JS async await fetch 捕获后端500错误详细信息

客户端异步请求抛出的500友好错误信息，处理起来很困难，需要把后端代码逐个检查下。

若是抛出500错误详细信息，就可以快速把问题处理掉。

如果是响应的是文本，就用 response.text() 方法，JSON 就用 response.json() 方法。

async function bigFileUpload_click(sef) {
	const chunk_size = 1024*1024; // 1MB per chunk
	const file = sef.files[0];
	const fileSize = file.size;
	const totalChunks = Math.ceil(fileSize/chunk_size); // 向上取整
	const progressBar = $$('#progressBar');
	
	console.log(  file.name );return;
	
	for (let chunkIndex = 0; chunkIndex < totalChunks; chunkIndex++) {
		const start = chunkIndex * chunk_size;
		const end = Math.min(fileSize, start+chunk_size);
		const chunk = file.slice(start, end);
		const formData = new FormData();
		formData.append('act', 'bigFileUpload_click');
		formData.append('file', chunk);
		formData.append('fileName', file.name);
		formData.append('chunkIndex', chunkIndex);
		formData.append('totalChunks', totalChunks);
		
		progressBar.innerHTML = Math.ceil(chunkIndex/fileSize*100);
 
		await fetch(location.pathname,{
			method: 'POST',
			body: formData
		})
			.then(response => {
				if (response.status !== 200) {
					console.log('response:', response, response.text());
				}
			})
			.catch(error => {
				console.error('Error:', error)
			});
		;
	}
	
	progressBar.innerHTML = '100%';
}