我创建了一个测试页面，用于学习如何上传图像。我试图将图像保存到我创建的名为image的文件夹中，然后在我的数据库中存储该图像的文件名，以帮助处理空间。现在图像文件名不存储，而是存储单词Array。我确实得到了错误，以及其他错误。点击上传后，我收到以下错误：
警告：move_uploaded_file（image / picturetest.jpg）：无法打开流：第34行/home4/fdfsfs/public_html/example.com/img_test.php中没有此类文件或目录< / p>
警告：move_uploaded_file（）：无法移动＆＃39; / tmp / phpUg7p4D＆＃39;到＆＃39; image / picturetest.jpg＆＃39;在/home4/fdsfafa/public_html/example.com/img_test.php第34行
There was an error!
**注意：第59行/home4/fdsfaf/public_html/example.com/img_test.php中的数组到字符串转换
第34行是：
if (move_uploaded_file($tmp_name, $destinationFolder.$filename)) {
第59行是：
$stmt->execute();
完整脚本：
$filename = $_FILES['file']['name'];
//$filesize = $_FILES['file']['size'];
//$filetype = $_FILES['file']['type'];
$tmp_name = $_FILES['file']['tmp_name'];
$file_error = $_FILES['file']['error'];
if(isset($_POST['create'])){
$file = $filename;
$file = $_FILES['file'];
//$file = "productpics/". $_FILES['file']['name']; // save the filename
}else {
echo "error!";
}
if (isset($filename )) {
if (!empty($filename)) {
$destinationFolder = 'image/';
if (move_uploaded_file($tmp_name, $destinationFolder.$filename)) {
echo 'Uploaded!';
} else {
echo 'There was an error!';
}
} else {
echo 'Please choose a file.';
}
}
//Connection
$con = mysqli_connect("localhost","","","");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if ($stmt = mysqli_prepare($con, "INSERT INTO image (img) VALUES (?)")) {
/* bind parameters for markers */
$stmt->bind_param('s', $file);
/* execute query */
$stmt->execute();
//if(!$stmt->execute()){trigger_error("there was an error....".$con->error, E_USER_WARNING);}
/* close statement */
mysqli_stmt_close($stmt);
echo "Success!";
} else {
echo "Failed!";
}
$result = mysqli_query($con,"SELECT * FROM image");
if($row = mysqli_fetch_array($result)) {
if($row['img'] == ""){
echo "<img src='images/default_pic.png' alt='No Picture'>";
} else {
echo "<img src='images/".$row['img']."' alt='Profile Picture'>";
}
echo "<br><br><br><br>";
}
?>
<form action="" method="POST" enctype="multipart/form-data">
<input type="file" name="file" class="inputbarfile">
<input type="submit" name="create" id="signinButton" value="Upload">
</form>
有人看到我做错了吗???
答案 0 :(得分：1)
因此，为了“重做”你所拥有的东西，我把它分解为部分（函数）。它基本相同，但是每个部分都被分解，因此1）更容易管理每个部分2）更容易排除故障3）更容易添加错误处理。
<?php
// If you make a file function, you can change where things are saved
// You can also change the destination (for portability)
function UploadFile($fileArray = array(), $destinationFolder = 'image/')
{
$filename = $fileArray['file']['name'];
$tmp_name = $fileArray['file']['tmp_name'];
$filesize = $fileArray['file']['size'];
$file_error = $fileArray['file']['error'];
$file = $fileArray['file'];
// Save all the default data.
// Success and error should be set by default to fail
$return['error'] = true;
$return['success'] = false;
$return['file']['dest'] = $destinationFolder.$filename;
$return['file']['size'] = $filesize;
if($file_error == 0)
$return['error'] = false;
// I added a directory creation function so you don't have to
// manually make folders. This will do it for you.
if(!is_dir($destinationFolder))
mkdir($destinationFolder,0755,true);
// If your filename is not empty, return success or fail of upload
if (!empty($filename))
$return['success'] = (move_uploaded_file($tmp_name, $destinationFolder.$filename));
return $return;
}
// Create a function that quickly returns your connection
function Connection()
{
//Connection
$con = mysqli_connect("localhost","","","");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
return $con;
}
// Create a save-to-database function so it's easier and reusable
function SaveToDb($con,$filename = false)
{
// Return fail immediately if the connection is false
// or the image name is invalid
if(empty($filename) || !$con)
return false;
if ($stmt = mysqli_prepare($con, "INSERT INTO image (img) VALUES (?)")) {
$stmt->bind_param('s', $filename);
$stmt->execute();
mysqli_stmt_close($stmt);
return true;
}
return false;
}
// This just gets the image from the destination column. Not the most
// efficient, but you can change it to fetch by id or whatever is unique
function getPhoto($con,$dest)
{
$result = mysqli_query($con,"SELECT * FROM `image` where `img` = '$dest'");
if($row = mysqli_fetch_array($result))
return $row;
return 0;
}
使用：
// Make sure all functions above are include here
// Get the database connection
$con = Connection();
// Check for post
if(isset($_POST['create'])) {
// Try uploading
$upload = UploadFile($_FILES);
// If upload fails
if(!$upload['success'])
echo '<h3>Sorry, an error occurred</h3>';
else {
// You could add error handling here based on the results of
// each function's success or failure below.
// Try to save it
$saveToDb = SaveToDb($con,$upload['file']['dest']);
// Get the profile from image name
$profPic = ($saveToDb)? getPhoto($con,$upload['file']['dest']) : false; ?>
<img src="<?php echo (!empty($profPic) && $profPic != 0)? $profPic['img'] : "default_pic.png"; ?>" alt="<?php echo (!empty($profPic) && $profPic != 0)? "Profile Picture" : "No Picture"; ?>" />
<?php
}
}
?>
答案 1 :(得分：0)
你需要将数组转换成字符串，检查出来......