我有一个blob触发的Azure函数,我想实现一个重试机制来处理函数在执行过程中由于特定错误而失败的场景。
要求如下:
如何在 Azure Functions 中设置这样的重试机制?
任何指导或示例将不胜感激!
各个触发器扩展的内置重试行为
按照以下步骤在 Java Azure Function 中设置重试机制:
代码片段:
public class BlobTriggerJava {
@FunctionName("BlobTriggerJava")
@StorageAccount("storageConnection")
@FixedDelayRetry(maxRetryCount = 4, delayInterval = "00:00:10")
public void run(
@BlobTrigger(name = "content", path = "testkp/{name}", dataType = "binary") byte[] content,
@BindingName("name") String name,
final ExecutionContext context
) {
context.getLogger().info("Processing blob: " + name);
try {
if (name.contains("error")) {
throw new Exception("Simulated error");
}
} catch (Exception e) {
context.getLogger().severe("Error processing blob: " + e.getMessage());
throw new RuntimeException(e);
} }
}
host.json
文件中定义重试策略,用于控制 Azure Function 的隐式重试。例如,您可以使用固定延迟或指数退避策略进行重试,请参阅 @JesseSquire 的 MSDOC 和 SO 答案。主机.json:
{
"version": "2.0",
"extensions": {
"blobs": {
"poisonBlobThreshold": 1,
"maxDegreeOfParallelism": 4
}
},
"retry": {
"strategy": "fixedDelay",
"maxRetryCount": 3,
"delayInterval": "00:00:10"
}
}
控制台输出:
Functions:
BlobTriggerJava: blobTrigger
For detailed output, run func with --verbose flag.
[2024-12-23T07:35:43.571Z] Host lock lease acquired by instance ID '000000000000000000000000F72731CC'.
[2024-12-23T07:35:46.945Z] Executing 'Functions.BlobTriggerJava' (Reason='New blob detected(LogsAndContainerScan): test/Hello World.txt', Id=b2ee3ba8-fc36-46ae-b8f5-31fc8e95201a)
[2024-12-23T07:35:46.947Z] Trigger Details: MessageId: 440a3afb-45a5-44a3-8b01-1ebe946874e6, DequeueCount: 1, InsertedOn: 2024-12-23T07:35:42.000+00:00, BlobCreated: 2024-12-23T06:47:20.000+00:00, BlobLastModified: 2024-12-23T06:56:54.000+00:00
[2024-12-23T07:35:47.236Z] Processing blob: Hello World.txt
[2024-12-23T07:35:47.236Z] Function "BlobTriggerJava" (Id: b2ee3ba8-fc36-46ae-b8f5-31fc8e95201a) invoked by Java Worker
[2024-12-23T07:35:47.262Z] Executed 'Functions.BlobTriggerJava' (Succeeded, Id=b2ee3ba8-fc36-46ae-b8f5-31fc8e95201a, Duration=1341ms)