Redis/Jedis - 按模式删除？

通常,我得到密钥集,然后使用外观删除每个键/值对.

是否可以通过模式删除所有键？

即:

Del sample_pattern:*

iCodeLikeImD.. 18

看来,对于Jedis来说,"按模式删除"基本上是获取特定模式的所有键然后循环遍历它.

即

Set keys = jedis.keys(pattern);
for (String key : keys) {
    jedis.del(key);
} 

扫描更好.但是你可以通过`Set keys = jedis.keys(pattern)来减少对jedis的调用次数; jedis.del(keys.toArray(new String [keys.size()]));` (3认同)

小智.. 11

不建议使用KEYS,因为它在生产中使用效率低下.请参阅https://redis.io/commands/keys.相反,最好使用SCAN.另外,比重复调用jedis.del()更有效的调用是对jedis进行一次调用以删除匹配的键,传入要删除的键数组.下面介绍一种更有效的解决方案:

Set matchingKeys = new HashSet<>();
ScanParams params = new ScanParams();
params.match("sample_pattern:*");

try(Jedis jedis = jedisPoolFactory.getPool().getResource()) {
    String nextCursor = "0";

    do {
        ScanResult scanResult = jedis.scan(nextCursor, params);
        List keys = scanResult.getResult();
        nextCursor = scanResult.getStringCursor();

        matchingKeys.addAll(keys);

    } while(!nextCursor.equals("0"));

    if (matchingKeys.size() == 0) {
      return;
    }

    jedis.del(matchingKeys.toArray(new String[matchingKeys.size()]));
}

小智.. 5

你应该尝试使用eval.我不是Lua专家,但这段代码有效.

private static final String DELETE_SCRIPT_IN_LUA =
    "local keys = redis.call('keys', '%s')" +
    "  for i,k in ipairs(keys) do" +
    "    local res = redis.call('del', k)" +
    "  end";

public void deleteKeys(String pattern) {
  Jedis jedis = null;

  try {
    jedis = jedisPool.getResource();

    if (jedis == null) {
      throw new Exception("Unable to get jedis resource!");
    }

    jedis.eval(String.format(DELETE_SCRIPT_IN_LUA, pattern));  
  } catch (Exception exc) {
    if (exc instance of JedisConnectionException && jedis != null) {
      jedisPool.returnBrokenResource(jedis);
      jedis = null;
    }

    throw new RuntimeException("Unable to delete that pattern!");
  } finally {
    if (jedis != null) {
      jedisPool.returnResource(jedis);
    }
  }
}

然后打电话:

deleteKeys("temp:keys:*");

这保证了一个服务器端调用,多个删除操作.