gpt4 book ai didi

PHP 为什么我的 INNER JOIN 返回错误?

转载 作者:行者123 更新时间:2023-11-29 08:00:02 24 4
gpt4 key购买 nike

我有这行代码:

$query = mysqli_query($connection, "SELECT `id`, `image`, `link`, `order` FROM `galleryImages` INNER JOIN `galleries` ON `galleryImages`.`galleryId` = `galleries`.`id`");

它返回此错误:

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result

完整代码:

$query = mysqli_query($connection, "SELECT `id`, `image`, `link`, `order` FROM `galleryImages` INNER JOIN `galleries` ON `galleryImages`.`galleryId` = `galleries`.`id`");
$results = array();
while($row = mysqli_fetch_assoc($query)){
$results[] = $row;
}
return $results;

最佳答案

尝试:

"SELECT `id`, `image`, `link`, `order`
FROM `galleryImages` `g`
INNER JOIN `galleries` `g2` ON `g`.`galleryId` = `g2`.`id`"

关于PHP 为什么我的 INNER JOIN 返回错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24069858/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com