Php 提交表格，文件输入留空

我正在提交一份表格，更新

州id

，

城市名称

和

城市形象

当我只更新图像时，它就工作了。 当我更新

州id

和

城市名称

并希望旧图像保持不变时，数据库中的照片字段变为空白

我的PHP代码如下所示：

 <?php 

if(isset($_POST) && $_POST['submit'] == "Update")
{
        extract($_POST);
        if($_FILES['photo'])
        {
        $cityimg = upload_file($_FILES['photo'],'cityimg/','image','N','true','thumb/', 100, 100);
        $sql = "UPDATE city SET mcid = '$mcid', city_name = '$city_name', photo = '$cityimg' WHERE cid = '$cid'";
        }
        else
        {
        $sql = "UPDATE city SET mcid = '$mcid', city_name = '$city_name' WHERE cid = '$cid'";
        }
        $result = mysql_query($sql);
        if($result)
        {
            $msg = "City Updated Successfully.";
        }
}

?>

---

您应该测试文件是否以不同的方式上载。例如，这将确保在文件上载过程中没有发生错误（如果有）：

如果您只想在未选择任何文件的情况下测试，则可以使用
上载\u ERR\u no\u file


这应该适合您：

<?php 

if(isset($_POST) && $_POST['submit'] == "Update")
{
        extract($_POST);

        // If image was uploaded
        if($_FILES['logo']['error'] == UPLOAD_ERR_OK)
        {
            $cityimg = upload_file($_FILES['photo'],'cityimg/','image','N','true','thumb/', 100, 100);
            $sql = "UPDATE city SET mcid = '$mcid', city_name = '$city_name', photo = '$cityimg' WHERE cid = '$cid'";
        }

        // If no image was uploaded
        else
        {
            $sql = "UPDATE city SET mcid = '$mcid', city_name = '$city_name' WHERE cid = '$cid'";
        }
        $result = mysql_query($sql);
        if($result)
        {
            $msg = "City Updated Successfully.";
        }
}

else{}
执行时，您的
photo
列中的该行内容会被清空吗？
<?php 

if(isset($_POST) && $_POST['submit'] == "Update")
{
        extract($_POST);

        // If image was uploaded
        if($_FILES['logo']['error'] == UPLOAD_ERR_OK)
        {
            $cityimg = upload_file($_FILES['photo'],'cityimg/','image','N','true','thumb/', 100, 100);
            $sql = "UPDATE city SET mcid = '$mcid', city_name = '$city_name', photo = '$cityimg' WHERE cid = '$cid'";
        }

        // If no image was uploaded
        else
        {
            $sql = "UPDATE city SET mcid = '$mcid', city_name = '$city_name' WHERE cid = '$cid'";
        }
        $result = mysql_query($sql);
        if($result)
        {
            $msg = "City Updated Successfully.";
        }
}