問題描述

我想知道是否有人可以提供幫助，因為我遇到了困難并且仍在學習 Laravel ORM.任何人都可以解釋為什么，當我運行時:

I wonder if anyone can help, as I've hit a wall and still learning Laravel ORM. Can anyone explain why, when I run:

public function locationTags(){
    return $this->hasMany('AppUserHasLocationTags', 'user_id')
        ->join('location_tags AS lt', 'lt.id', '=', 'location_tag_id');
}

我得到了這個結果集:(剪斷了...)

I get this result set: (snipped...)

{
    "id": 1,
    "created_at": "2015-05-13 13:04:56",
    "updated_at": "2015-05-13 13:04:56",
    "email": "REMOVED",
    "firstname": "REMOVED",
    "lastname": "REMOVED",
    "location_id": 0,
    "deleted_at": null,
    "permissions": [],
    "location_tags": [
        {
            "user_id": 1,
            "location_tag_id": 1,
            "id": 1,
            "created_at": "2015-05-13 13:06:28",
            "updated_at": "2015-05-13 13:06:28",
            "name": "Test Tag 0",
            "location_id": 1,
            "deleted_at": null
        },
        {
            "user_id": 1,
            "location_tag_id": 2,
            "id": 2,
            "created_at": "2015-05-13 11:40:21",
            "updated_at": "2015-05-13 12:56:13",
            "name": "Test Tag 123",
            "location_id": 1,
            "deleted_at": null
        }
    ]
}

這是王牌！但是，當我開始從 location_tags 連接中選擇我想要的列時，使用:

Which is ace! However, when I start to select the columns I want from the location_tags join, with:

public function locationTags(){
    return $this->hasMany('AppUserHasLocationTags', 'user_id')
        ->join('location_tags AS lt', 'lt.id', '=', 'location_tag_id')
        ->select('lt.id', 'lt.name');
}

我最終得到:

{
    "id": 1,
    "created_at": "2015-05-13 13:04:56",
    "updated_at": "2015-05-13 13:04:56",
    "email": "REMOVED",
    "firstname": "REMOVED",
    "lastname": "REMOVED",
    "location_id": 0,
    "deleted_at": null,
    "permissions": [],
    "location_tags": []
}

誰能解釋一下這是怎么回事?并可能指出我正確的方向來限制選擇?謝謝！

Can someone explain what's going on? And possibly point me in the right direction to limit the selects? Thanks!

更新我也試過:

        $query = AppUser::with(['permissions', 'locationTags' => function($query){
            $query->select('lt.id', 'lt.name');
        }]);

返回相同的結果:(

推薦答案

想通了.這里的關鍵是你必須包含至少一個 Laravel 可以用來映射結果集的鍵的 select() 值.就我而言，它是 user_id，如下所示:

Figured it out. The key here was that you must include a select() value of at least one key that Laravel can use to map the result set. In my case it was user_id, like so:

public function locationTags(){
    return $this->hasMany('AppUserHasLocationTags', 'user_id')
        ->join('location_tags AS lt', 'lt.id', '=', 'location_tag_id')
        ->select('user_id', 'lt.name', 'location_tag_id');
}

然后返回一個更好的結果集:

Which then returns a much nicer results set:

{
    "id": 1,
    "created_at": "2015-05-13 13:04:56",
    "updated_at": "2015-05-13 13:04:56",
    "email": "REMOVED",
    "firstname": "REMOVED",
    "lastname": "REMOVED",
    "location_id": 0,
    "deleted_at": null,
    "permissions": [],
    "location_tags": [
        {
            "user_id": 1,
            "name": "Test Tag 0",
            "location_tag_id": 1
        },
        {
            "user_id": 1,
            "name": "Test Tag 123",
            "location_tag_id": 2
        }
    ]
}

希望這對未來的人有所幫助，因為它讓我猜了好幾個小時.

Hope this helps someone out in the future, because it kept me guessing for a good couple of hours.

這篇關于Laravel 多對多 - 意外的結果集 ->select()的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！