我想将calory
作为fruits
的第一个值,但我做不到,有人可以帮忙吗?
$sql = 'INSERT INTO fruits VALUES('', ?, ?, ?)'
SELECT calory
FROM diet
WHERE fruit = ?
';
$this->db->query($sql, array($a, $b, $c, $d));
最佳答案
正确的语法是:
INSERT INTO "table1" ("column1", "column2", ...)
SELECT "column3", "column4", ...
FROM "table2"
在您的情况下,应为:
INSERT INTO fruits (calory)
SELECT calory
FROM diet
WHERE fruit = ?
(如果“卡路里”是表“水果”中列的名称)