php count() 参数必须是数组或在laravel中实现countable的对象
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count() parameter must be an array or an object that implements countable in laravel
提问by faraz
This is code here:
这是这里的代码：
protected function credentials(Request $request)
{
$admin=admin::where('email',$request->email)->first();
if(count($admin))
{
if($admin->status==0){
return ['email'=>'inactive','password'=>'You are not an active person, Please contact to admin'];
}
else{
return ['email'=>$request->email,'password'=>$request->password,'status'=>1];
}
}
return $request->only($this->username(), 'password');
}
When i run the code this error become:
当我运行代码时，此错误变为：
"count(): Parameter must be an array or an object that implements Countable"
“count(): 参数必须是一个数组或一个实现 Countable 的对象”
回答by Jericho Manalo
This is my solution:
这是我的解决方案：
count(array($variable));
hope it works!
希望它有效！
回答by Dmytro Huz
It happens because of in PHP 7.2 NULL in count() return Warning. You can try to change
发生这种情况是因为在 PHP 7.2 中 count() 返回警告为 NULL。你可以尝试改变
count($admin)
to
到
count((is_countable($admin)?$admin:[]))
回答by Teoman T?ng?r
Note that here, When you use the count()method, there should be countable element, like an array or object.
注意这里，当你使用count()方法时，应该有可数元素，比如数组或对象。
Admin::where('email',$request->email)->first();
But the first()method give you single element, not a collection or array. The get()method returns you countable a collection with found elements
但是该first()方法为您提供单个元素，而不是集合或数组。该get()方法返回一个包含找到元素的可数集合
Instead of using count you can directly check variable itself is it defined or null
您可以直接检查变量本身是否已定义或为空，而不是使用计数
if($admin){
// do something here
}
or you can use is_null()method
或者你可以使用is_null()方法
if(!is_null($admin)){
// do something here
}
回答by Ali Farhoudi
$adminvariable is neither array nor object that implements countable. When you use first()the result will be a model object if record is found else it will be null. For this condition you can use:
$admin变量既不是数组也不是实现可数的对象。当您使用时first()，如果找到记录，结果将是一个模型对象，否则它将为空。对于这种情况，您可以使用：
if (!empty($admin)) {
//
}
Just replace if (count($admin))with if (!empty($admin)).
只需替换if (count($admin))为if (!empty($admin)).
And when you use get()method to get multiple records you can check by:
当您使用get()方法获取多条记录时，您可以通过以下方式进行检查：
if ($admins->count() > 0) {
//
}
回答by rajesh
$admin = null;
var_dump(count($admin));
output: Warning: count(): Parameter must be an array or an object that implements Countable in … on line 12 // as of PHP 7.2
输出：警告：count()：参数必须是一个数组或一个对象，在第 12 行中实现了 Countable // 自 PHP 7.2 起
if condition should be like:
如果条件应该是这样的：
if(isset($admin) && count($admin))
回答by sonu pokade
Well,
$admin=Admin::where('email',$request->email)->first();
//It will always return an **object**.
And make sure you included Admin model in your controller like as.
Use App\Admin;
at the same time check that you will have to mention which field of table needs to be fillable like in your model such as
protected $fillable = [
'first_name',
'last_name'
];
whatever data you will going to save in your database.
and then check object is null or not
I mean is.
if($admin && $admin!==null){
//do whatver you want to do.
}
回答by nakov
You should check if it is null instead of count, because you ask for one result with first()just this
您应该检查它是否为空而不是计数，因为您first()只需要一个结果
if($admin)
will do it.
会做的。
if you use return a collection using ->get()then you can check $admin->count().
如果您使用返回一个集合，->get()那么您可以检查$admin->count().
回答by paranoid
Use isset($admin->id)instead of count($admin)
使用isset($admin->id)代替count($admin)
Try this :
尝试这个 ：
protected function credentials(Request $request)
{
$admin=admin::where('email',$request->email)->first();
if(isset($admin->id)))
{
if($admin->status==0){
return ['email'=>'inactive','password'=>'You are not an active person, Please contact to admin'];
}
else{
return ['email'=>$request->email,'password'=>$request->password,'status'=>1];
}
}
return $request->only($this->username(), 'password');
}
回答by Arif Nofriadi
add this your controler this code:
将此代码添加到您的控制器中：
$user = User::where('email',$request->email)->first();
if ($user){
return redirect()->back()->with('errors','We cant find a user with that e-mail address.');
}else{
$user->password = bcrypt($request->new_password);
$user->update();
return redirect()->back()->with('success','Success');
}
