1 回答
TA贡献1998条经验 获得超6个赞
我已经想通了,感谢所有评论,我不得不将其附加fileName到我的FormData()并编辑我的 PHP 脚本,如下所示:
JS:
function uploadFile() {
var dbPath = '<?php echo $DATABASE_PATH ?>';
var file = $("#file")[0].files[0];
var fileName = file.name;
console.log('FILE NAME: ' + fileName);
Swal.fire({icon: 'success', title: 'Loading...', showConfirmButton: false })
var data = new FormData();
data.append('file', file);
data.append('fileName', fileName); <!-- ADDED THIS LINE
$.ajax({
url : dbPath + "upload-file.php?fileName=" + fileName, <!-- ADDED THIS LINE
type: 'POST',
data: data,
contentType: false,
processData: false,
mimeType: "multipart/form-data", <!-- ADDED THIS LINE
success: function(data) {
Swal.close();
var fileURL = dbPath + data;
console.log('FILE UPLOADED TO: ' + fileURL);
// error
if (data.includes("ERROR:")) {
Swal.fire({ icon: 'error', title: 'Oops...', text: data, });
// show file data
} else {
$("#fileURL").val(fileURL);
$("#viewButton").attr("href", fileURL);
$("#viewButton").css("display", "block");
}
// error
}, error: function(e) {
Swal.fire({ icon: 'error', title: 'Oops...', text: 'Something went wrong: ' + e.message, });
}});
}
我的upload-file.php脚本:
<?php include 'config.php';
if ($_FILES["file"]["error"] > 0) {
echo "Error: " .$_FILES["file"]["error"]. "<br>";
} else {
// Check file size
if ($_FILES["file"]["size"] > 20485760) { // 20 MB
echo "ERROR: Your file is larger than 20 MB. Please upload a smaller one.";
} else { uploadImage(); }
}// ./ If
// UPLOAD IMAGE ------------------------------------------
function uploadImage() {
// generate a unique random string
$randomStr = generateRandomString();
$filePath = "uploads/".$randomStr;
// upload image into the 'uploads' folder
move_uploaded_file($_FILES['file']['tmp_name'], $filePath);
// echo the link of the uploaded image
echo $filePath;
}
// GENERATE A RANDOM STRING ---------------------------------------
function generateRandomString() {
$characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
$charactersLength = strlen($characters);
$randomString = '';
for ($i = 0; $i<20; $i++) {
$randomString .= $characters[rand(0, $charactersLength - 1)];
}
return $randomString."_".$_POST['fileName']; <!-- ADDED THIS LINE
}
?>
现在我终于可以上传任何类型的文件并在 URL 末尾获取其名称 + 扩展名,它可以完美运行。
- 1 回答
- 0 关注
- 164 浏览
添加回答
举报