关于php：尝试加入一个空表没有返回

Trying to JOIN an empty table nothing returns

尝试将空表(注释表)联接到现有的已准备语句时出现问题。

这非常有效：

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// prepare images
if ($stmt = $mysqli->prepare(" SELECT uu.*, m.*,
(
SELECT COUNT(*)
FROM img_likes AS t
WHERE t.img_id = uu.imgID AND t.user_id = ?
) AS user_likes,
(
SELECT COUNT(*)
FROM img_likes AS t
WHERE t.img_id = uu.imgID
) AS total_likes
FROM user_uploads AS uu
INNER JOIN members AS m ON m.id = uu.user_id
ORDER BY up_time DESC")) {
$stmt->bind_param('i', $user_id);
$stmt->execute(); // get imgs

// foreach print images
// working as expected
}

我不知道为什么如果我加入另一个空的表格(img_comments)，图像就不会被打印出来…如果我在表格中添加一行并刷新页面，就会打印一个图像…

我正在尝试但不起作用的声明是：

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SELECT uu.*, m.*, ic.*,
(
SELECT COUNT(*)
FROM img_likes AS t
WHERE t.img_id = uu.imgID AND t.user_id = ?
) AS user_likes,
(
SELECT COUNT(*)
FROM img_likes AS t
WHERE t.img_id = uu.imgID
) AS total_likes
FROM user_uploads AS uu
INNER JOIN members AS m ON m.id = uu.user_id
INNER JOIN img_comments AS ic ON ic.img_id = uu.imgID
ORDER BY up_time DESC

号

为什么只打印基于表行数的图像？我也试过左进，但我不太熟悉这个。我只在其他脚本中使用内部联接，而且从未遇到过这样的问题。

我希望对我的查询进行任何优化。