Question

我相信我会对我的工作方式有很多评论;我这是兼职并尝试使用MVC开发新系统。这是我当前的问题：当我加载表单（视图）时，我的控制器调用我的模型文件，该文件立即实例化视图的新模型并从MySQL数据库中检索旧的表单值。该视图有一堆下拉列表。我将db中的值加载到模型私有变量中，这些变量与视图中的名称相同，但是值没有传递到视图中，因此下拉列表都显示其默认值 - 它们应该显示值存储在db中。下面的代码，但只是相关的片段：

<?php
require_once("../controller/db_conn_ctlr.php");

class dom_leader_model {
    // Variables
    private $ldr1_rec;  // leader 1 record
    private $ldr1_role; // leader 1 role
    private $ldr2_rec;  // leader 2 record
    private $ldr2_role; // leader 2 role
    private $ldr3_rec;  // leader 3 record
    private $ldr3_role; // leader 3 role
    private $ldr4_rec;  // leader 3 record
    private $ldr4_role; // leader 3 role    
    private $trail_no;  // TRAIL # of plan

    // constructor
    public function __construct() {
        $trail_no = $this->getTrailNumber();
        $this->getOldLeaderInfo($trail_no);
    }

    // get TRAIL number
    public function getTrailNumber() {
        return $this->trail_no = $_SESSION['trail_no'];
    }

    // get leader info from database and put it on the form
    public function getOldLeaderInfo($trail_no) {

        $fiscal_yr = substr($trail_no, 0, 2);
        $trip_id = substr ($trail_no, 2, 3);
        $section_id = substr($trail_no, 5, 1);

        $query20 = "
            SELECT * FROM leader_info
            WHERE fiscal_yr = '$fiscal_yr' AND trip_id = '$trip_id'
                AND section_id = '$section_id'";

        $connect = db_connection();    
        $result20 = mysqli_query($connect, $query20) or die("Query20 failed: 
             " . mysqli_error($connect));
        $connect->close();

        if($result20) {
            $row = $result20->fetch_array(MYSQLI_ASSOC);
            $this->ldr1_rec = $row['leader1_record'];         
            $this->ldr1_role = $row['leader1_role'];
            $this->ldr2_rec = $row['leader2_record'];
            $this->ldr2_role = $row['leader2_role']; 
            $this->ldr3_rec = $row['leader3_record'];
            $this->ldr3_role = $row['leader3_role'];
            $this->ldr4_rec = $row['leader4_record'];
            $this->ldr4_role = $row['leader4_role'];            
            return 1;
        } else {
            return 0;
        }        
    }

[from controller]
<?php
require_once '../model/dom_leader_model.php';
require_once '../utilities/check_info_saved.php';
$session_start = session_start();

$leaderInfoModel = new dom_leader_model();
$trail_no = $leaderInfoModel->getTrailNumber();


if (isset($_POST['save']) || isset($_POST['next'])) {   

[from view]
<p>Leader 1 Information:</p>
<?php
    if (empty($ldr1_rec)) {
        $ldr1_switch = 'Select Leader';
        echo '<select name="ldr1_rec"><option selected 
            value="">',$ldr1_switch,'</option>',
            require("../../lists/leader_list.txt"),'</select>';
    } else {
        $ldr1_switch = leader_switch($ldr1_rec, 'dom');         
        echo '<select name="ldr1_rec"><option selected 
            value="',$ldr1_rec,'">',$ldr1_switch,'</option>',
            require("../../lists/leader_list.txt"),'</select>'; 
    }

    if (empty($ldr1_role)) {
        $role1 = "Select Role";
        echo '<select name="ldr1_role"><option selected 
            value="">',$role1,'</option>',
            require("../../lists/staff_role_list.txt"),'</select>';
    } else {    
        $role1 = role_switch($ldr1_role);
        echo '<select name="ldr1_role"><option selected 
            value="',$ldr1_role,'">',$role1,'</option>',
            require("../../lists/staff_role_list.txt"), '</select>&nbsp;';
     }
?><p><hr>

就像我说的那样，我做兼职，我相信很多人会看到更好的编写代码的方法，但我真正想要的是为什么我从数据库中检索到的值没有得到的答案加载时的形式。

提前致谢。

John C.

Answer 1

您应该使用selected="selected"而不仅仅是select 即：<option value="',$ldr1_role,'" selected="selected">',$role1,'</option>

尽管select对于浏览器来说是正确的，但它有时可能会被拒绝（这在几周之前发生在我身上，帮助/处理某人的其他代码）。

selected="selected"对AFAIK的所有文档都有效。

PHP：查看未从模型中获取选择值

1 个答案: