我用jquery ajax代码来运行搜索功能。jquery传递值到单选按钮
,这里是我的ajax至今:
$('button[type="search"]').click(function(e) {
$.ajax({
url: "{{ route('fine.search') }}",
type: "POST",
data: {
'_token' : '{{csrf_token() }}',
'driver_id' : $('select[name="driver_id"]').val(),
'fine_date' : $('input[name="fine_date"]').val(),
},
success: function(data) {
if(data.status == true) {
var result= $('#search-result');
$.each(data.getCarbyDriver, function(i, data) {
PlateEle = $('<input/>').attr({ type: 'radio', name:'rad'});
$("#search-result").html(data.plate_no);
StartEle = $('<div />').html(data.start_time);
EndEle = $('<div />').html(data.end_time);
});
$('#search-result').append(PlateEle, StartEle, EndEle);
}
},
error: function(data) {
}
});
});
和返回的数据像这样我的网络(inpect元)上:
car_id:5
driver_id:1
end_time:"2016-11-16 18:00:00"
plate_no:"DFE82846J"
start_time:"2016-11-16 08:00:00"
working_date:"2016-11-16 00:00:00"
这里是到目前为止的形式刀片代码:
<div class="row top">
<div class="col-xs-4 col-sm-4 col-md-4">
<div class="form-group">
<label class="control-label">Driver Name:</label>
{!! Form::select('driver_id', $driver, null, array('class' => 'form-control')) !!}
</div>
</div>
<div class="col-xs-4 col-sm-4 col-md-4">
<div class="form-group">
<label>Fine Date:</label>
{!! Form::text('fine_date', null, array('id' => 'datetimepicker', 'class' => 'form-control')) !!}
</div>
</div>
<div class="col-xs-4 col-sm-4 col-md-4">
<div class="form-group filter-btn">
<button class='btn btn-info' type='search'>Search</button>
</div>
</div>
</div>
<div class="row mid ">
<div class="col-xs-4 col-sm-4 col-md-4">
<div id="search-result"></div>
</div>
</div>
问题是我无法保存。它会抛出错误car_id'不能为空'。
我如何通过car_id到我的电台,所以一旦我的单选按钮,点击它节省plate_no
IM使用laravel car_id 5.3
你的ajax调用在哪里有'car_id'? –
@SandrinaPereira多数民众赞成我的问,如何将car_id传递给电台? –
但你想保存car_id在哪里?在无线电价值?在阿贾克斯呼叫(数据)我不理解你的问题... –