我有一个具有以下结构的日志文件:
输入:
$logLine = "[2017-12-23 19:15:59:634187] [INFO] User SIMDesign successfully logged in."
我希望得到一个数组(使用preg_split或preg_match_all),如下所示:
预期产量:
[
0 => '[2017-12-23 19:15:59:634187]',
1 => '[INFO]',
2 => 'User SIMDesign successfully logged in.'
]
或这个:
[
0 => '2017-12-23 19:15:59:634187',
1 => 'INFO',
2 => 'User SIMDesign successfully logged in.'
]
尝试:
我已经尝试了几个小时找到一个正则表达式模式,但不幸的是我没有找到适合我的解决方案。
preg_split("/[][]/", $logLine);
[
0 => '', //this is empty
1 => '2017-12-23 19:15:59:634187',
2 => ' ', //there is a space
3 => 'INFO',
4 => ' User SIMDesign successfully logged in.',
]
输出没问题但不完美。我不需要数组的空元素([0],[2])。我需要在字符串“用户SIMDesign成功登录”之前删除空格。
preg_split("/[][]\s\S/", $logLine);
[
0 => '[2017-12-12 12:12:12:132123',
1 => 'INFO',
2 => 'ser SIMDesign successfully logged in.' //The first character of the string was removed (why?)
]
提前致谢
首先,你想要第一组括号之间的信息,但由于这些是正则表达式的特殊字符,我们逃避它们:[(.*?)]
- > \[(.*?)\]
$logLine = "[2017-12-23 19:15:59:634187] [INFO] User SIMDesign successfully logged in."
^--------------------------^
然后是第二个括号,相同的逻辑:[(.*?)]
- > \[(.*?)\]
$logLine = "[2017-12-23 19:15:59:634187] [INFO] User SIMDesign successfully logged in."
^----^
而“其余的直到最后”是(.*)$
$logLine = "[2017-12-23 19:15:59:634187] [INFO] User SIMDesign successfully logged in."
^------------------------------------^
他们之间有空格(\s
):\[(.*?)\]\s\[(.*?)\]\s(.*)$
,有neat little demo here
我更喜欢上面写的答案,但preg_split页面还记录了一些可以与preg_split()一起传递的标志,其中一个是以下内容。我不是粉丝的原因是因为你“隐藏”了问题而不是修复它。
PREG_SPLIT_NO_EMPTY 如果设置了此标志,preg_split()将仅返回非空片段。