java NumberFormatException 错误 (parseInt)
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NumberFormatException error (parseInt)
提问by Alex
Hopefully a very simple query, but it's left me scratching my head.
希望是一个非常简单的查询，但它让我摸不着头脑。
I have a string, which is just a single integer, and I'm trying to then get that integer out as an int. This on the face of it shouldn't be a problem.
我有一个字符串，它只是一个整数，然后我试图将该整数作为 int 输出。从表面上看，这应该不是问题。
// this is how I create the string (it's the playload from a UDP datagram packet,
// thought I don't think the origins hugely important - it's juts a test run so the
// stringMessage is always 1 (created by a seperate client process)
...
recvSoc.receive(pac);
String stringMessage = new String(pac.getData());
port = pac.getPort();
System.out.println("RECEIVED: " + stringMessage + " on port: " + port);
processMessage(stringMessage);
...
// Then in processMessage
public void processMessage(String data) {
int message;
message = Integer.parseInt(data);
...
This always crashes with a NumberFormatException error. I cannot for the life of me figure out what's causing this, any ideas greatly appreciated. I haven't coded much in Java (recently) so might simply be forgetting something critical or what not.
这总是因 NumberFormatException 错误而崩溃。我一生都无法弄清楚是什么导致了这种情况，任何想法都非常感谢。我（最近）用 Java 编写的代码不多，所以可能只是忘记了一些重要的东西或不重要的东西。
Exception in thread "main" java.lang.NumberFormatException: For input string: "1"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:481)
at java.lang.Integer.parseInt(Integer.java:514)
at udp.UDPServer.processMessage(UDPServer.java:85)
at udp.UDPServer.run(UDPServer.java:52)
at udp.UDPServer.main(UDPServer.java:156)
采纳答案by Joachim Sauer
Note that DatagramPackate.getData()returns the whole buffer!
请注意，DatagramPackate.getData()返回整个缓冲区！
The data you received is only a part of it:
您收到的数据只是其中的一部分：
The data received or the data to be sent starts from the
offsetin the buffer, and runs forlengthlong.
接收到的数据或要发送的数据从
offset缓冲区中的开始，并运行length很长时间。
So to convert the data to a Stringyou should use this constructor:
所以要将数据转换为 aString你应该使用这个构造函数：
String message = new String(pac.getData(), pac.getOffset(), pac.getLength(), "UTF-8");
Note that I specify the UTF-8 encoding here, as not specifying an encoding would result in the platform default encoding to be used, which is generally not what you want.
请注意，我在这里指定了 UTF-8 编码，因为不指定编码会导致使用平台默认编码，这通常不是您想要的。
回答by Bozho
If the string is really 1, the exception can't happen. So I would say the string is not actually 1.
如果字符串是真的1，则不会发生异常。所以我会说字符串实际上不是1.
do a data.toCharArray()and print each character's code (cast to int). It may turn out that there is a hidden character before the digit, for example. (edit: it appears iluxa mentioned this option in a comment while I was writing the answer)
执行 adata.toCharArray()并打印每个字符的代码（转换为int）。例如，结果可能是在数字之前有一个隐藏字符。（编辑：似乎 iluxa 在我写答案时在评论中提到了这个选项）
Try data = data.trim()before passing it to parseInt(..)
data = data.trim()在将其传递给之前尝试parseInt(..)
