问题描述
当需要对象时，此查询返回null.
This query is returning null when an object is expected.
$vow = DB::table('media_featured')->where('is_video_of_the_week', 1)->
where('video_of_week_expired', '!=', 1)->first();
CREATE TABLE `media_featured` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`media_id` int(10) unsigned DEFAULT NULL,
`is_video_of_the_week` tinyint(1) DEFAULT NULL,
`is_featured` tinyint(1) DEFAULT NULL,
`video_of_week_expired` tinyint(1) DEFAULT NULL,
`featured_expired` tinyint(1) DEFAULT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL,
`deleted_at` timestamp NULL DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `media_featured_media_id_foreign` (`media_id`),
CONSTRAINT `media_featured_media_id_foreign` FOREIGN KEY (`media_id`) REFERENCES `media` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
一条记录可能具有is_video_of_the_week = 1和video_of_week_expired = NULL，但以上查询返回空值.
A record might have is_video_of_the_week = 1 and video_of_week_expired = NULL but the above query returns null.
有什么想法吗?
推荐答案
NULL值不等于或不等于其他任何值.
NULL values are not equal or not equal to anything else.
所以column != NULL和column = NULL一样总是虚假的
要检查列是否包含NULL值，您需要使用IS NULL运算符.
To check if a column contains NULL value you need to use IS NULL operator.
在使用laravel db查询生成器的情况下，您可以使用
In case of laravel db query generator you could use
->whereNull('video_of_week_expired')
方法.
PS:如果假定video_of_week_expired是类似于标志的列，则最好将其设置为NOT NULL并使用0/1值，而不是NULL/1
PS: if video_of_week_expired is assumed to be a flag-alike column, you better make it NOT NULL and use 0/1 values, instead of NULL/1
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