所以我试图只获得active = 1的行,但出于某种原因,它给了我这个错误。
Call to a member function fetch_object() on boolean
$db2 = new mysqli('localhost', 'root', '', '123');
$imagesQuery = $db2->query("
SELECT
images.id,
images.caption,
images.active,
COUNT(images_likes.id) AS likes
FROM images WHERE active = 1
LEFT JOIN images_likes
ON images.id = images_likes.image
GROUP BY images.id
");
while($row = $imagesQuery->fetch_object()) {
$images[] = $row;
}
echo '<pre style="color: #fff">', print_r($images, true) , '</pre>';
答案 0 :(得分:0)
SELECT
images.id,
images.caption,
images.active,
COUNT(images_likes.id) AS likes
FROM images
LEFT JOIN images_likes
ON images.id = images_likes.image
WHERE active = 1
GROUP BY images.id