如何将此日期字符串 select c.id, c.nume,c.localitate,l.masina_id, i.nr_inmatriculare, i.an_fabricatie,
i.rafinarie, m.marca from clienti c inner join livrari l on c.id = l.id inner join incarcari I on l.incarcare_id = l.livrari_id inner join masina m on i.id_marca = m.id, sum(select quantity, deliver_quantity) as total_quantity group by an_fabricatie having quantity >1000 order by marca;
Incarcari table
Id|livrari_id|id_marca|nr_inmatriculare|an_fabricatie|rafinarie|aviz_incarcare|quantity|
Livrari table
Id|masina_id|client_id|incarcare_id|deliver_quantity|aviz_livrare
Masini table
Id|numar_inmatriculare|marca|an_fabricatie|
Clienti table
Id|nume|localitate|date_add|date_upd|
转换为19/04/2015:21:43:47.40
个对象。 Date
返回无效日期。
答案 0 :(得分:1)
要绝对确定我会将字符串拆分为任何不具有正则表达式\D+
的数字字符。然后你有一个包含所有部分的数组,你可以按正确的顺序将它传递给new Date()
:
var aParts = '19/04/2015:21:43:47.40'.split(/\D+/);
document.write(new Date(aParts[2], parseInt(aParts[1], 10)-1, aParts[0], aParts[3], aParts[4], aParts[5], aParts[6]));