Question

我正在提交一份表单，我正在更新state id，city name和city image。

当我只更新图像时，它可以工作。当我更新state id和city name并希望我的旧图片保持不变时，照片字段在数据库中变为空白。

我的PHP代码是这样的：

 <?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.";
        }
}

?>

我认为我的循环存在一些问题。

Answer 1

您应该测试文件是否以不同的方式上传。例如，这将确保在文件上传期间没有错误发生（如果有的话）：

if($_FILES['photo']['error'] == UPLOAD_ERR_OK){
    // A file was uploaded and there is no error
}

如果您只想测试如果没有选择文件，则可以使用UPLOAD_ERR_NO_FILE。

More info on file upload error messages

这应该适合你：

<?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.";
        }
}

查看this post了解详情

提交表格，文件输入留空

1 个答案: