Laravel Eloquent 不只更新插入
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/17154210/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Laravel Eloquent doesn't update only inserting
提问by kim larsen
I am facing some problems with PHP Laravel update function. It doesn't update, only insert.
我在 PHP Laravel 更新功能方面遇到了一些问题。它不更新，只插入。
public function post_rate(){
$o = Input::all();
//has user already rated?!
$query = DB::table("ratings")->select("id")->where("user", "=", Auth::user()->id)->where("story_id", "=", $o['story_id'])->get();
foreach ( $query as $d):
$theID = $d->id;
endforeach;
if ( empty($query)): //User hasn't rated!
$z = new Rating();
else:
$z = new Rating($theID); //user has rated, tell which to update
-----------------------------^ that cause the problem!
endif;
$z->story_id = $o['story_id'];
$z->user = Auth::user()->id;
$z->rating = $o['rating'];
$z->save();
echo "OK";
}
It works when no rows founded, but when i will use new Rating(@something) it fails.
它在没有建立行时工作，但是当我使用 new Rating(@something) 时它会失败。
the column id in "ratings" table is primary and auto_increment.
“ratings”表中的列id是primary和auto_increment。
In my model i have also set up
在我的模型中，我还设置了
public static $key = 'id';
The output "$theID" also contains the correctly ID for the mysql row.
输出“$theID”还包含 mysql 行的正确 ID。
回答by Half Crazed
Try: $z = Rating::find($theID)with ->first()or ->get()instead
尝试：$z = Rating::find($theID)用->first()或->get()代替
回答by coffe4u
$z = Rating::find($theID);
or
或者
$z = Rating::where('id', '=', $theID)->first();
Both are equivalent.
两者是等价的。
回答by saran banerjee
In order to update in laravel simply use the command
为了在 Laravel 中更新，只需使用命令
Rating::where('id','=',$theID)->update(array('name'=> $name));
I hope this can be of some help.
我希望这能有所帮助。
回答by Luiz Alb Silva
The dude !
伙计！
I was taking the same problem. Checking out the source code, I found that the correct field to set the primary key name is 'primarykey'.
我遇到了同样的问题。查看源代码，我发现设置主键名称的正确字段是“primarykey”。
For example:
例如：
class User extends Eloquent {
protected $table = 'users';
protected $primaryKey = 'ID';
public $timestamps = false;
}
回答by Mehdi Maghrooni
Take it easy!
take a look at this , when user hasn't rated you just save it as you did before !
把它容易！
看看这个，当用户没有评分时，您只需像以前一样保存它！
if ( empty($query)){ //User hasn't rated!
$z = new Rating();
$z->story_id = $o['story_id'];
$z->user = Auth::user()->id;
$z->rating = $o['rating'];
$z->save();
}
and in the updatepart ! do as follows :
并在更新部分！做如下：
else{
$rates = array(
'story_id' => $o['story_id'],
'user' => Auth::user()->id,
'rating' => $o['rating'],
);
$old = Rating::find($theID);
$old->fill($rates);
$old->save();
}
回答by Gilberto Alexandre
For registration, Eloquent was not running the update because the method I used (see below) did not set the exists attribute to true. This is because I was getting all the fields returned ($ request-> all) from the request and creating a model, but that does not enable the object to update.
对于注册，Eloquent 没有运行更新，因为我使用的方法（见下文）没有将 exists 属性设置为 true。这是因为我从请求中获取了返回的所有字段（$ request-> all）并创建了一个模型，但这并不能使对象更新。
This did not work
这没有用
$produto = new Produto($request->all());
$produto->update();
This also did not work
这也不起作用
$produto = new Produto($request->all());
$produtoToUpdate = Produto::find($this->id);
$produtoToUpdate->update(get_object_vars($produto));
Never use get_object_var, Eloquent Model has virtual attribute (you get with getAttributes), so get_object_vardid not return your fields.
永远不要使用get_object_var，Eloquent模型有虚拟属性（你用 getAttributes 得到），所以get_object_var没有返回你的字段。
This work!!
这个作品！！
$produto = new Produto($request->all());
if ($this->id <= 0 )
{
$produto->save();
} else
{
$produto->exists = true;
$produto->id = $this->id;
$produto->update();
}
回答by Ravi Ranjan
I also faced same problem, so it turns out to be eloquent model issue.
我也遇到了同样的问题，所以结果证明是雄辩的模型问题。
First of all add a protected array name fillable in your model and define the attributes.Like
首先在模型中添加一个可填充的受保护数组名称并定义属性。比如
protected $fillable = array('story_id', 'rating');
There are many way to update table in Laravel in your case.
在您的情况下，有很多方法可以在 Laravel 中更新表。
1. $requestedRating = Rating::findOrFail($id)->first()->update($request->all()); // No need for first if the id is unique.
2. $requestedRating = Rating::findOrFail($id)->get()->update($request->all()); // fetch records from that model and update.
3. $requestedRating = Rating::whereIn('id', $id)->update($request->all());
In place of update you can also use fill like.
您还可以使用填充来代替更新。
1. $requestedRating = Rating::findOrFail($id)->first()->fill($request->all())->save();
There is always DB::table method but use this only when no option is available.
总是有 DB::table 方法，但只有在没有选项可用时才使用它。
