IBM C2090-421

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IBM InfoSphere DataStage v8.5 Sample Questions:

1. Which three statements are true about characteristics of shared containers? (Select three)

A) Containers allow you to implement recurring logic.
B) Containers allow you to provide pre-configured job design logic.
C) Containers are used to execute jobs.
D) Containers allow multiple users to access a single job.
E) Containers are a group of stages and links.

2. You are processing groups of rows in a Transformer. The first row in each group contains "1" in the Flag column and "0" in the remaining rows of the group. At the end of each group you want to sum and output the QTY column values. Which three techniques will enable you to retrieve the sum of the last group? (Choose three.)

A) Output the sum that you generated up to the previous row each time you process a row with a "1" in the Flag column. Use the LastRow() function to determine when the last group is done.
B) Output a running total for each group for each row. Follow the Transformer stage by an Aggregator stage. Take the MAX of the QTY column for each group.
C) Output the sum that you generated up to the previous row each time you process a row with a "1" in the Flag column.
D) Within each group sort the Flag column in ascending order. Output the sum each time you process the row with a "1" in the Flag column.
E) Output the sum that you generated each time you process a row for which the LastRow() function returns True.

3. Which three data types can the ODBC Enterprise stage handle? (Choose three.)

A) SQL_VARBINARY
B) SQL_CLOB
C) SQL_BIGINT
D) SQL_TIMESTAMP
E) SQL_SMALLDATETIME

4. Which job design technique can be used to give unique names to sequential output files that are used in multi-instance jobs?

A) Use DSJobInvocationID to generate a unique filename.
B) Use parameters to identify file names.
C) Generate unique file names by using a macro.
D) Use a Transformer stage variable to generate the name.

5. Your job uses the MQ connector stage to read messages from an MQ queue. The job should retrieve the message ID into the MessageID field and parse the payload into two fields: Name is to get the first ten characters, Description is to get the remaining characters.
What will accomplish this?

A) First column is MessageID; select the WSMQ.MSGID data element for the Message ID field; second column is Name as Binary 10; third column is Description as VarBinary 200.
B) First column is MessageID; select the WSMQ.MSGID data element for the Message ID field; column is Description as VarBinary 200; third column is Name as Binary 10.
C) First column is MessageID as Binary 24; second column is Name as Binary 10; select WSMG. MSPAYLOAD data element; third column is Description as VarBinary 200; select WSMG.MSPAYLOAD data element.
D) First column is MessageID; select the WSMQ.MSGID data element for the Message ID field; second column is Name; select WSMG.MSPAYLOAD data element; third column is Description; select WSMG.MSPAYLOAD data element.

Solutions: