我在laravel 5.4视图中有这样的表格:
<form action="#" method="POST" class="form-inline" role="form">
{{ csrf_field() }}
<div class="form-group">
<select id="tdcategory" class="selectpicker show-tick show-menu-arrow" onchange="this.form.submit()">
@foreach($categories as $category)
<option value="{{ $category->id }}">{{ $category->name }}</option>
@endforeach
</select>
</div>
<div class="form-group">
<select id="tdlocation" class="selectpicker show-tick show-menu-arrow" onchange="this.form.submit()">
@foreach($locations as $location)
<option value="{{ $location->id }}">{{ $location->name }}</option>
@endforeach
</select>
</div>
</form>
我有两个问题:
1)如何将表单操作动态设置为
类别/ {类别 - &GT; ID} /位置/ {位置 - &GT; ID}
其中,category-&gt; id和location-&gt; id都将基于用户当前选择的选项的值
2)由于在用户更改选项时提交了表单,因此会重新加载页面并重置<select>
标记。如何跟踪以前选择的选项,当页面加载时,我可以向用户显示他/她在提交表单之前选择了哪一个?
答案 0 :(得分:1)
1)您可以通过JS执行此操作:从选择中删除“onchange”属性,并添加初始代码,如下所示:
$('#tdcategory,#tdlocation').change(function(){
var category = $('#tdcategory').val();
var location = $('#tdlocation').val();
if (category && location) {
$(this).closest('form')
.attr('action', '/categories/'+ category +'/locations/' + location)
.submit();
}
});
2)您可以从请求中获取提交的值,并为每个选择添加一个插入selected =“selected”属性的条件,如下所示:
<select id="tdcategory" name="tdcategory" class="selectpicker show-tick show-menu-arrow">
@foreach($categories as $category)
<option value="{{ $category->id }}" @if($category->id == request()->get('tdcategory'))selected="selected"@endif>{{ $category->name }}</option>
@endforeach
</select>
<select id="tdlocation" name="tdlocation" class="selectpicker show-tick show-menu-arrow">
@foreach($locations as $location)
<option value="{{ $location->id }}" @if($location->id == request()->get('tdlocation'))selected="selected"@endif>{{ $location->name }}</option>
@endforeach
</select>