PHP和MySQLi - 在下拉列表中显示两次选定的值

Question

我创建了我的个人资料页面,显示了MySQL中的所有表格数据.所有数据都在表单和下拉列表中正确显示.但问题是选项值在选项列表中显示两次.

这是我的代码:

<select class="form-control" name="country" id="country">
     <option value="">Select Country
         <?php
            //Get country list from Country master
            $qry = "select * from country_master";
            //Execute query
            $result = mysqli_query($conn, $qry);
            //Assigned fetched array to $Country
            while($country = mysqli_fetch_array($result))
            {
              echo "<option value='$country[1]'>$country[1]</option>";
              //Compare User Country with country list. $row[4] is the country column in user table
              if($row[4] == $country[1])
                 echo "<option value='$country[1]' selected='selected'>$country[1]</option>";
            }
         ?>
      </option>
</select>

Answer 1

您需要更改while以下代码: -

while($country = mysqli_fetch_array($result)){
    //Compare User Country with country list. $row[4] is the country column in user table
    if($row[4] == $country[1]){
        echo "<option value='$country[1]' selected='selected'>$country[1]</option>";
    }else{
        echo "<option value='$country[1]'>$country[1]</option>";
    }
}

注:在创建代码中第一个选项,然后检查条件,这就是为什么两次,它会显示所选择的选项.