【发布时间】:2019-05-21 14:34:52
【问题描述】:
我一直在使用 ajax post 使用 Laravel 5.2 成功更新我的数据库信息。视图、控制器和模型已经到位,但是当我使用 Laravel 5.7 实现它时,我收到了这个错误
Symfony\Component\HttpKernel\Exception\> MethodNotAllowedHttpException 没有消息。
使用 Laravel 5.7 是否需要进行任何更改或调整?请帮忙。谢谢
查看:
<form class="update-member-form" id="update-member-form" enctype="multipart/form-data">
{{ csrf_field() }}
<input type="text" name="id" class="form-control" id="primary_id" value="">
<div class="form-group">
<label for="recipient-name" class="col-form-label">Member ID:</label>
<input type="text" name="member_id" class="form-control" id="member_id" value="" required>
</div>
<div class="form-group">
<label for="recipient-name" class="col-form-label">First Name:</label>
<input type="text" name="fname" class="form-control" id="editMember_name" value="" required>
</div>
<div class="form-group">
<label for="recipient-name" class="col-form-label">Middle Name:</label>
<input type="text" name="mname" class="form-control" id="editMember_mname" value="" required>
</div>
<div class="form-group">
<label for="recipient-name" class="col-form-label">Last Name:</label>
<input type="text" name="lname" class="form-control" id="editMember_lname" value="" required>
</div>
</form>
<div class="modal-footer">
<button type="button" class="btn btn-secondary" data-dismiss="modal" onClick="window.location.reload()" >Close</button>
<button type="button" class="btn btn-primary btn-update-member">Save updates</button>
</div>
Javascript
$(".btn-update-member").click(function(e){
e.preventDefault();
$.post("{{ url('/updatemember') }}", $("#update-member-form").serialize(), function(data){
if(data.notify == "Success"){
swal({
title: "Record successfully updated",
text: "Message will close in 2 seconds",
type: "success",
timer: 2000
});
} else{
console.log(data.notify);
}
},"json");
});
web.php
Route::post('/updatemember', 'MembersController@update');
控制器
public function update(Request $request, $id)
{
//
$updateMember = Member::where( 'id', $request['id'] )
->update( $request->all() );
if( $updateMember ){
$notification = "Success";
} else{
$notification = "Failed";
}
return json_encode( array( 'notify'=>$notification ) );
}
【问题讨论】:
标签: javascript ajax laravel laravel-5.7