具有内连接的 Laravel 范围
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Laravel scope with inner join
提问by GluePear
I have 2 tables, projectsand jobs. jobshas a column called project_id. In Laravel 5.2 I want to run a search that will return all jobs which belong to a project of a given search term. This SQL works:
我有 2 张桌子，projects和jobs. jobs有一个名为project_id. 在 Laravel 5.2 中，我想运行一个搜索，该搜索将返回属于给定搜索词的项目的所有工作。此 SQL 有效：
SELECT jobs.*, projects.name FROM jobs INNER JOIN projects ON jobs.project_id = projects.id WHERE projects.name LIKE "%$keyword%"
In my Job.phpmodel I have created a scope method, which errors:
在我的Job.php模型中，我创建了一个 scope 方法，该方法出错：
public function scopeSearch($query, $keyword)
{
if ($keyword != '') {
$query->where(function($query) use ($keyword) {
$query->where('projects.name', 'LIKE', '%' . $keyword . '%')->join('projects', 'jobs.project_id', '=', 'projects.id');
});
}
return $query;
}
This produces the error:
这会产生错误：
Column not found: 1054 Unknown column 'projects.name' in 'where clause' (SQL: select * from
jobswhere (projects.nameLIKE %test%))
列未发现：在1054未知列'projects.name 'where子句'（SQL：SELECT * FROM
jobs哪里（projects。nameLIKE％测试％））
In my JobsController.phpI have:
在我的JobsController.php我有：
$searchResults = Job::Search($searchTerm)->get();
回答by xmhafiz
The parameter $queryin where(function($query)is not the $querythat you passed in public function scopeSearch($query, $keyword)
该参数$query在where(function($query)不是$query你传入public function scopeSearch($query, $keyword)
You can either remove it with just the query like below (as @Rob mentioned)
您可以仅使用如下查询将其删除（如@Rob 所述）
public function scopeSearch($query, $keyword)
{
if ($keyword != '') {
$query->where('projects.name', 'LIKE', '%' . $keyword . '%')->join('projects', 'jobs.project_id', '=', 'projects.id');
}
return $query;
}
or you need to include the $queryin use()
或者你需要包括$queryin use()
$query->where(function() use ($keyword, $query)
