异常：java.lang.NumberFormatException 将 String 转换为 long 时
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Exception : java.lang.NumberFormatException When convert String to long
提问by N Sharma
I am converting the string which have '1520056800` to long to have date. But I am getting NumberFormatException to convert this
我正在将具有 '1520056800` 的字符串转换为 long 以获取日期。但我得到 NumberFormatException 来转换这个
Please help me.
请帮我。
long expiryDateMS = Long.parseLong(responseArray[0].replaceAll(" ", ""));
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-mm-dd");
Date date = new Date(expiryDateMS);
Stack trace
堆栈跟踪
09-02 00:52:28.984: E/AndroidRuntime(12025): Caused by: java.lang.NumberFormatException: 1520056800
09-02 00:52:28.984: E/AndroidRuntime(12025): at java.lang.Long.parse(Long.java:353)
09-02 00:52:28.984: E/AndroidRuntime(12025): at java.lang.Long.parseLong(Long.java:344)
09-02 00:52:28.984: E/AndroidRuntime(12025): at java.lang.Long.parseLong(Long.java:311)
09-02 00:52:28.984: E/AndroidRuntime(12025): at com.example.astrill_openvpn.MainOnOffActivity.onCreate(MainOnOffActivity.java:99)
09-02 00:52:28.984: E/AndroidRuntime(12025): at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1047)
09-02 00:52:28.984: E/AndroidRuntime(12025): at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:1615)
回答by Martijn Courteaux
This code should do the job. So I'm guessing you have got some encoding/special character problem. Try this to verify you have a real ASCII encoded number:
这段代码应该可以完成这项工作。所以我猜你有一些编码/特殊字符问题。试试这个来验证你有一个真正的 ASCII 编码数字：
String str = responseArray[0].replaceAll(" ", "");
for (int i = 0; i < str.length(); ++i)
{
char a = str.charAt(i);
if (!('0' <= a && a <= '9')) System.out.println(a + " is not a valid digit!");
}
