Question

我不确定是否可以通过一个查询在SQL本身中完成它，但这就是我要问的原因。

有没有办法获得表中出现次数最少的列值？

表格如下：

var counter = [0,1,2];
var count = counter[0];
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console.log(questions);
questions = $.parseJSON(questions);

if (questions != 0){
    $('.question').html(questions[count]['question']);
    var options = (questions[count]['options']);
    options.forEach(function (item) {
        $('.options').html(item);
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    var index = counter.indexOf(1);
    questions = questions.splice(index, 1);
    console.log(questions);
    localStorage.removeItem('questions);
    counter = counter.splice(index, 0);

现在有一个出现次数最少的“Carrot”值，SQL如何获得具有最少出现次数的值？有点混乱，也许你第一次尝试就不会理解。感谢。

Answer 1

让我们重新创建您的测试用例：

sqlite> create table vegetables(id int, value text);
sqlite> insert into vegetables values('1', 'Banana');
sqlite> insert into vegetables values('2', 'Carrot');
sqlite> insert into vegetables values('3', 'Apple');
sqlite> insert into vegetables values('4', 'Apple');
sqlite> insert into vegetables values('5', 'Banana');
sqlite> insert into vegetables values('6', 'Apple');
sqlite> insert into vegetables values('7', 'Apple');
sqlite> insert into vegetables values('8', 'Banana');
sqlite> insert into vegetables values('9', 'Apple');

检查数据与您的数据相同：

sqlite> select * from vegetables;
1|Banana
2|Carrot
3|Apple
4|Apple
5|Banana
6|Apple
7|Apple
8|Banana
9|Apple

如果您想要出现次数最少的蔬菜：

sqlite> select value from vegetables group by value order by count(value) limit 1;
Carrot

如果您想要出现次数最多的蔬菜：

sqlite> select value from vegetables group by value order by count(value) desc limit 1;
Apple

Answer 2

这是一种方法：

select value
from t
group by value
order by count(*)
fetch first 1 row only;

fetch first 1 row only是ANSI标准语法，由多个数据库支持。有些拼写为limit或select top 1。

Answer 3

您也可以尝试使用此方法，即使它更加复杂且不太美观：

integers

选择出现次数最少的列值

3 个答案: