php - Select2 Ajax不匹配结果

Question

我的 Select2 表单使用 ajax 从数据库加载结果，但在前端，select2 搜索返回所有结果，而不是特定于搜索词的结果。经过一些研究后，我意识到匹配器必须位于服务器端，但我似乎无法让它工作。这是我的代码

查询数据库并返回json数据的代码

<?php

  $servername = "localhost";
  $username = "myusername";
  $password = "mypassword";

   try {
$conn = new PDO("mysql:host=$servername;dbname=mydatabase", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
   }
    catch(PDOException $e)
   {
   echo "Connection failed: " . $e->getMessage();
   }



  // strip tags may not be the best method for your project to apply extra 
   layer of security but fits needs for this tutorial 
   $search = strip_tags(trim($_GET['q'])); 

 // Do Prepared Query
   $query = $conn->prepare("
   SELECT * FROM (
    SELECT wp_terms.name
   FROM wp_terms
   JOIN wp_term_taxonomy
    ON wp_term_taxonomy.term_id = wp_terms.term_id
    WHERE taxonomy = 'job_listing_region'
    AND count = 0
    ) as T"
     );

    // Add a wildcard search to the search variable
     $query->execute(array(':search'=>"%".$search."%"));


   // Do a quick fetchall on the results
    $list = $query->fetchall(PDO::FETCH_ASSOC);

   // Make sure we have a result
   if(count($list) > 0){
    foreach ($list as $key => $value) {
    $data[] = array('id' => $value['name'], 'text' => $value['name']);              
   } 
    } else {
   $data[] = array('id' => '0', 'text' => 'No Products Found');
   }


// return the result in json
echo json_encode($data);

我的jquery

<script>
 jQuery(function($) {
 $(document).ready(function() {
     function matchStart (term, text) {
  if (text.toUpperCase().indexOf(term.toUpperCase()) == 0) {
    return true;
  }

  return false;
}
$( "#regions1" ).select2({        
    ajax: {
        url: "/ajax/connect.php",
        dataType: 'json',
        delay: 250,
        data: function (params) {
            return {
                q: params.term // search term
            };
        },
        processResults: function (data) {
    return {
        results: $.map(data, function(obj) {
            return { id: obj.id, text: obj.text };

            })

        };

        }


},
  minimumInputLength: 3,
  width: "300px",
  placeholder: "Please Select Up To Four Additional Cities"

 });

    });
 });
</script>

这是我的 json 结果的示例

[{"id":"Abanda, AL","text":"Abanda, AL"},{"id":"Abbeville, AL","text":"Abbeville, AL"},{"id":"Adak, AK","text":"Adak, AK"},

Answer 1

您在 sql 语句中没有提到 like 条件。

你的sql应该是这样的

$query = $conn->prepare("
   SELECT * FROM (
    SELECT wp_terms.name
   FROM wp_terms
   JOIN wp_term_taxonomy
    ON wp_term_taxonomy.term_id = wp_terms.term_id
    WHERE taxonomy = 'job_listing_region'
    AND count = 0 AND search_column LIKE ?    <==== this line
    ) as T"
     );

$query->execute(array("%".$search."%"));