Question

我有类别的数据库，其中我有id和名称行

<select name="category_id">
    <option value="<?php if($category_id === 1) { echo 'selected';} ?>">Electronics</option>
    <option value="<?php if($category_id === 2) { echo 'selected';} ?>">Automotive</option>
    </select>

出于某种原因，这并未显示在尝试编辑提交的帖子时选择了哪一个

我确实像这样检索category_id：

$res6 = mysql_query("SELECT * FROM `posts` WHERE `id` = '" . $id . "' LIMIT 1"); 

    if(mysql_num_rows($res6) > 0){
        while($row6 = mysql_fetch_assoc($res6)){
            $id = $row6['id'];
            $category_id = $row6['category_id'];
            $price = $row6['price'];
            $cover = $row6['cover'];
            $title = $row6['title'];
            $description = $row6['description'];
            }
    }

Answer 1

我认为所选的选项参数是

<option selected="selected"> </option>

Answer 2

我相信MySQL结果总是以字符串形式返回，因此$category_id === 1将为false。

尝试$category_id === "1"或$category_id == 1

此外，您需要回显selected="selected属性

之外的value

Answer 3

您也可以尝试这样做。

<select name="category_id">
<?php 
    $cats = array("1"=>"Electronics", "2"=>"Automotive");
    foreach($cats as $k=>$cat){
         if( $k == $category_id )
             echo "<option value='{$k}' selected='selected'>{$cat}</options>";
         else
             echo "<option value='{$k}' >{$cat}</options>";
    }
?>
</select>

如何从帖子编辑的下拉列表中获取所选值

3 个答案: