ErrorException strtotime()期望参数1为字符串,给定数组 [英] ErrorException strtotime() expects parameter 1 to be string, array given
问题描述
有人可以帮助我吗?这是我的控制器,当我尝试存储购买"数据时,它显示此消息 strtotime()期望参数1为字符串,给定数组 错误发生在 $ purchase-> date = date('Ym-d',strtotime($ request-> date));
和 $ purchaseDetail-> date = date('Ym-d',strtotime($ request-> date));
Can anyone help me? this is my controller, when i try to store 'purchase' data it show this message strtotime() expects parameter 1 to be string, array given
The error was on the $purchase->date = date('Y-m-d', strtotime($request->date));
and $purchaseDetail->date = date('Y-m-d', strtotime($request->date));
public function store(Request $request){
if($request->category_id == null){
return redirect()->back()->with('error', 'Please Purchase The Product');
} else{
// Multipale Data Insert start //
$purchase = new purchase();
$purchase->purchase_no = $request->purchase_no;
$purchase->date = date('Y-m-d', strtotime($request->date));
$purchase->description = $request->description;
$purchase->status = '0';
$purchase->created_by = Auth::user()->id;
DB::transaction(function() use($request,$purchase) {
if($purchase->save()) {
// Purchase Details Insert Start //
$category_id = count($request->category_id);
for ($i=0; $i < $category_id; $i++) {
$purchaseDetail = new purchaseDetail();
$purchaseDetail->date = date('Y-m-d', strtotime($request->date));
$purchaseDetail->purchase_id = $purchase->id;
$purchaseDetail->supplier_id = $request->supplier_id[$i];
$purchaseDetail->category_id = $request->category_id[$i];
$purchaseDetail->product_id = $request->product_id[$i];
$purchaseDetail->buying_qty = $request->buying_qty[$i];
$purchaseDetail->unit_price = $request->unit_price[$i];
$purchaseDetail->buying_price = $request->buying_price[$i];
$purchaseDetail->discount_amount = $request->discount_amount[$i];
$purchaseDetail->ppn = $request->ppn[$i];
$purchaseDetail->status = '0';
$purchaseDetail->save();
}
}
});
}
// Redirect
return redirect()->route('purchase.view')->with('success', 'Purchase Added Successfully');
}
你好,我尝试将 $ purchase-> date = date('Ym-d',strtotime($ request-> date));
更改为 $ purchase-> date = $ request-> date;
但是结果是 TypeError传递给Illuminate \ Database \ Grammar :: parameterize()的参数1必须为数组类型,给定字符串,在D:\ Project Laravel \ alc-pos \ vendor \ laravel \ framework \ src \ Illuminate \ Database \ Query \中调用第869行的Grammars \ Grammar.php
Hello, i try to change $purchase->date = date('Y-m-d', strtotime($request->date));
into $purchase->date = $request->date;
But the result was TypeError
Argument 1 passed to Illuminate\Database\Grammar::parameterize() must be of the type array, string given, called in D:\Project Laravel\alc-pos\vendor\laravel\framework\src\Illuminate\Database\Query\Grammars\Grammar.php on line 869
推荐答案
对于我的意见,在保存到数据库时不得格式化日期.相反,您可以在将其从数据库输出到刀片模板时对其进行格式化.
As for my opnion you must not format your date when saving to database. Instead you can format it when outputit from database into your blade templates.
无论如何都要检查 $ request-> date
,似乎是一个数组.
Anyway make a check on $request->date
, seems to be an array.
您可以在模型中使用Laravel访问器设置日期格式(例如):
You can format your date using Laravel accessor in your model like this (example):
public function getFormattedStartDateAttribute()
{
return $this->start_date->format('d-m-Y');
}
,然后像这样(例如)在您的刀片服务器模板中调用它:
and then call it in your blade template like this (example):
<input type="text" name="start_date" class="form-control" value="{{$trip->formatted_start_date}}">
有关Laravel Mutators&的更多信息访问者: Laravel文档
More info about Laravel Mutators & Accessors : Laravel Docs
您还可以做另一件事.在您的 schema :: create
make:
Also you can do another thing. On your schema::create
make:
$table->date('date_name_column');
然后将其保存,如下所示:
and then just save it like follow:
$purchaseDetail->date = $request->input('date');
这样,数据库中的日期将被保存为:YYYY-MM-DD
This way the date in your database will be save as follow : YYYY-MM-DD
希望有帮助!
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