2017-04-12 17 views
0

我试图将数据插入到我的数据库中,但似乎无法获取此出生日期形式以便接受。这里是我的表格是一个jsfiddle https://jsfiddle.net/10990b6t/2/然而,当我尝试发送这个数据到我的数据库它只是返回到注册页面。我以前有一个纯文本框,它工作得很好,但显然没有使JavaScript形式好多了。Laravel - 努力将已被JavaScript更改的数据插入到数据库中

--old文本框,在形式worked--

<label for="title">Date of Birth</label> 
<input type="text" name="dateofbirth" id="dateofbirth"> 
</div> 

- 新的尝试并不work-- registerForm.blade.php

<input 
    type="text" 
    name="date" 
    id="dateofbirth" 
    placeholder="dd/mm/yyyy" 
    onkeyup=" 
     var v = this.value; 
     if (v.match(/^\d{2}$/) !== null) { 
      this.value = v + '/'; 
     } else if (v.match(/^\d{2}\/\d{2}$/) !== null) { 
      this.value = v + '/'; 
     }" 
    maxlength="10" 
> 

AuthController.php

function registerUser(Request $request) 
    { 
     $this->validate($request, [ 
      'firstname' => 'required|min:3|max:25', 
      'surname' => 'required', 
      'address' => 'required', 
      'postcode' => 'required', 
      'email' => 'required|min:5|max:30|unique:users', 
      'password' =>'required|min:5|max:30|', 
      'dateofbirth' => 'required', 
     ]); 
     //create a User object 
     $user = new User(); 
     $user->firstname = $request->firstname; 
     $user->surname = $request->surname; 
     $user->address = $request->address; 
     $user->postcode = $request->postcode; 
     $user->email = $request->email; 
     $user->password = $request->password; 
     $user->dateofbirth = $request->dateofbirth; 
     $user->role = 1; 

     $user->save(); 
     return redirect('all'); 
    } 

回答

1

您期待的字段名为dateofbirth,并且您的输入具有name =“date”,它应该是name =“date分娩“

<input 
    type="text" 
    name="dateofbirth" 
    id="dateofbirth" 
    placeholder="dd/mm/yyyy" 
    onkeyup=" 
     var v = this.value; 
     if (v.match(/^\d{2}$/) !== null) { 
      this.value = v + '/'; 
     } else if (v.match(/^\d{2}\/\d{2}$/) !== null) { 
      this.value = v + '/'; 
     }" 
    maxlength="10" 
> 
+0

尴尬的错误,谢谢! – TF120

+0

没问题:p考虑接受我的答案,当你能够:) – WilomGfx

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