我一直在研究php文件,并希望显示文件是否已上传
我尝试过:
if (move_uploaded_file($file_tmp, $file)) {
echo "<script type='text/javascript'>alert('File sucessfully uploaded');</script>";
} else {
echo "<script type='text/javascript'>alert('Upload failed');</script>";
}
但是它不会产生弹出窗口。但是我可以在开发人员选项的响应下看到它。知道我该如何解决吗?
答案 0 :(得分:0)
您必须在alert();
内打印valid html
。参见以下示例:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Upload</title>
</head>
<body>
<form name="form" method="post" enctype="multipart/form-data" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>">
Upload File: <input type="file" size="30" id="userfile" name="userfile">
<?php
$upload_dir = $_SERVER['DOCUMENT_ROOT'] . "/upload/";
if (!is_dir($upload_dir)) {
@mkdir($upload_dir, 0755, true);
}
if (isset($_FILES['userfile'])) {
$temp_name = $_FILES['userfile']['tmp_name'];
$file_name = $_FILES['userfile']['name'];
$file_path = $upload_dir.$file_name;
}
if ((isset($_FILES['userfile'])) && (is_uploaded_file($_FILES['userfile']['tmp_name']))) {
if (@move_uploaded_file($temp_name, $file_path)) {
@chmod($file_path,0755);
echo "<script type='text/javascript'>alert('File sucessfully uploaded');</script>";
} else {
echo "<script type='text/javascript'>alert('Upload failed');</script>";
}
}
?>
<input type="submit" name="submit" value="Upload">
</form>
</body>
</html>
这个例子对我来说很好。我希望这会有所帮助。