass = logindat.substring(startPass, endPass);

Suppose you have file like this:

假设你有这样的文件：

---

---

user:pass.

用户：通过。

---

---

so:

所以：

• startUsnm = 0 //first char
• tendUsnm = 4 // index of :char
• endUsnm = 4 // sum? this is probably not what you intent to do
• startPass = 5 // pchar
• tendPass = 9 // .char
• endPass = 14 // sum again? Those integers are indexes - will throw IndexOutOfBounds

• startUsnm = 0 //第一个字符
• tendUsnm = 4 //:字符索引
• endUsnm = 4 // 总和？这可能不是你想要做的
• startPass = 5 //p字符
• tendPass = 9 //.字符
• endPass = 14 // 再求和？这些整数是索引 - 将抛出 IndexOutOfBounds

And consider also that indexOfwill return -1value if there is no such match in given String.

并考虑如果在 given 中没有这样的匹配，那indexOf将返回-1值String。

回答by Denys Séguret

A lot of things can be wrong. You'll have to check what is printed.

很多事情都可能是错误的。您必须检查打印的内容。

An exemple :

一个例子：

int startUsnm = logindat.indexOf(usnm);
String logdat = logindat.substring(startUsnm, logindat.indexOf("."));

If you don't have "Username: " in logindat, it produces the exception you see.

如果您在 logindat 中没有“用户名：”，它会产生您看到的异常。

EDIT : Note that if you want to allow "username:" (and not only "Username:"), you can do this :

编辑：请注意，如果您想允许“用户名：”（而不仅仅是“用户名：”），您可以这样做：

 int startUsnm = logindat.toLowerCase().indexOf("username:");

And test that startUsnm is >= 0.

并测试 startUsnm 是否 >= 0。