我試圖使用PHP和MySQL上傳的圖像,並且下面是所使用的代碼..爲什麼move_upload_file在給定的代碼中總是返回false?
的index.php
<form action="submit.php" enctype="multipart/form-data" method="post">
<table style="border-collapse: collapse; font: 12px Tahoma;" border="1" cellspacing="5" cellpadding="5">
<tbody><tr>
<td>
<input name="uploadedimage" type="file">
</td>
</tr>
<tr>
<td>
<input name="Upload Now" type="submit" value="Upload Image">
</td>
</tr>
</tbody></table>
</form>
submit.php
<?php
include("mysqlconnect.php");
function GetImageExtension($imagetype) {
if (empty($imagetype))
return false;
switch ($imagetype) {
case 'image/bmp': return '.bmp';
case 'image/gif': return '.gif';
case 'image/jpeg': return '.jpg';
case 'image/png': return '.png';
default: return false;
}
}
if (!empty($_FILES["uploadedimage"]["name"])) {
$file_name = $_FILES["uploadedimage"]["name"];
$temp_name = $_FILES["uploadedimage"]["tmp_name"];
$imgtype = $_FILES["uploadedimage"]["type"];
$ext = GetImageExtension($imgtype);
$imagename = date("d-m-Y") . "-" . time() . $ext;
$target_path = "images/" . $imagename;
if (move_uploaded_file($_FILES['uploadedimage']['tmp_name'], $target_path)) {
$detail = date("Y-m-d");
$sql = "INSERT INTO `image_upload`(`id`, `image`, `detail`) VALUES (NULL,'$target_path','$detail')";
if (!$conn->query($sql)) {
echo $conn->error;
} else {
echo "Successfully inserted. ";
}
} else {
exit("Error While uploading image on the server");
}
}
?>
表結構:
# Name Type Collation Attributes Null Default Extra
1 id(Primary) int(11) No None AUTO_INCREMENT
2 image blob Yes NULL
3 detail varchar(500) utf8_general_ci Yes NULL
每當我執行這個它總是顯示我「錯誤在服務器上傳圖片時「我不明白爲什麼。有人可以讓我知道我哪裏錯了,我該如何改進我的實施? 在此先感謝。
檢查您的目錄中存在與否,並有寫權限給予 –
權限目錄已經它仍然不執行。 –