如何在每个选项标签中显示多个值？

Question

我试图在每个选项标记中显示多个值。我阅读了一些问题，并回答了我的问题，例如this。但这不适合我。选项标签的数据将通过mysql动态创建。

如何解决这一问题？提前致谢！

E.g.

<option value="data 1,data 2, data 3....">Text 1</option> 
<option value="data 6,data 7, data 8....">Text 2</option> 

我目前的形式：

$results_streets = $wpdb->get_results('SELECT DISTINCT streets,stores FROM table WHERE streets IN ("Street 1") ORDER BY streets ASC', OBJECT); 

<form action="" name="formName"> 
<div> 
<select> 
    <option name="default" class="filter_by" value="Select by Street">Select by Street</option> 
    <?php 
    foreach($results_streets as $option){ 
     if(isset($_POST['streets_list']) && $_POST['streets_list'] == $option->streets) 
      echo '<option name="streets_list" class="filter_by" selected value="'. $option->stores .'">'. $option->streets .'</option>'; 
    }; 
    ?> 
</select> 
</div> 
<input type="submit" value="Submit"/> 
</form> 

Answer 1

我认为对于选项标记的属性是不values。尝试代替value，所以这样的：

<select name="test"> 
    <option value="data 1,data 2, data 3....">Text 1</option> 
    <option value="data 6,data 7, data 8....">Text 2</option> 
</select> 

UPDATE：

$results_streets = $wpdb->get_results('SELECT DISTINCT streets,stores FROM table WHERE streets IN ("Street 1") ORDER BY streets ASC', OBJECT); 

<form action="" name="formName"> 
<div> 
<select name='streets_list'> 
    <option class="filter_by" value="Select by Street">Select by Street</option> 
    <?php 
    $values = ""; 
    foreach($results_streets as $option){ 
     if(isset($_POST['streets_list']) && $_POST['streets_list'] == $option->streets){ 
      $values.=$option->stores."|"; 
     } 
    } 
    echo '<option class="filter_by" selected value="'. $values .'">Street 1</option>'; 
    ?> 
</select> 
</div> 
<input type="submit" value="Submit"/> 
</form>