Hi Team,
Does anyone know in the experimental scenario,
JAVA runtime 8 or 11
1. Kafka's topic -> Testing_topic
2. Kafka's topic partition count -> 1
3. Kafka's topic partition offset -> 2 [two messages are available, each weight about 1 mb ]
4. Bandwidth of 1 kb and Kafka's topic offset's message of 1 mb is available each
5. In consumer scenario
* While polling the consumer using consumer.poll(10000L) deprecated method supporting the long argument
* With properties (exactly used properties)
i. props.put(ConsumerConfig.BOOTSTRAP_SERVERS_CONFIG, bootstrap.server.config.url);
ii. props.put(ConsumerConfig.GROUP_ID_CONFIG, group-1);
iii. props.put(ConsumerConfig.KEY_DESERIALIZER_CLASS_CONFIG, "org.apache.kafka.common.serialization.StringDeserializer");
iv. props.put(ConsumerConfig.VALUE_DESERIALIZER_CLASS_CONFIG, "org.apache.kafka.common.serialization.StringDeserializer");
v. props.put(ConsumerConfig.MAX_POLL_RECORDS_CONFIG, 1000);
vi. props.put(ConsumerConfig.ENABLE_AUTO_COMMIT_CONFIG, false);
vii. props.put("request.timeout.ms", "60000");
Question
1. in poll(10000L ) given , whether this is in seconds or milliseconds or minutes (Deprecated method) ?
2. so in bandwidth of 1 kb for experimental purpose -> while polling assuming the 10000L is milliseconds so if I kafka consumer doesn't get the offset message which of 1 mb
* what will happens in brief details ?
* whether any errors/exceptions comes up ?
* consumerRecords<String, String> records = consumer.poll(10000L);
* whether this records count will be 0 in this case ?
* if poll gets 500kb from topic in first poll within 10s after processing assuming no errors comes up and records count is 0 then in next poll whether next 500kb is fetched or it will attempt to again fetch full message ?
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Does anyone know in the experimental scenario,
JAVA runtime 8 or 11
1. Kafka's topic -> Testing_topic
2. Kafka's topic partition count -> 1
3. Kafka's topic partition offset -> 2 [two messages are available, each weight about 1 mb ]
4. Bandwidth of 1 kb and Kafka's topic offset's message of 1 mb is available each
5. In consumer scenario
* While polling the consumer using consumer.poll(10000L) deprecated method supporting the long argument
* With properties (exactly used properties)
i. props.put(ConsumerConfig.BOOTSTRAP_SERVERS_CONFIG, bootstrap.server.config.url);
ii. props.put(ConsumerConfig.GROUP_ID_CONFIG, group-1);
iii. props.put(ConsumerConfig.KEY_DESERIALIZER_CLASS_CONFIG, "org.apache.kafka.common.serialization.StringDeserializer");
iv. props.put(ConsumerConfig.VALUE_DESERIALIZER_CLASS_CONFIG, "org.apache.kafka.common.serialization.StringDeserializer");
v. props.put(ConsumerConfig.MAX_POLL_RECORDS_CONFIG, 1000);
vi. props.put(ConsumerConfig.ENABLE_AUTO_COMMIT_CONFIG, false);
vii. props.put("request.timeout.ms", "60000");
Question
1. in poll(10000L ) given , whether this is in seconds or milliseconds or minutes (Deprecated method) ?
2. so in bandwidth of 1 kb for experimental purpose -> while polling assuming the 10000L is milliseconds so if I kafka consumer doesn't get the offset message which of 1 mb
* what will happens in brief details ?
* whether any errors/exceptions comes up ?
* consumerRecords<String, String> records = consumer.poll(10000L);
* whether this records count will be 0 in this case ?
* if poll gets 500kb from topic in first poll within 10s after processing assuming no errors comes up and records count is 0 then in next poll whether next 500kb is fetched or it will attempt to again fetch full message ?
The information in this e-mail and any attachments is confidential and may be legally privileged. It is intended solely for the addressee or addressees. Any use or disclosure of the contents of this e-mail/attachments by a not intended recipient is unauthorized and may be unlawful. If you have received this e-mail in error please notify the sender. Please note that any views or opinions presented in this e-mail are solely those of the author and do not necessarily represent those of TEMENOS. We recommend that you check this e-mail and any attachments against viruses. TEMENOS accepts no liability for any damage caused by any malicious code or virus transmitted by this e-mail.
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