Question

这是我前面问过的一个问题的延伸，其中@eHussain是nice enough to help out with。

我有一个表单，它将各种细节插入MySQL表并上传一个文件（其名称也在数据库中注册）。这很好用。当我更新名称而不是图像时会出现问题。在这种情况下，图像名称被覆盖为“空白”，正确地将其写为文件字段中的值。

更新代码：

<?php
error_reporting(E_ALL^E_NOTICE);
define('INCLUDE_CHECK',true);
include "connect.php";

$target = "../uploads/";
$target = $target . basename( $_FILES['photo']['name']);

//This gets all the other information from the form
$name=$_POST['name'];
$url=$_POST['url'];
$description=$_POST['description'];
$pic=($_FILES['photo']['name']);
$author=$_POST['author'];
$company=$_POST['company'];
$published=$_POST['published'];
$dashboardID=$_POST['dashboardID'];

//Writes the information to the database
mysql_query("UPDATE dashboard SET name='$name', url='$url', description='$description', documentName='$pic', author='$author', company='$company', publish='$published' WHERE dashboardID='$dashboardID'");

//Writes the photo to the server

if(isset($_FILES['photo']['tmp_name'])) //  check if any file is uploaded
{
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{

header("Location: ../dashboard.php?success=2"); } else {

header("Location: ../dashboard.php?success=0"); }
}

?>

我理解'isset'是为了避免在没有选择文件的情况下产生错误，但我不明白如何扩展它以避免更新具有空值的字段。

Answer 1

在运行查询之前检查$_FILES数组。

从那里，您可以动态构建查询（包括或排除documentName字段），也可以获取当前值并将其分配给$pic。

例如（未经测试）

$values = array(
    'name' => $name,
    'url'  => $url,
    // etc
);

if (isset($_FILES['photo']['name'])) {
    $values['documentName'] = $_FILES['photo']['name']
}

// mysql functions are naff, use PDO

$query = 'UPDATE dashboard SET %s WHERE dashboardID = :dashboardID';
$set = array();
foreach (array_keys($values) as $col) {
    $set[] = sprintf('`%s` = :%s', $col, $col);
}
$stmt = $pdo->prepare(sprintf($query, implode(', ', $set)));
$values['dashboardID'] = $dashboardID;

$stmt->execute($values);

Answer 2

@rrfive，请尝试以下方法，希望它能运作，

//first put all post variable in an array 
$post_data = compact($_POST);    
$pic=($_FILES['photo']['name']);
//now push pic name in `$post_data`
if (!empty($pic) ) { array_push( $post_data,$pic ) }

//now use UPDATE query using `vsprintf` . but first check the order of `$post_data` @Thanks Phill

$stmt = "UPDATE dashboard SET 
                            name='%s', 
                            url='%s',
                            description='%s',                           
                            author='%s',
                            company='%s',
                            publish='%s'";
$stmt .=(!empty($pic)) ? documentName='%s', : "";
$stmt .= "WHERE dashboardID=%d";
// To check the complete query before execute. uncomment below 2 lines
//print vsprintf($stmt,$post_data);
//die;
mysql_query( vsprintf($stmt,$post_data) );

<强>参考
- compact
- vsprintf

Answer 3

以下代码可以解决问题。

 if(isset($_FILES)){
      ...stuff...
 }

PHP更新不更新空白的表单字段

3 个答案: