INSERT INTO ..... SELECT FROM

Question

我想把它calory作为第一个价值fruits,我做不到,任何人都可以帮忙吗？

   $sql = 'INSERT INTO fruits VALUES('', ?, ?, ?)'
          SELECT calory
          FROM diet
          WHERE fruit = ?
         ';

   $this->db->query($sql, array($a, $b, $c, $d));

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"calory"是表"fruits"中列的名称)