javascript - 无法获取数组结果以填充 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);

如果我自己拉出该页面，我会以数组格式获得正确的结果：

[["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>

但是当您在 Select2 对象的搜索框中键入内容时，该页面什么也没有显示...

标签： javascriptphpjquerymysqlijquery-select2

Answer 1

感谢imvain2关于查看输出格式的建议。确切的解决方案......而不是

$response = mysqli_fetch_all($result);

我循环如下，保持其他一切相同：

while($row = mysqli_fetch_array($result)) {
$response[] = array("id"=>$row['id'],"text"=>$row['text'])
;}

而且效果很好。