Question

我使用Datatable作为我的候选人列表，并使用区和党名搜索候选人。我正在使用ajax。问题是当搜索没有任何结果时我无法显示“找不到结果”。我该怎么做？码：单击搜索按钮时：

$(document).on("click", "#submit_filter", function(){
        $(this).attr('disabled', 'disabled');
        $('input#loading').css('display', 'block');
        var districtId = $("select#filter_district option:selected").val();
        var partyId = $("select#filter_party option:selected").val();
        var langId = $("input#hidden_lang").val();
        var table = $.fn.dataTable.fnTables(true);
        if ( table.length > 0 ) {
          var oTable = $('#candidates_table').dataTable();
          oTable.fnDestroy();
        }
        $("#candidates_table").DataTable({
            "bSort": false,
            "bProcessing": true,
            "sAjaxSource": "<?php echo base_url('search_candidates') ?>"+"/"+districtId+"/"+partyId+"/"+langId,
            "fnServerData": function ( sSource, aoData, fnCallback ) {
                $.ajax({
                    "dataType": 'json', 
                    "type": "POST", 
                    "url": sSource, 
                    "data": aoData, 
                    "success": fnCallback
                });
            }

在search_candidates中，控制器代码为：

function index($districtId,$partyId,$langId){
$result = $this->search->SearchCandidates($langId, $districtId, $partyId);
            $json_array = array();
            $res = array("aaData"=>'');
            if(isset($result) && !empty($result) && is_array($result)){
                foreach($result as $ind=>$val){
$nos = $ind+1;
                        $data['S.Nos.'] = $nos;
                        $data['District'] = $val->district;
                        $data['Election Area Nos'] = $val->election_nos;
                        $data['Candidate Name'] = $val->candidate_name;
                        $data['Political Party'] = $val->party_name;
                        $data['Age'] = $val->age;
                        $data['Gender'] = $val->gender;
                        $array = array($nos, $district, $election_nos, $val->candidate_name, $val->party_name, $val->age, $val->gender);
                        array_push($json_array, $array);
}
$res['aaData'] = $json_array;
}
echo json_encode($res);
}

通过此获得的数据是：

{"aaData":[[1,"Kathmandu","1","Prakash Man Singh","Nepali Congress","58","M"],[2,"Kathmandu","2","Madhav Kumar Nepal","Nepal Communist Party(Markswadi-Leninwadi)","60","M"],[3,"Kathmandu","3","Rameshwor Fuyal","Nepal Communist Party (Ekikrit Markswadi-Leninwadi)","51","M"],[4,"Kathmandu","4","Gagan Kumar Thapa","Nepali Congress","37","M"],[5,"Kathmandu","5","Narhari Acharya","Nepali Congress","60","M"],[6,"Kathmandu","6","Bhimsen Das Pradhan","Nepali Congress","59","M"],[7,"Kathmandu","7","Ram Bir Manandhar","Nepal Communist Party (Ekikrit Markswadi-Leninwadi)","50","M"],[8,"Kathmandu","8","Nabindra Raj Joshi","Nepali Congress","52","M"],[9,"Kathmandu","9","Dhyan Govinda Ranjit","Nepali Congress","66","M"],[10,"Kathmandu","10","Rajendra Kumar K.C.","Nepali Congress","55","M"]]}

我如何解决这个问题？欢迎任何帮助/建议。

Answer 1

Datatables负责这一点。您所要做的就是传递一个空数组[]。从代码中我看到你有$res = array("aaData"=>'');。这可能就是为什么你没有得到＆＃34;没有找到结果＆＃34;信息。我不知道php所以我无法建议确切的更改，但要获取消息json输出应该是这样的

{"aaData":[]}

干杯!!

Answer 2

您将返回一个json值到页面，检查页面 decode_json($res); 和

if(empty($res)){
echo "No data found";
}

检查数组是否为空

array['']

如何在数据表中显示错误消息

2 个答案: