bugfix: reduce unnecessary competition and remove missing locks #3448
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Codecov Report
@@ Coverage Diff @@
## develop #3448 +/- ##
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- Coverage 51.57% 51.55% -0.02%
+ Complexity 3379 3373 -6
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Files 619 619
Lines 20468 20467 -1
Branches 2563 2564 +1
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- Hits 10556 10552 -4
Misses 8852 8852
- Partials 1060 1063 +3
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funky-eyes
added
module/server
| if (success.length > 0) { | ||
| jedis.del(success); | ||
| if (success.size() > 0) { | ||
| jedis.del(success.toArray(new String[0])); |
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纯del 无需lua
我是指这里整个加锁动作，封装到一个lua里，这样就不会存在del操作了
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纯del 无需lua
我是指这里整个加锁动作，封装到一个lua里，这样就不会存在del操作了
这个得后续优化了，把操作写成纯lua来原子执行
| for (int i = 0; i < partitions.size(); i++) { | ||
| String key = readyKeys.get(i); | ||
| if (partitions.get(i).contains(FAILED)) { | ||
| if (Objects.equals(partitions.get(i).get(0),FAILED)) { |
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partitions.get(i) 里面放了七个hset的结果。contains可以发现任何一个错误，如果是partitions.get(i).get(0),那等于是只比较pipeline.hsetnx(key, XID, value.getXid())操作的返回结果了。有没有可能，hset xid这个操作成功了，但是后面的hset失败了。那这样比较就出问题了。
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有hsetnx可能失败,hset没理由失败,除非redis突然宕机了,如果算突然宕机即使都是hsetnx也可能出现这个问题
| @@ -119,29 +119,29 @@ public boolean acquireLock(List<RowLock> rowLocks) { | |||
| pipeline.hsetnx(key, XID, value.getXid()); | |||
| pipeline.hsetnx(key, TRANSACTION_ID, value.getTransactionId().toString()); | |||
| pipeline.hsetnx(key, BRANCH_ID, value.getBranchId().toString()); | |||
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Hsetnx doesn't work. Why don't we replace it with hset
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如果锁已经被其他的事务持有，那么如果用hset会导致后续解锁时失败，会串改锁持有者的信息，而下面的4个信息无论持有者是谁都没有影响，是固定的 @ph3636
Ⅰ. Describe what this PR did
1.当我抢到了1个锁时，失败了2个，而succes[0]这下标只会记录一个key，导致会有锁被遗漏删除
2.仅需竞争key rowkey，如果失败时del会一并删除，所以没必要让redis处理其他field的竞争上
3.因为指令是有序响应，仅需get(0)拿到setnx的信息比较即可
Ⅱ. Does this pull request fix one issue?
Ⅲ. Why don't you add test cases (unit test/integration test)?
Ⅳ. Describe how to verify it
Ⅴ. Special notes for reviews