laravel 雄辩的 Where 子句中的类似通配符的语法？
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Wildcard-like syntax in an eloquent Where clause?
提问by Robert Brax
Here is a where clause I have:
这是我有的 where 子句：
->where( 'post_type', '=', 'blog' )
Is there any way to replace 'blog' by a wildcard, so it will match any 'post_type' ?
有没有办法用通配符替换 'blog' ，这样它就会匹配任何 'post_type' ？
Full query:
完整查询：
$db = ( new DbSql() )->db()->getConnection();
$postsCount = $db->table( 'posts' )
->where( 'status', '=', 'published' )
->where( 'post_type', '=', 'blog' )
->where( 'alerts', '<', Settings::CONTENT_ALERT_NUMBER_ALLOWANCE )
->count() ?? null;
采纳答案by Pankit Gami
I assume that you are storing blogin any variable, say $blog.
我假设您将博客存储在任何变量中，例如$blog.
if($blog) {
$query->where('post_type', '=', $blog);
}
If blog is empty, do not add the where condition.
如果博客为空，则不添加 where 条件。
Edit :
编辑 ：
I assume you are using Laravel query builder.
我假设您正在使用 Laravel 查询构建器。
$postsQuery = $db->table( 'posts' )
->where( 'status', '=', 'published' );
if($blog){
$postQuery->where( 'post_type', '=', $blog );
}
$postCount = $postQuery->where( 'alerts', '<', Settings::CONTENT_ALERT_NUMBER_ALLOWANCE )
->count() ?? null;
回答by Harsha W
Try this
尝试这个
$query->where($field, 'like', $input . '%');
