插入到.....选择 [英] INSERT INTO.....SELECT FROM

问题描述

我想将卡路里作为 fruits 的第一个值，我做不到，任何人都可以帮忙吗?

i want to put calory as the first value of fruits, i couldn't do it, can anyone help?

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

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

推荐答案

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)

(if "calory" is the name of the column in the table "fruits")

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