Question

不确定这是否是一个简单的查询，但是由于某种原因无法弄清楚……

我有状态记录的降序记录（从最旧到最新）：

#include <iostream>
#include <string>
using namespace std;
class StackNode
{
public:
    StackNode *  topPtr = NULL;
    StackNode* next;
    string item;
bool push( string newItem) {

   // create a new node
   StackNode *newPtr = new StackNode;

   // set data portion  of new node
   newPtr->item = newItem;

   // insert the new node
   newPtr->next = topPtr;
   topPtr = newPtr;

   return true;
}
bool pop() {

   if (topPtr == NULL)
      return false;

   // stack is not empty; delete top
   else{
      StackNode *temp = topPtr;
      topPtr = topPtr->next;

      // return deleted node to system
      temp->next = NULL;  // safeguard
      delete temp;
      return true;
   }
}
int ope(string op, string val1,string val2)
{

    int vaL1 = stoi(val1);
    int vaL2 = stoi(val2);

    int res = 0;
    if( op == "*")
        res = vaL1 * vaL2;
if( op == "/")
        res = vaL1 / vaL2;
if( op == "-")
        res = vaL1 - vaL2;
if( op == "+")
        res = vaL1 + vaL2;
return res;
}
int cal(string pre_exp[],int len)
{
    int numb = 0;

    for(int i = len -1;i>=0;i--)
    {

       if ( pre_exp[i] == "*" || pre_exp[i] == "/" || pre_exp[i] == "+" || pre_exp[i] == "-")
        {

       string op1 = topPtr->item;
       pop();
       string op2 = topPtr->item;
       pop();

      numb = numb + ope(pre_exp[i],op1,op2);
    }
    else
    {
       push( (pre_exp[i]));
    }
}
  return numb;
}
int main()
{
StackNode nbr;
string eyoo[] = {"+","-","2","3","9"};
cout<< nbr.cal(eyoo,5)<<endl;
return 0;
}

但是我只想将唯一更改返回到Name Status Time A Up 2018-06-21 00:07:00.000000 A Up 2018-05-21 00:07:00.000000 A Down 2018-04-21 00:07:00.000000 A Down 2018-03-21 00:07:00.000000 A Up 2018-02-21 00:07:00.000000 A Down 2018-01-21 00:07:00.000000列，因此为我提供了状态更改的时间列表。因此答案将是这样：

Status

请注意结果是如何返回3月和5月而不是4月和6月的条目的，因为那是状态从不同状态更改的时间。

我的查询当前正在进行一些联接，所以我也不确定如何/在哪里将其适合查询：

  Name       Status               Time
   A          Up        2018-05-21 00:07:00.000000
   A         Down       2018-03-21 00:07:00.000000
   A          Up        2018-02-21 00:07:00.000000
   A         Down       2018-01-21 00:07:00.000000

Answer 1

理想情况下，您需要lag()并从中进行选择。让我假设您是MySQL的v8之前的版本。

一种方法是相关子查询：

select t.*
from (select t.*,
             (select t2.status
              from t t2
              where t2.time < t.time
              order by t2.time desc
              limit 1
             ) as prev_status
      from t
     ) t
where prev_status is null or prev_status <> status;

在较新版本的MySQL中，您只需执行以下操作即可：

select t.*
from (select t.*,
             lag(status) over (order by time) as prev_status
      from t
     ) t
where prev_status is null or prev_status <> status;

Answer 2

我对外部查询通过public class KeyValueClass { public int Age { get; set; } public string Name { get; set; } } private void DoTheJob() { var myList = new List<KeyValueClass> { new KeyValueClass {Age = 21, Name = "Carl"}, new KeyValueClass {Age = 23, Name = "Vladimir"}, new KeyValueClass {Age = 25, Name = "Bob"}, new KeyValueClass {Age = 21, Name = "Olivia"}, new KeyValueClass {Age = 21, Name = "Carl"}, new KeyValueClass {Age = 30, Name = "Jacob"}, new KeyValueClass {Age = 23, Name = "Vladimir"}, }; var myDistinctList = myList.GroupBy(x => new { x.Age, x.Name }) .Select(c => c.First()).ToList(); }逐行匹配子查询的结果进行了一些通用的尝试，原因是输出是行未按某种方式排序的方式。可以检查一下是否适合您。

#include <fstream>
#include <iostream>
#include <iterator>
#include <sstream>
#include <string>
#include <vector>

using namespace std;

int main() {
  ifstream file("data.txt");
  string str = "";
  string::size_type sz;

  uint32_t line_number = 4;

  while (std::getline(file, str)) {
    cout << str << endl;

    istringstream buf(str);
    istream_iterator<string> beg(buf), end;

    vector<string> tokens(beg, end);

    for (auto &s : tokens)
      cout << atof(s.c_str()) << " " << flush;

    cout << endl;
  }
}

Answer 3

另一种方法使用伪临时表（由于MySql不允许对一个子表发出多个子查询，因此不能使用实际的临时表），该表是原始表，并在其中附加了“行号”作为主键： / p>

SET @row_number = 0;
create table temp as
  SELECT (@row_number:=@row_number + 1) AS row_num, t.* FROM t order by time;
ALTER TABLE temp ADD PRIMARY KEY (row_num);

然后查询变得简单：

select name, status, time from temp temp1
    where temp1.row_num = 1
    or temp1.status <>
       (select temp2.status from temp temp2 where temp2.row_num = temp1.row_num - 1);

drop table temp; /* no longer need this table */

MySQL仅显示列的唯一更改

3 个答案: