php - 这个 MySQL 查询似乎不起作用

Question

SELECT * FROM dislikes WHERE pid = $post AND uid = $userid AND like = 1;

这个查询不起作用(我不知道为什么)。一旦我使用 mysqli 执行此查询并尝试查找 num 行，我得到

Notice: Trying to get property of non-object in C:\wamp\www\include\like.php on line 13

只是想让你知道，表 dislikes 目前是空的，有 4 列:id、pid、uid 和 like。

编辑:

echo "SELECT * FROM dislikes WHERE pid = $post AND uid = $userid AND like = 1;";
$result = $mysqli->query("SELECT * FROM dislikes WHERE pid = $post AND uid = $userid AND like = 1;");
$row_cnt = $result->num_rows;

这是PHP代码。第一行输出的 echo :

SELECT * FROM dislikes WHERE pid = 2 AND uid = 3 AND like = 1;

Answer 1

LIKE 是 reserved word在 SQL 中。

您必须分隔与 SQL 保留字冲突的列名:

SELECT * FROM dislikes WHERE pid = $post AND uid = $userid AND `like` = 1;

但对您的错误的解释是 mysqli_query() 返回了 false 而不是有效的查询结果。您需要检查函数的返回值以确保它是一个资源而不是 false。

另见 Reference - What does this error mean in PHP?