[英]Laravel Eloquent doesn't update only inserting
I am facing some problems with PHP Laravel update function. 我正面临着PHP Laravel更新功能的一些问题。 It doesn't update, only insert.
它不会更新,只会插入。
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. 当没有成立行时它会起作用,但是当我使用新的等级(@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,我也设置了
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。
尝试: $z = Rating::find($theID)
,而不是->first()
或->get()
$z = Rating::find($theID);
or 要么
$z = Rating::where('id', '=', $theID)->first();
Both are equivalent. 两者都是等价的。
In order to update in laravel simply use the command 要在laravel中更新,只需使用该命令即可
Rating::where('id','=',$theID)->update(array('name'=> $name));
I hope this can be of some help. 我希望这可以提供一些帮助。
The dude ! 老兄!
I was taking the same problem. 我也遇到了同样的问题。 Checking out the source code, I found that the correct field to set the primary key name is 'primarykey'.
检查源代码,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 $timestamps = false;
}
Take it easy ! 别紧张 !
take a look at this , when user hasn't rated you just save it as you did before ! 看看这个,当用户没有评价你只是像以前一样保存它!
if ( empty($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 update part ! 并在更新部分! do as follows :
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);
$old->save();
}
For registration, Eloquent was not running the update because the method I used (see below) did not set the exists attribute to true. 对于注册,Eloquent没有运行更新,因为我使用的方法(见下文)没有将exists属性设置为true。 This is because I was getting all the fields returned ($ request-> all) from the request and creating a model, but that does not enable the object to update.
这是因为我从请求中获取了所有返回的字段($ request-> all)并创建了一个模型,但这并不能使对象更新。
This did not work 这没用
$produto = new Produto($request->all());
$produto->update();
This also did not work 这也行不通
$produto = new Produto($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_var did not return your fields.
永远不要使用get_object_var , Eloquent Model有虚拟属性(你得到getAttributes),所以get_object_var没有返回你的字段。
This work!! 这项工作!!
$produto = new Produto($request->all());
if ($this->id <= 0 )
{
$produto->save();
} else
{
$produto->exists = true;
$produto->id = $this->id;
$produto->update();
}
I also faced same problem, so it turns out to be eloquent model issue. 我也遇到了同样的问题,所以事实证明这是一个雄辩的模型问题。
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
protected $fillable = array('story_id', 'rating');
There are many way to update table in Laravel in your case. 在您的情况下,有很多方法可以更新Laravel中的表。
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. 代替更新,您也可以使用填充。
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. 始终存在DB :: table方法,但仅在没有可用选项时才使用此方法。
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