我试图使用PHP和MySQL上传的图像,并且下面是所使用的代码..为什么move_upload_file在给定的代码中总是返回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");
}
}
?>
表结构:
# 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
每当我执行这个它总是显示我“错误在服务器上传图片时“我不明白为什么。有人可以让我知道我哪里错了,我该如何改进我的实施? 在此先感谢。
检查您的目录中存在与否,并有写权限给予 –
权限目录已经它仍然不执行。 –