当许多用户使用相同的功能时，Firebase 云函数截止时间错误

问：

每当用户必须在存储中发布图像时，我不会直接使用 firebase 存储函数来执行此操作，而是使用 onCall 云函数，通过向其传递 base64 图像、修改它（5 次）并将其发布到存储中。

功能如下：

exports.uploadImage = functions.https.onCall(async (data, context) => {
    

   var bucket = admin.storage().bucket(); 

   // Convert the base64 string back to an image 
   var base64EncodedImageString = data.image,
   mimeType = 'image/jpeg',
   fileName = 'testName',
   imageBuffer = Buffer.from(base64EncodedImageString, 'base64');

   if (!fs.existsSync(os.tmpdir()+"/myfolder")){
      fs.mkdirSync(os.tmpdir()+"/myfolder");
  }

   const tempFilePath = path.join(os.tmpdir(),"myfolder", fileName+".jpg");


   fs.writeFileSync(tempFilePath,base64EncodedImageString,'base64',function(err){
      functions.logger.log("file scritto in"+tempFilePath);
   })
  
   await bucket.upload(tempFilePath, {
      destination: 'test/'+fileName,
      metadata: { contentType: mimeType,
         metadata: {
            firebaseStorageDownloadTokens: uuid()
         } 
       },
    });


   const tempFilePath_25 = path.join(os.tmpdir(),"myfolder", fileName+"_25.jpg");
   spawnSync('convert', [tempFilePath, '-scale', '10%','-scale','1000%>', tempFilePath_25]);
   await bucket.upload(tempFilePath_25, {
      destination: 'test/'+fileName+"_25.jpg",
      metadata: { contentType: mimeType,
         metadata: {
            firebaseStorageDownloadTokens: uuid()
         } 
       },
    });
    fs.unlinkSync(tempFilePath_25);
      
   const tempFilePath_50 = path.join(os.tmpdir(),"myfolder",  fileName+"_50.jpg");
   spawnSync('convert', [tempFilePath, '-scale', '5%','-scale','2000%>', tempFilePath_50]);
   await bucket.upload(tempFilePath_50, {
      destination: 'test/'+fileName+"_50.jpg",
      metadata: { contentType: mimeType,
         metadata: {
            firebaseStorageDownloadTokens: uuid()
         } 
       },
    });
    fs.unlinkSync(tempFilePath_50);

   const tempFilePath_75 = path.join(os.tmpdir(),"myfolder",  fileName+"_75.jpg");
   spawnSync('convert', [tempFilePath, '-scale', '3%','-scale','3333%>', tempFilePath_75]);
   await bucket.upload(tempFilePath_75, {
      destination: 'test/'+fileName+"_75.jpg",
      metadata: { contentType: mimeType,
         metadata: {
            firebaseStorageDownloadTokens: uuid()
         } 
       },
    });
    fs.unlinkSync(tempFilePath_75);

   const tempFilePath_100 = path.join(os.tmpdir(),"myfolder",  fileName+"_100.jpg");
   spawnSync('convert', [tempFilePath, '-scale', '1%','-scale','10000%>', tempFilePath_100]);
   await bucket.upload(tempFilePath_100, {
      destination: 'test/'+fileName+"_100.jpg",
      metadata: { contentType: mimeType,
         metadata: {
            firebaseStorageDownloadTokens: uuid()
         } 
       },
    });
    fs.unlinkSync(tempFilePath_100);


});

我尝试每 2 秒使用 for 循环进行模拟，结果我得到了 60% 请求的截止时间错误。当我发布应用程序时，会有很多用户（希望）可以同时调用相同的函数来发布照片。我该如何解决这个问题？提前致谢。