我正在嘗試讓 jQuery SELECT2 從 mysqli 查詢中提取數據：[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($response);如果我自己打開那個頁面，我會得到正確的數組格式結果：[["1","Spring Revels"],["2","Trois Chansons"]]我將查詢插入 jQuery SELECT2 ...<fieldset> <legend>Title</legend> <div class='inputlabel'> <label for='title'>Title</label> <input type='text' size='64' id='title'> <label for 'subtitle'>Subtitle</label> <input type='text' size='64' id='subtitle'> <label for='series'>Series</label> <span><select class="js-example-basic-single" id='select2_series' style='width:32'> <option>Select/Search ...</option> </select> (Enter Opus, Collection, Volume, Libretto, Etc. or select below...)</span> <label for='sequence'>Sequence</label> <span><input type='number' min='1' max='99' step='1' size='4' id='sequence'> (Enter the sequence number of this item in the above series)</span></fieldset><script type="text/javascript">$(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>
2 回答
肥皂起泡泡
TA貢獻1829條經驗 獲得超6個贊
根據規范，正確的格式是：
{
? "results": [
? ? {
? ? ? "id": 1,
? ? ? "text": "Option 1"
? ? },
? ? {
? ? ? "id": 2,
? ? ? "text": "Option 2"
? ? }
? ],
? "pagination": {
? ? "more": true
? }
}
因此，您需要遍歷結果并創建一個與該格式匹配的數組。
qq_遁去的一_1
TA貢獻1725條經驗 獲得超8個贊
$response = mysqli_fetch_all($result);
我按如下方式循環，其他一切保持不變：
while($row = mysqli_fetch_array($result)) {
$response[] = array("id"=>$row['id'],"text"=>$row['text'])
;}
它完美地工作。
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