我试图自己输入一个完整的表,更改city变量,但是当我创建一个临时表来存储数据时,insert语句将数据插入到原始语句中并不起作用。以下是代码 insert into from_php select * from menuinstant where city='Kota';
update from_php set id = replace(id,'Kota',
pictures(name, description, url, users_id) VALUES ('try','try','try', 12345)但是如果我使用下面的php/archives/create-a-rest-api-with-php/ {
// these could bei also tried with '