Question

我正在尝试创建一个查询，该查询在执行时将显示日期，状态（有多个状态选项）和该日期的事件数-按状态区分。

我能够创建一个查询，该查询显示我想要的所有数据，但是我得到的是重复的日期。我认为做到这一点的方法是使用别名，但是我找不到方法。我也尝试了案例陈述，但也没有成功。

SELECT cast(event_date AS date) AS date, count(event_status) AS amount, status
FROM events
GROUP BY date, status

现在这就是我得到的：

    date    |  amount    |   status   
---------------------------------------
2019-05-07  |    5       |     YES
2019-05-07  |    1       |      NO
2019-05-06  |    4       |     YES
2019-05-05  |    3       |     YES
2019-05-04  |    6       |     YES
2019-05-04  |    2       |    MAYBE

这就是我想要得到的：

    date    |  POSITVE   |   Negative    
---------------------------------------
2019-05-07  |    5       |      1
2019-05-06  |    4       |      0
2019-05-05  |    3       |      0
2019-05-04  |    6       |      2

其中任何不等于“是”的状态均为负。有什么建议么？只是要澄清一下：我想要的输出只是选项之一，我知道可以用不同的方法来解决我的问题，但这很适合我。谢谢

Answer 1

使用case和Sum，例如：

SELECT cast(event_date AS date) AS date, 
SUM(CASE WHEN status='YES' THEN 1 ELSE 0 END) as POSITIVE,
SUM(CASE WHEN status !='YES' THEN 1 ELSE 0 END) as NEGATIVE
FROM events
GROUP BY date

Answer 2

您可以进行条件聚合：

for i in l:
    print(i)

但是，('2', '40', '1', '69', '42') ('10', '67', '3', '101', '43') ('1', '122', '2', '153', '45') ('10', '151', '3', '183', '44') ('2', '186', '4', '216', '47') ('5', '214', '5', '245', '46')具有速记版本：

select cast(event_date AS date) AS date,
       sum(case when  status = 'Yes' then 1 else 0 end) as Positive,
       sum(case when  status <> 'Yes' then 1 else 0 end) as Negative
from events e
group by cast(event_date AS date);

Answer 3

MariaDB与MySQL一样，在数字上下文中将布尔值视为整数。我将其写为：

const app = document.getElementById('root')
const container = document.createElement('div')
container.setAttribute('class', 'container')
var prev = ''
app.appendChild(container)
var opt = []
var request = new XMLHttpRequest()
request.open('GET', 'https://findfalcone.herokuapp.com/planets', true)
request.onload = function() {
  // Begin accessing JSON data here
  var data = JSON.parse(this.response)
  data.forEach(element=>{
      opt.push(element)
  });
console.log(opt) 
populateDrop1()

}
function populateDrop1(){
    const dropdown = document.getElementById('sel1')


    var i =0
    opt.forEach(element => {
    const option = document.createElement('option')
    option.setAttribute('value','')
    option.textContent = element.name
    i++
    dropdown.appendChild(option)
    //console.log(document.getElementById("id").value)

  });  

}
function populateDrop2(){



    //console.log(opt)
    const dropdown = document.getElementById('sel2')


    var i =0
    opt.forEach(element => {
    const option = document.createElement('option')
    option.setAttribute('value','')
    option.textContent = element.name
    i++
    dropdown.appendChild(option)
    //console.log(document.getElementById("id").value)

  });  


}
function populateDrop3(){



  //console.log(opt)
  const dropdown = document.getElementById('sel3')


  var i =0
  opt.forEach(element => {
  const option = document.createElement('option')
  option.setAttribute('value','')
  option.textContent = element.name
  i++
  dropdown.appendChild(option)
  //console.log(document.getElementById("id").value)

});  


}

function populateDrop4(){



  //console.log(opt)
  const dropdown = document.getElementById('sel4')


  var i =0
  opt.forEach(element => {
  const option = document.createElement('option')
  option.setAttribute('value','')
  option.textContent = element.name
  i++
  dropdown.appendChild(option)
  //console.log(document.getElementById("id").value)

});  


}
function show1(ele){
     prev = ele.selectedIndex-1
     //prev.push(ele.options[ele.selectedIndex].text)
     opt.splice(prev,1)    

     populateDrop2()

}

function show2(ele){
  prev = ele.selectedIndex-1
  //prev.push(ele.options[ele.selectedIndex].text)
  opt.splice(prev,1)    

  populateDrop3()

}

function show3(ele){
  prev = ele.selectedIndex-1
  //prev.push(ele.options[ele.selectedIndex].text)
  opt.splice(prev,1)    

  populateDrop4()

}
request.send()

如何避免我的mysql查询重复？

3 个答案: