0

我有下表，我想选择所有（通过 php 查询）但只选择与 course1 匹配的行，我尝试了以下但它不起作用。

$comments = array();
$result = mysql_query("select * from comments where course1='$post_id' order by id ASC");

什么是正确的查询？

+----+-----------+----------+-----+-------------------------+---------------+-------------
| id | post_id   | name     | url | email              | body               | dt         |
+----+-----------+----------+-----+-------------------------+---------------+-------------
|  1 | course2   | john     |     | john@john.com      | comments by john   | 2012-11-16 |
|  2 | course1   | wiki     |     | wiki@wiki.com      | comments by wiki   | 2012-11-16 |
|  3 | course2   | daniel   |     | daniel@gmail.com   | comments by daniel | 2012-11-16 |
|  4 | course2   | ram      |     | ram@ram.com        | comments by ram    | 2012-11-16 |
|  5 | course1   | velu     |     | velu@velu.com      | comments by velu   | 2012-11-16 |
+----+-----------+----------+-----+-------------------------+---------------+-------------
4

3 回答 3

0

您需要引用要搜索的字段名称：

$result = mysql_query("select * from comments where post_id='$post_id' order by id ASC");

您的 $post_id 字符串变量包含值“course1”。

于 2012-11-17T11:32:30.470 回答
0

试试这个：

$result = mysql_query("select * from comments where post_id='$post_id' order by id ASC");
于 2012-11-17T11:33:55.147 回答
0

试试这个：SELECT * FROM comments WHERE post_id='course1' order by id ASC

我认为您在 sql 中混淆了列名和搜索值

于 2012-11-17T11:37:40.643 回答