php - 插入…………从中选择

Question

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));

Answer 1

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)