我想把 calory
作为 fruits
的第一个值,我做不到,有人能帮忙吗?
$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"
在你的情况下,这应该是:
INSERT INTO fruits (calory)
SELECT calory
FROM diet
WHERE fruit = ?
(如果“卡路里”是“水果”表中列的名称)
关于php - 插入…………从中选择,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13603374/