mysql - 应根据下拉列表显示值

Question

如果我从数据库中选择一个下拉值，之后应该显示标题值，例如我从下拉列表中选择“关于我们”页面，之后应该显示标题值帮助我任何人

查看页面

<form class="form-horizontal" action="#" method="POST">
    <?php foreach ($seo_tag_update as $rows) { ?>
        <div class="form-group">
            <label class="col-sm-4 control-label">Select Pages:</label>
            <div class="col-sm-6">
                <select class="form-control" name="select_update_page" id="select_update_page">
                    <option value=""> -- select pages -- </option>
                    <?php
                    echo '<option value="' . $rows->page_id . '">' . $rows->page_name . '</option>';
                    ?>  
                </select>

            </div>
        </div>

        <div class="form-group">
            <label class="col-sm-4 control-label">Title:</label>
            <div class="col-sm-6">   
                <input type="text" class="form-control" name="update_title" id="update_title " value="<?php echo $rows->page_title; ?>">
            </div>
        </div>

    </div><?php } ?>
<div class="form-group">
    <div class="col-md-offset-4 col-sm-6">
        <button class="btn btn-sumit">SUBMIT</button>
    </div>
</div>
</form> 

Controller

public function update() {
$this->load->helper('url');
$data['seo_tag_update'] = $this->Question_Insertion_Model->seo_tag_Update();
echo "$posted_id";
$this->load->view('update', $data);
}

型号

public function seo_tag_Update() {
$this->db->select('*');
$this->db->from('kas_all_pages s');
$this->db->join('kas_all_pages_tags q','s.page_id=q.page_fk');
$this->db->where('s.page_id', 1);
$query = $this->db->get();
return $result = $query->result();
}

Answer 1

问题是您正在循环并创建多个 <select> (和 <input> )元素，而不是创建元素并仅生成 <option>元素。

<form class="form-horizontal" action="#" method="POST">
    <!-- CREATE SELECT ELEMENT -->
    <div class="form-group">
        <label class="col-sm-4 control-label">Select Pages:</label>
        <div class="col-sm-6">
            <select class="form-control" name="select_update_page" id="select_update_page">
                <option value=""> -- select pages -- </option>
       <!-- ADD OPTIONS TO DROPDOWN -->
       <?php 
       foreach ($seo_tag_update as $rows) { 
                    echo '<option value="' . $rows->page_id . '">' . $rows->page_name . '</option>';

       } 
       ?>  
       <!-- CLOSE SELECT ELEMENT -->      
              </select>
            </div>
        </div>
       <!-- CREATE TITLE ELEMENT --> 
        <div class="form-group">
            <label class="col-sm-4 control-label">Title:</label>
            <div class="col-sm-6">   
                <input type="text" class="form-control" name="update_title" id="update_title " value="">
            </div>
        </div>
<div class="form-group">
    <div class="col-md-offset-4 col-sm-6">
        <button class="btn btn-sumit">SUBMIT</button>
    </div>
</div>
</form> 

然后您将需要一些 JavaScript 来更新更改时的值。有多种方法可以在“普通”JavaScript 中执行此操作，或者如果您已经包含 jQuery，则可以查看 .change()函数(docs)。