使用正则表达式拆分日志文本，忽略第一个匹配项

Question

我有以下一段日志文本，我想使用 # User@Host 的正则表达式对其进行拆分。我正在使用 Java 正则表达式库函数。

# Time: 160204  1:56:31
# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000142  Lock_time: 0.000000 Rows_sent: 1  Rows_examined: 0
SET timestamp=1454579791;
SELECT DATABASE();
# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.001254  Lock_time: 0.000000 Rows_sent: 1  Rows_examined: 0
use test;
SET timestamp=1454579791;
# administrator command: Init DB;
# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000441  Lock_time: 0.000077 Rows_sent: 4  Rows_examined: 4
SET timestamp=1454579791;
show databases;
# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000207  Lock_time: 0.000074 Rows_sent: 1  Rows_examined: 1
SET timestamp=1454579791;
show tables;
# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000537  Lock_time: 0.000000 Rows_sent: 0  Rows_examined: 0
SET timestamp=1454579791;
;

如果我这样做，我会得到以下 6 个字符串。

字符串 1：

# Time: 160204  1:56:31

字符串 2：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000142  Lock_time: 0.000000 Rows_sent: 1  Rows_examined: 0
SET timestamp=1454579791;
SELECT DATABASE();

字符串 3：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.001254  Lock_time: 0.000000 Rows_sent: 1  Rows_examined: 0
use test;
SET timestamp=1454579791;
# administrator command: Init DB;

字符串 4：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000441  Lock_time: 0.000077 Rows_sent: 4  Rows_examined: 4
SET timestamp=1454579791;
show databases;

字符串 5：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000207  Lock_time: 0.000074 Rows_sent: 1  Rows_examined: 1
SET timestamp=1454579791;
show tables;

字符串 6：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000537  Lock_time: 0.000000 Rows_sent: 0  Rows_examined: 0
SET timestamp=1454579791;
;

因此，使用 # User@Host 由正则表达式拆分会返回 6 个字符串。我实际上只对五个字符串感兴趣，这两个字符串首先组合在一起。所以，结果应该是这样的

字符串 1：

# Time: 160204  1:56:31
# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000142  Lock_time: 0.000000 Rows_sent: 1  Rows_examined: 0
SET timestamp=1454579791;
SELECT DATABASE();

字符串 2：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.001254  Lock_time: 0.000000 Rows_sent: 1  Rows_examined: 0
use test;
SET timestamp=1454579791;
# administrator command: Init DB;

字符串 3：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000441  Lock_time: 0.000077 Rows_sent: 4  Rows_examined: 4
SET timestamp=1454579791;
show databases;

字符串 4：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000207  Lock_time: 0.000074 Rows_sent: 1  Rows_examined: 1
SET timestamp=1454579791;
show tables;

字符串 5：

# User@Host: root[root] @ localhost []  Id:     3
# Query_time: 0.000537  Lock_time: 0.000000 Rows_sent: 0  Rows_examined: 0
SET timestamp=1454579791;
;

我怎样才能做到这一点？

Answer 1

您可以在拆分后附加第一个和第二个元素：

String string1 = splitArray[0] + splitArray[1];

当然，这只有在您知道格式始终是您列出的格式时才有效。为确保是这种情况，您可以添加如下所示的检查：

if(splitArray[0].startsWith("# Time:"){
    String string1 = splitArray[0] + splitArray[1];
}

我确信有更优雅的方法可以实现这一点，但这会奏效;)