我需要在下拉列表更改时更新HTML表。我正在使用Ajax来执行此操作。 这是HTML代码和Ajax代码 HTML代码
<form role="form" action="/surveys/viewsurveys/{{ survey_id }}" method="post" >
<div class="form-group">{% csrf_token %}
<input type="hidden" id="Surey_ID" name="Surey_ID" value="{{ survey_id }}">
<label>Select Department or Section</label>
<select id="dept" name="dept" class="form-control" onchange="allusers();">
{% for items in dept_data %}
<option value="{{ items.id }}">{{ items.dept_name }}</option>
{% endfor %}
</select>
</div>
<div class="box-header">
<h3 class="box-title">All  <small class="badge pull-right bg-green"> {{ counts.hr_count }}</small></h3>
</div><!-- /.box-header -->
<div class="box-body table-responsive">
<table id="example2" class="table table-bordered table-hover">
<thead>
<tr>
<th> </th>
<th>First Name</th>
<th>Last Name</th>
<th>Email</th>
<th>Username</th>
<th>Data join</th>
<th>last login</th>
<th>Is Active</th>
</tr>
</thead>
<tbody>
{% for items in allusers %}
<tr >
<td>
<div class="input-group">
<span class="input-group-addon">
<input name="{{ items.id }}" type="checkbox">
</span>
</div><!-- /input-group -->
</td>
<td>{{ items.first_name }}</td>
<td>{{ items.last_name }}</td>
<td>{{ items.email }}</td>
<td>{{ items.username }}</td>
<td>{{ items.date_join }}</td>
<td>{{ items.last_login }}</td>
<td>{{ items.is_active }}</td>
<td class="text-center"><a class='btn btn-info btn-xs' href="/accounts/editusers/{{ items.id }}"><span class="glyphicon glyphicon-edit"></span> Edit</a>
</td>
</tr>
{% endfor %}
</tbody>
<tfoot>
<tr>
<th> </th>
<th>First Name</th>
<th>Last Name</th>
<th>Email</th>
<th>Username</th>
<th>Data join</th>
<th>last login</th>
<th>Is Active</th>
</tr>
</tfoot>
</table>
</div><!-- /.box-body -->
</form>
Ajax代码
function allusers()
{
$.ajax({
url : "/surveys/ajaxalldeptusers",
type : "POST",
dataType: "json",
data : {
csrfmiddlewaretoken: document.getElementsByName('csrfmiddlewaretoken')[0].value,
dept: $('#dept').val(),
Surey_ID:$('#Surey_ID').val(),
},
success : function(json) {
//$('#result').append( 'Server Response: ' + json.server_response);
$('#box').html(json.message),
console.log('my message' + json)
},
error : function(xhr,errmsg,err) {
alert(xhr.status + ": " + xhr.responseText);
}
});
}
后端
def ajax_all_dept_users(request):
try:
request.session['user_login_data']
dept_data=SuUserDepartment.objects.filter(org=request.session['user_login_data']['org'])
dept=request.POST.get('dept','')
survey_id=request.POST.get('Surey_ID','')
responce_data={}
if request.method == 'POST':
surey_data=SuSurey.objects.get(id=survey_id)
allusers=SuUser.objects.filter(dept_id=dept)
responce_data['allusers']=allusers
responce_data['dept_data']=dept_data
responce_data['surey_data']=surey_data
responce_data['survey_id']=survey_id
data = serializers.serialize('json', responce_data)
return HttpResponse(json.dumps(data),content_type="application/json")
#return HttpResponse(json.dumps({'allusers':allusers,'dept_data':dept_data,'surey_data':surey_data, 'survey_id':survey_id }),content_type="application/json")
#return render_to_response("pages/forms/publish.html",{'allusers':allusers,'dept_data':dept_data,'surey_data':surey_data, 'survey_id':survey_id },context_instance=RequestContext(request))
except KeyError, e:
messages={'alert':'You need to loging'}
return render(request, 'index.html',{'messages': messages},context_instance=RequestContext(request))
我尝试调试并尝试找出问题。我发现当来到serializers.serialize()时会出现此错误
我该如何修复此错误?需要快速帮助
答案 0 :(得分:1)
这是serializers.serialize
的常见错误您正在尝试序列化不是模型查询集的对象。
您的数据对象是字典,因此您不需要对其进行序列化,只需使用json dump返回该字典。
答案 1 :(得分:0)
responce_data['allusers']=allusers
responce_data['dept_data']=dept_data
responce_data['surey_data']=surey_data
responce_data['survey_id']=survey_id
data = serializers.serialize('json', responce_data)
此处您的调用将dict response_data
传递给序列化程序,而不是您应该传递要序列化的对象列表。
因此,要创建对象列表,也不要在列表中指定要序列化的字符串survey_id
。
将您的代码更新为
dept_data=list(SuUserDepartment.objects.filter(org=request.session['user_login_data']['org']))
allusers=list(SuUser.objects.filter(dept_id=dept))
responce_data = [ surey_data, ] + dept_data + allusers
data = serializers.serialize('json', responce_data)