处理数组中的元素

Question

所以我有一个名为match的数组，其中match[0] =

[2014-05-30 15:21:20,781] DEBUG [scheduler-4] (DiamondSchedulerRunner.java:41) Current node is not a manager of:publishEmail in tenant:0 
[2014-05-30 15:21:20,781] DEBUG [scheduler-1] (DiamondSchedulerRunner.java:41) Current node is not a manager of:readEmail in tenant:0 
[2014-06-11 10:18:51,370] ERROR [executorWithPoolSizeRange-5] (PushNotificationsServiceBean.java:140) cannot send notification: service Whaler is not available for user himanshu.taunk@hp.com on device Desktop.device.uuid813374 
[2014-05-30 15:21:50,608] DEBUG [scheduler-3] (DiamondSchedulerProxy.java:96) Diamond Scheduler name:index Mode:TENANT_NODE runOnTenant:true split:true 
[2014-05-30 15:21:50,624] DEBUG [scheduler-3] (DiamondSchedulerRunner.java:41) Current node is not a manager of:index in tenant:0 
[2014-06-11 10:18:49,124] ERROR [executorWithPoolSizeRange-5] (PushNotificationsServiceBean.java:140) cannot send notification: service Whaler is not available for user himanshu.taunk@hp.com on device Desktop.device.uuid798084 
[2014-05-30 15:21:50,780] DEBUG [scheduler-5] (DiamondSchedulerProxy.java:96) Diamond Scheduler name:snooze Mode:TENANT_NODE runOnTenant:true split:true 
[2014-05-30 15:21:50,780] DEBUG [scheduler-5] (DiamondSchedulerRunner.java:41) Current node is not a manager of:snooze in tenant:0 
[2014-05-30 15:21:50,796] ERROR [scheduler-4] (DiamondSchedulerProxy.java:96) Diamond Scheduler name:publishEmail Mode:TENANT_NODE runOnTenant:true split:true 

match[1]是时间戳，match[2]是状态，match[3]是线程名，依此类推。有没有什么办法可以颠倒第一个索引中的元素，match[0]，使得第一行成为最后一行，反之亦然，而不会扰乱其他索引？

更好的是，我可以逐行读取第一个索引到另一个数组中，然后反转后一个数组吗？

Answer 1

你能用分隔符拆分匹配，\n然后反转并加入吗？即

match[0] = match[0].split('\n').reverse().join('\n')