來自mysql / JSON的twitter bootstrap typeahead源碼

Question

所以基本上，我希望我的typeahead從mysql表獲取源代碼。

以下是我的代碼：

 <script src="../../js/bootstrap-typeahead.js"></script>
 <script type="text/javascript">
  var patients = [HERE I WANT TO PUT THE SUGGEST_PATIENTS.PHP];
       $('#search_bar').typeahead({source: patients})

 </script> 

suggest_patient.php

 <?php
include('../../db.php');

       $query = $conn->prepare('SELECT * FROM patients WHERE fname LIKE ?');
       $query->execute(array('value%'));

           $output_string = '';

           for($i=0; $row = $query->fetch(); $i++){
                $fname = $row['fname'];
                $lname = $row['lname'];
                $mname = $row['mname'];
                $bday = $row['bday'];
                $religion = $row['religion'];
                $occupation = $row['occupation'];
                $gender = $row['gender'];
                $phoneno = $row['phoneno'];
                $address = $row['address'];
                $type = $row['type'];    
           }

           $output_string = $fname;
      echo json_encode($output_string);
 ?>

但它沒有獲得先行效應。 你能幫我弄清楚我的代碼中缺少什么嗎？ 非常感謝幫助。

Answer 1

如果您閱讀文檔，則說明source應該是數組或javascript函數。