Php 在codeigneter上选择时出现未定义变量错误有更好的方法吗？

如果$personal_info['reg_civil_status']和$u POST['reg_civil_status']为空，有没有更好的方法在不出错的情况下执行此操作？因为我有一个“未定义变量错误”

很抱歉，我还是一名初级程序员，我是PHP新手。无论如何，谢谢你。

                    <b>Civil Status: </b>
                    <select class="controler-1" name="reg_civil_status">
                        <?php 
                            foreach($civil_status_opt as $civil_status_opt){
                                if($personal_info['reg_civil_status'] == $civil_status_opt || $_POST['reg_civil_status'] == $civil_status_opt) {
                                    echo "<option selected='selected' value='$civil_status_opt'>$civil_status_opt</option>";
                                }
                                else {
                                    echo "<option value='$civil_status_opt'>$civil_status_opt</option>";
                                }
                            }
                        ?>
                    </select>

公民身份：
您需要对三元运算符使用
isset
，以避免错误

<b>Civil Status: </b>
<select class="controler-1" name="reg_civil_status">
    <?php 
        $reg_civil_status = isset($personal_info['reg_civil_status']) ? $personal_info['reg_civil_status'] : "";
        $posted_reg_civil_status = isset($_POST['reg_civil_status']) ? $_POST['reg_civil_status'] : "";

        foreach($civil_status_opt as $civil_status_opt){
            if($reg_civil_status == $civil_status_opt || $posted_reg_civil_status == $civil_status_opt) {
                echo "<option selected='selected' value='$civil_status_opt'>$civil_status_opt</option>";
            }
            else {
                echo "<option value='$civil_status_opt'>$civil_status_opt</option>";
            }
        }
    ?>
</select>

公民身份：
设置
或
！空的
在询问类似的问题之前，请做一些研究。我做了，但我似乎找不到类似的问题。谢谢你顺便说一句：）你的字典里有语法错误（你用
=
而不是
：
）。谢谢你，这真的帮助了我。我是php新手（对不起：）