我正在查看此示例代码:
// Download file from bucket.
const bucket = gcs.bucket(fileBucket);
const tempFilePath = path.join(os.tmpdir(), fileName);
return bucket.file(filePath).download({
destination: tempFilePath
}).then(() => {
console.log('Image downloaded locally to', tempFilePath);
// Generate a thumbnail using ImageMagick.
return spawn('convert', [tempFilePath, '-thumbnail', '200x200>', tempFilePath]);
}).then(() => {
console.log('Thumbnail created at', tempFilePath);
// We add a 'thumb_' prefix to thumbnails file name. That's where we'll upload the thumbnail.
const thumbFileName = `thumb_${fileName}`;
const thumbFilePath = path.join(path.dirname(filePath), thumbFileName);
// Uploading the thumbnail.
return bucket.upload(tempFilePath, {destination: thumbFilePath});
// Once the thumbnail has been uploaded delete the local file to free up disk space.
}).then(() => fs.unlinkSync(tempFilePath));
我的问题具体在这一行:
return bucket.upload(tempFilePath, {destination: thumbFilePath});
为什么我们在这里提供destination
参数的完整文件路径?据我了解,destination
参数表示上传到存储桶后将采用的文件名。
所以我的猜测就像"thumb_Qsdflsdfa.png"
而不是"/tmp/../thumb_Qsdflsdfa.png"
答案 0 :(得分:1)
根据文档here,第二个参数是可选的。
如果您不介意图像名称与存储桶中的文件名相同,则可以将其留空。例如 -
bucket.upload('/local/path/image.png')
- 您的广告位名称为image.png
但是,如果你想根据你的项目命名其他有意义的东西,你可以传递第二个参数,如下所示 -
bucket.upload('/local/path/image.png', { destination: 'thumb_image.png' })
- 您的广告管理名称现在为thumb_image.png
希望这是有道理的。
这是一个明确的截图 -