PHP为什么我的内部连接返回错误?
我有一行代码:PHP为什么我的内部连接返回错误?,php,mysql,inner-join,Php,Mysql,Inner Join,我有一行代码: $query = mysqli_query($connection, "SELECT `id`, `image`, `link`, `order` FROM `galleryImages` INNER JOIN `galleries` ON `galleryImages`.`galleryId` = `galleries`.`id`"); 它将返回此错误: Warning: mysqli_fetch_assoc() expects parameter 1 to be mysql
$query = mysqli_query($connection, "SELECT `id`, `image`, `link`, `order` FROM `galleryImages` INNER JOIN `galleries` ON `galleryImages`.`galleryId` = `galleries`.`id`");
它将返回此错误:
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 $results;
尝试:
这应该行得通
SELECT
`galleries`.`id`,
`galleries`.`image`,
`galleries`.`link`,
`galleries`.`order`
FROM
`galleryImages`
INNER JOIN `galleries`
ON `galleryImages`.`galleryId` = `galleries`.`id`
在您的代码中,我们没有看到任何
mysqli\u fetch\u assoc()
。在第一行之后是var\u dump($query),以确保它包含您认为它包含的内容。运行查询后,请使用mysqli\u error($connection)了解更多信息<代码>变量转储($query)返回bool(false)
;如果我去掉了内部连接,它就工作了。请检查表字段,可能是相反的。我唯一的挑剔是删除了“周围的一切,但每个都有自己的”。
SELECT
`galleries`.`id`,
`galleries`.`image`,
`galleries`.`link`,
`galleries`.`order`
FROM
`galleryImages`
INNER JOIN `galleries`
ON `galleryImages`.`galleryId` = `galleries`.`id`