我是SQL新手,我在Udemy独自学习。我遇到了很多问题,我正在努力解决其中一个问题:
此SQL查询有什么问题?
SELECT sport, count(*)
FROM activities
WHERE username IN ‘tony’
GROUP BY 1;
我有两个假设: 1 - 如果活动中的字段'sport'填充了字符串值,那么我们就不能使用count。 2 - 最后一个陈述应该是:
WHERE username in:‘tony’ GROUP BY 1;
我很乐意收到您对该问题的反馈并向您学习! 感谢
答案 0 :(得分:5)
搜索IN将使用with()圆括号
display: inline-block;
答案 1 :(得分:1)
为tony添加括号, 例如:如果你想检查多个名字('tony','stark')
{
const createDiv = clubs => {
const $div = document.createElement(`div`);
$div.classList.add(`club-info`);
document.querySelector(`.clubs`).appendChild($div);
const $club = document.createElement(`p`);
$club.classList.add(`.clubname`);
$club.textContent = `${clubs.name}`;
const $origin = document.createElement(`p`);
$origin.classList.add(`.origin`);
$origin.textContent = `${clubs.origin}`;
const $championships = document.createElement(`p`);
$championships.classList.add(`.championships`);
$championships.textContent = `${clubs.championships}`;
document.querySelector(`.club-info`).appendChild($club);
document.querySelector(`.club-info`).appendChild($origin);
document.querySelector(`.club-info`).appendChild($championships);
};
const makeDivs = clubs => {
clubs.forEach(club => {
createDiv(club);
});
};
const parse = clubs => {
makeDivs(clubs);
};
const init = () => {
const url = `./assets/data/data.json`;
fetch(url)
.then(r => r.json())
.then(json => parse(json.clubs));
};
init();
}