才3w数据, 做一张多对多的数据表去关联查都比直接加redis好. 不要慌!
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3w还是查出来 放到内存处理吧 之前在mybatis里递归查 卡的要死 优化后直接提速100倍
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那肯定的
redis+数据库的一致性问题，无非就是前删、后删、双删，高级一点还可以整整canal这些，但是我的建议是数据量不大直接从数据库里面取就行了，遇到性能问题的时候再考虑优化，不然事情都做完了后面做什么
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之前我同事也遇到过这个问题 我们是单层数据用父子关系组成一棵树 用广度优先遍历去做比纯递归性能好点 不知道能不能给你提供帮助 很久之前写的了 说实话今天不是看到你这个帖子 我都已经快忘记了 如果已经解决了 也请分享一下解决思路哈哈
@Data
@AllArgsConstructor
public class AuthCorpDeptUserRelationDTO {
private String deptId;
private String userid;
}
@Data
@AllArgsConstructor
public class CorpTreeDTO {
private String deptId;
private String departmentId;
private String parentId;
private List<CorpTreeDTO> corpTreeDTOList;
private List<String> userList;
}
@Data
@AllArgsConstructor
public class CustomerDTO {
private String name;
private String customerId;
private String customerParentId;
private List<CustomerDTO> childList;
}
private static List<CustomerDTO> getTree(){
List<CustomerDTO> res = new ArrayList<>();
List<CustomerDTO> customerDTOS = new ArrayList<>();
for(int i = 0; i < 10000; i++){
customerDTOS.add(new CustomerDTO("测试",String.valueOf(i) , null, null));
}
for (int i = 10000; i < 20000; i++){
customerDTOS.add(new CustomerDTO("测试",String.valueOf(i) , String.valueOf(i - 10000), null));
}
for (int i = 20000; i < 30000; i++){
customerDTOS.add(new CustomerDTO("测试",String.valueOf(i) , String.valueOf(i - 10000), null));
}
for (int i = 30000; i < 40000; i++){
customerDTOS.add(new CustomerDTO("测试",String.valueOf(i) , String.valueOf(i - 10000), null));
}
for (int i = 40000; i < 50000; i++){
customerDTOS.add(new CustomerDTO("测试",String.valueOf(i) , String.valueOf(i - 10000), null));
}
for (int i = 50000; i < 60000; i++){
customerDTOS.add(new CustomerDTO("测试",String.valueOf(i) , String.valueOf(i - 10000), null));
}
//递归实现构建树
long time3 = System.currentTimeMillis();
List<CustomerDTO> resultList = new ArrayList<>();
for (int i = 0; i < customerDTOS.size();i++){
if (StringUtils.isEmpty(customerDTOS.get(i).getCustomerParentId())){
resultList.add(customerDTOS.get(i));
}
}
findParent(resultList, customerDTOS);
long time4 = System.currentTimeMillis();
// System.out.println(JSONObject.toJSONString(resultList));
//队列实现构建树
long time1 = System.currentTimeMillis();
Queue<CustomerDTO> indexQueue = new ArrayDeque<>();
Queue<CustomerDTO> childQueue = new ArrayDeque<>();
for (int i = 0 ; i < customerDTOS.size(); i++){
customerDTOS.get(i).setChildList(new ArrayList<>());
if (StringUtils.isEmpty(customerDTOS.get(i).getCustomerParentId())) {
indexQueue.add(customerDTOS.get(i));
res.add(customerDTOS.get(i));
}else {
childQueue.add(customerDTOS.get(i));
}
}
while(indexQueue.peek() != null && childQueue.size() > 0){
CustomerDTO poll = indexQueue.poll();
int size = childQueue.size();
for(int i = 0; i < size; i++){
CustomerDTO childPoll = childQueue.poll();
if (poll.getCustomerId().equals(childPoll.getCustomerParentId())){
poll.getChildList().add(childPoll);
indexQueue.add(childPoll);
}else{
childQueue.add(childPoll);
}
}
}
long time2 = System.currentTimeMillis();
// System.out.println(JSONObject.toJSONString(res));
System.out.println("递归构建: " + (time4 - time3) + "ms");
System.out.println("队列构建: " + (time2 - time1) + "ms");
