更新 Hive 中指定时间范围的列值 table

Question

一个Hivetable"Employee"包含一列"timerange"，数据为

timerange
1:10
1:13
1:17
1:21
1:26

如果最后一位数字范围介于 (0 & 4) 之间，则数据必须更新为 0。如果最后一位数字范围介于 (5 & 9) 之间，则必须更新为 5。

预期输出为

timerange
1:10
1:10
1:15
1:20
1:25

我该怎么做？

Answer 1

您可以创建通用 UDF (GenericUDF)。

这是一个示例 UDF：

import org.apache.hadoop.hive.ql.exec.UDFArgumentException;
import org.apache.hadoop.hive.ql.exec.UDFArgumentLengthException;
import org.apache.hadoop.hive.ql.metadata.HiveException;
import org.apache.hadoop.hive.ql.udf.generic.GenericUDF;
import org.apache.hadoop.hive.serde2.objectinspector.ObjectInspector;
import org.apache.hadoop.hive.serde2.objectinspector.ObjectInspectorUtils;
import org.apache.hadoop.hive.serde2.objectinspector.primitive.IntObjectInspector;
import org.apache.hadoop.hive.serde2.objectinspector.primitive.PrimitiveObjectInspectorFactory;
import org.apache.hadoop.hive.serde2.objectinspector.primitive.StringObjectInspector;

public class TimeRangeConverter GenericUDF {

    @Override
    public ObjectInspector initialize(ObjectInspector[] arguments) throws UDFArgumentException {
        if (arguments.length != 1) {
            throw new UDFArgumentLengthException("The function time_range_converter(time_rage) requires 1 argument.");
        }

        ObjectInspector timeRangeVal = arguments[0];

        if (!(timeRangeVal instanceof StringObjectInspector)) {
            throw new UDFArgumentException("First argument must be of type String (time_range as String)");
        }
        return PrimitiveObjectInspectorFactory.writableStringObjectInspector;
    }

    @Override
    public Object evaluate(DeferredObject[] arguments) throws HiveException {
        String timeRangeVal = (String) ObjectInspectorUtils.copyToStandardJavaObject(arguments[0].get(),
                PrimitiveObjectInspectorFactory.javaStringObjectInspector);

        char[] characters = timeRangeVal.toCharArray();

        if (characters[characters.length - 1] > '5') {
            characters[characters.length - 1] = '5';
        } else {
            characters[characters.length - 1] = '0';
        }

        return String.valueOf(characters);
    }

    @Override
    public String getDisplayString(String[] arguments) {
        assert (arguments.length == 1);
        return "time_range_converter(" + arguments[0] + ")";
    }
}

像这样调用 Hive 更新语句：

CREATE TEMPORARY FUNCTION time_range_converterAS 'TimeRangeConverter';

UPDATE 
    Employee 
SET 
    timerange = time_range_converter(timerange);

Answer 2

您可以通过内置的字符串操作来做到这一点：

SELECT CASE WHEN SUBSTRING(timerange, LENGTH(timerange)) < "5"
            THEN CONCAT(SUBSTRING(timerange, 1, LENGTH(timerange) - 1), "0")
            ELSE CONCAT(SUBSTRING(timerange, 1, LENGTH(timerange) - 1), "5")
       END AS timerange
FROM Employee;

更新 Hive 中指定时间范围的列值 table

Update column value for a specified time range in Hive table

hadoop

hive