无法获取数组结果来填充 jquery select2

Question

我正在尝试让 jQuery SELECT2 从 mysqli 查询中提取数据：

[fxn/fxn_select2_series.php]

if(!isset($_POST['searchTerm'])){ 

    $qry_select2_series = 

    "SELECT DISTINCT series_id as 'id', series_title as 'text' 

     FROM `series` WHERE series_title IS NOT NULL" ;

}else{ 

    $search = $_POST['searchTerm'];  

    $qry_select2_series = 

    "SELECT DISTINCT series_id as 'id', series_title as 'text' 

     FROM `series` WHERE series_title IS NOT NULL and series_title LIKE '%".$search."%'" ;

   }

$cxn = new mysqli('localhost', $user, $pass, $db);

$result = mysqli_query($cxn,$qry_select2_series);

$response = mysqli_fetch_all($result);

echo json_encode($response);

如果我自己打开那个页面，我会得到正确的数组格式结果：

[["1","Spring Revels"],["2","Trois Chansons"]]

我将查询插入 jQuery SELECT2 ...

<fieldset>

    <legend>Title</legend>

    <div class='inputlabel'>

        <label for='title'>Title</label>

        <input type='text' size='64' id='title'>

        <label for 'subtitle'>Subtitle</label>

        <input type='text' size='64' id='subtitle'>

        <label for='series'>Series</label>

        <span><select class="js-example-basic-single" id='select2_series' style='width:32'>

            <option>Select/Search ...</option>

        </select>

        (Enter Opus, Collection, Volume, Libretto, Etc. or select below...)</span>

        <label for='sequence'>Sequence</label>

        <span><input type='number' min='1' max='99' step='1' size='4' id='sequence'>

        (Enter the sequence number of this item in the above series)</span>

</fieldset>

<script type="text/javascript">

$(document).ready(function(){

   $("#select2_series").select2({

      ajax: {

        url: "fxn/fxn_select2_series.php",

        type: "post",

        dataType: 'json',

        delay: 250,

        data: function (params) {

           return {

              searchTerm: params.term // search term

           };

        },

        processResults: function (response) {

           return {

              results: response

           };

        },

        cache: true

      }

   });

});

</script>

Answer 1

根据规范，正确的格式是：{  "results": [    {      "id": 1,      "text": "Option 1"    },    {      "id": 2,      "text": "Option 2"    }  ],  "pagination": {    "more": true  }}因此，您需要遍历结果并创建一个与该格式匹配的数组。

Answer 2

$response = mysqli_fetch_all($result);我按如下方式循环，其他一切保持不变：while($row = mysqli_fetch_array($result)) {$response[] = array("id"=>$row['id'],"text"=>$row['text']);}它完美地工作。