我有这样的
MySQL查询：
SELECT cp.plan_name,cp.plan_time FROM courses c INNER JOIN course_to_plan cpl ON cpl.course_id = c.course_id INNER JOIN courseplans cp ON cp.plan_id = cpl.plan_id WHERE cpl.course_id = '$course_id';
这将输出数据,例如：
+----------------------------+-----------+ | plan_name | plan_time | +----------------------------+-----------+ | Plan number one name | 6 | | Plan number two name | 6 | | Plan number three name | 10 | +----------------------------+-----------+
我希望将这些行插入表单提交的新表中.
如何继续编写我的update.PHP代码以使其在表newtable中插入值？
if (isset($_POST['submit'])) {
$course_id = $_POST['course_id'];
$course_result = MysqL_query
("SELECT cp.plan_name,cp.plan_time FROM courses c
INNER JOIN course_to_plan cpl ON cpl.course_id = c.course_id
INNER JOIN courseplans cp ON cp.plan_id = cpl.plan_id
WHERE cpl.course_id = '$course_id'");
/* I want the result of the above rows to be inserted in the table
newtable which has the columns plan_name,plan_time */
我不愿意承认我在PHP和MysqL中完全没用,但我正在努力学习.我想我必须创建某种数组来存储结果然后循环插入但我不知道如何.
您必须知道的一件事是查询返回的列数必须与要插入的列数相匹配
"INSERT INTO NewTable(plan_name,plan_time)
SELECT cp.plan_name,cp.plan_time
FROM courses c
INNER JOIN course_to_plan cpl ON cpl.course_id = c.course_id
INNER JOIN courseplans cp ON cp.plan_id = cpl.plan_id
WHERE cpl.course_id = '$course_id'"
Warning: watch out for sql injection through
$course_id.
请注意,我在INSERT语句中指定了2列,因为SELECT查询返回2列
如果表中的列数与查询返回的列数完全匹配,则无需指定列.
