我想用包含4-5个字段的弹出表单对帖子发表评论。我想在弹出表单中提供要评论的帖子的帖子ID,但是当通过隐藏字段发布此ID时,它总是导致最后一个帖子ID。如何生成动态不同的ID,该ID可为我提供确切的帖子ID?
<form id="forms">
<label for="email">Facility :</label>
<input type="text" class="form-control" id="facility" name="facility" maxlength="50">
//printing{{$customerpost->id}} this value gives correct post id
<input type="hidden" name="hidden" value="{{$customerpost->id}}" id="hidden">
<button type="submit" class="btn btn-lg btn-primary btn-block" id ="submit">Post It! </button>
</form>
$( 'form' ).submit(function ( e ) {
var data;
data = new FormData();
$.ajax({
headers: {
'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content'),
authorization: $('meta[name="csrf-token"]').attr('content'),
}
});
data.append('facility', $("#facility").val());
data.append('hidden', $("#hidden").val());
console.log(data);
alert(data);
$.ajax({
url: 'api/bidon/',
data: data,
processData: false,
contentType: false,
type: 'POST',
success: function ( data ) {
alert("sucess");
alert( data );
}
});
e.preventDefault();
});
当我登录到控制台时,它总是为我提供最后的帖子ID。如何使用Laravel框架解决此问题?
答案 0 :(得分:0)
点击即可简单执行
<!-- Add post id in attribute -->
<div data-id="12345" class="btn"> add comment </a>
<script type="text/javascript">
$('body').on('click','.btn',function(){
// get id from
var id = $(this).attr('data-id');
// set id to hidden attribute
// $('#hidden').attr('data-id', id);
// or set id to hidden input
$('#hidden').val(id);
// Show form
$('#forms').slideUp();
});
</script>