return res;
}
private static List<CorpTreeDTO> getListTree(){
List<CorpTreeDTO> mapperList = new ArrayList<>();
List<CorpTreeDTO> resList = new ArrayList<>();
List<AuthCorpDeptUserRelationDTO> userMapperList = new ArrayList<>();
Queue<CorpTreeDTO> indexQueue = new ArrayDeque<>();
Queue<CorpTreeDTO> childQueue = new ArrayDeque<>();
Queue<AuthCorpDeptUserRelationDTO> userQueue = new ArrayDeque<>();
//初始化数据
for (int i = 0; i < 100; i++){
mapperList.add(new CorpTreeDTO(String.valueOf(i),"","",null,null));
}
for(int j = 1; j < 40; j++){
for (int i = j * 100; i < j * 200; i++){
mapperList.add(new CorpTreeDTO(String.valueOf(i*j),"",String.valueOf(i - j * 100),null,null));
}
}
for (int i = 0; i < 600; i++){
for (int j = 0; j < 10; j++){
userMapperList.add(new AuthCorpDeptUserRelationDTO(String.valueOf(i), "userId" + j));
}
}
long time1 = System.currentTimeMillis();
mapperList.forEach(mapper ->{
mapper.setCorpTreeDTOList(new ArrayList<>());
mapper.setUserList(new ArrayList<>());
if (StringUtils.isEmpty(mapper.getParentId())){
indexQueue.add(mapper);
resList.add(mapper);
}else {
childQueue.add(mapper);
}
});
userMapperList.forEach(userMapper ->{
userQueue.add(userMapper);
});
while(indexQueue.peek() != null && childQueue.size() > 0){
CorpTreeDTO poll = indexQueue.poll();
int userQueueSize = userQueue.size();
for (int i = 0; i < userQueueSize; i++){
AuthCorpDeptUserRelationDTO relationPoll = userQueue.poll();
if (relationPoll.getDeptId().equals(poll.getDeptId())){
poll.getUserList().add(relationPoll.getUserid());
}else {
userQueue.add(relationPoll);
}
}
int size = childQueue.size();
for(int i = 0; i < size; i++){
CorpTreeDTO childPoll = childQueue.poll();
if (childPoll.getParentId().equals(poll.getDeptId())){
poll.getCorpTreeDTOList().add(childPoll);
indexQueue.add(childPoll);
}else{
childQueue.add(childPoll);
}
}
}
long time2 = System.currentTimeMillis();
// System.out.println(JSONObject.toJSONString(resList));
System.out.println("构建tree: " + (time2 - time1) + "ms");
return resList;
}
private static void findParent(List<CustomerDTO> resultList, List<CustomerDTO> customerDTOS){
for(int i = 0; i < resultList.size(); i++){
resultList.get(i).setChildList(new ArrayList<>());
for (int j = 0; j < customerDTOS.size(); j++){
if (!StringUtils.isEmpty(customerDTOS.get(j).getCustomerParentId()) && customerDTOS.get(j).getCustomerParentId().equals(resultList.get(i).getCustomerId())){
resultList.get(i).getChildList().add(customerDTOS.get(j));
}
}
findParent(resultList.get(i).getChildList(), customerDTOS);
}
}
public static void main(String[] args) {
getTree();
getListTree();
}
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3w数据你真的是大海里涮拖把了,sql就可以了,不想频繁的执行sql语句可以在orm框架中加上缓存的
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3W…直接放内存里也没多少吧
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直接数据库查，感觉会更快，毕竟你要有更新redis的操作，脱裤子放屁，直接查数据库，重新梳理设计按照层级设计几个表，搞好主键和索引，3万数据，不会慢多少的
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我好奇的是 为什么要把这个操作拿到代理商登录账号的时候做？代理商通过什么渠道知道是否他的用户结构发生变化？
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3w数据，查询慢，不是很理解
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3w这个数量应该还好吧，我寻思这个量是不是连redis都不用了
你需要具体分析下慢的原因，是数据库还是代码
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