Php 选择2部分加载数据

我正在使用select2插件显示我的下拉数据

我有问题，因为我的下拉菜单select2有18000个数据。 如何使数据部分显示，而不是加载所有数据。我使用Medoo数据库框架。我该怎么做

这是我的源代码

        <div class="form-group">
            <label for="nama_user"><?php _e('Nama User'); ?></label>
            <select class="form-control select2 select2-hidden-accessible" id="nama_user" name="nama_user" style="width: 100%;" tabindex="-1" aria-hidden="true">
                <option value="0"><?php _e('Nobody'); ?></option>
                <?php foreach ($ldap as $ldap) { ?>
                <option value='<?php echo $ldap['id']; ?>'><?php echo $ldap['nama_user']; ?></option>
                <?php } ?>
            </select>
       </div>

此函数用于使用Medoo调用数据

    case "gemilang/add":
    $contacts = getTable("contacts");
    if($isAdmin) { $assets = getTable("assets"); } else { $assets = getTableFiltered("assets","clientid",$liu['clientid']); }
    $clients = getTable("clients");
    $cabang_gemilang = getTable("cabang_gemilang");
    $ldap = getTable("ldap");
    $admins = getTableFiltered("people","type","admin");
    if($isAdmin) { $users = getTableFiltered("people","type","user"); } else { $users = getTableFiltered("people","type","user","clientid",$liu['clientid']); }
    break;

---

您可以按照我用来执行数据限制的代码，使用AJAX加载值？延迟：250？您不需要设置数据限制，只需设置“minimumInputLength=3”，这样，当您在selectbox中键入3个字符时。这会给你们所有的比赛结果谢谢jaydp，这超出了我的想象。

    case "gemilang/add":
    $contacts = getTable("contacts");
    if($isAdmin) { $assets = getTable("assets"); } else { $assets = getTableFiltered("assets","clientid",$liu['clientid']); }
    $clients = getTable("clients");
    $cabang_gemilang = getTable("cabang_gemilang");
    $ldap = getTable("ldap");
    $admins = getTableFiltered("people","type","admin");
    if($isAdmin) { $users = getTableFiltered("people","type","user"); } else { $users = getTableFiltered("people","type","user","clientid",$liu['clientid']); }
    break;