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我正在尝试使用 Parameter 实现 Hive UDF，因此我正在扩展GenericUDF类。

问题是我在字符串数据类型上找到的 UDF 作品，但是如果我在其他数据类型上运行它会引发错误。无论数据类型如何，我都希望 UDF 运行。

有人请让我知道以下代码有什么问题。

@Description(name = "Encrypt", value = "Encrypt the Given Column", extended = "SELECT Encrypt('Hello World!', 'Key');")
public class Encrypt extends GenericUDF {
    StringObjectInspector key;
    StringObjectInspector col;

    @Override
    public ObjectInspector initialize(ObjectInspector[] arguments) throws UDFArgumentException {
        if (arguments.length != 2) {
            throw new UDFArgumentLengthException("Encrypt only takes 2 arguments: T, String");
        }

        ObjectInspector keyObject = arguments[1];
        ObjectInspector colObject = arguments[0];

        if (!(keyObject instanceof StringObjectInspector)) {
            throw new UDFArgumentException("Error: Key Type is Not String");
        }

        this.key = (StringObjectInspector) keyObject;
        this.col = (StringObjectInspector) colObject;

        return PrimitiveObjectInspectorFactory.javaStringObjectInspector;
    }

    @Override
    public Object evaluate(DeferredObject[] deferredObjects) throws HiveException {
        String keyString = key.getPrimitiveJavaObject(deferredObjects[1].get());
        String colString = col.getPrimitiveJavaObject(deferredObjects[0].get());
        return AES.encrypt(colString, keyString);
    }

    @Override
    public String getDisplayString(String[] strings) {
        return null;
    }

}

错误

java.lang.ClassCastException: org.apache.hadoop.hive.serde2.objectinspector.primitive.JavaIntObjectInspector 不能转换为 org.apache.hadoop.hive.serde2.objectinspector.primitive.StringObjectInspector

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1 回答 1

0

我建议您替换StringObjectInspector col为PrimitiveObjectInspector col和相应的演员表this.col = (PrimitiveObjectInspector) colObject。那么有两种方法：

首先是处理所有可能的 Primitive 类型，像这样

    switch (((PrimitiveTypeInfo) colObject.getTypeInfo()).getPrimitiveCategory()) {
        case BYTE:
        case SHORT:
        case INT:
        case LONG:
        case TIMESTAMP:
            cast_long_type;
        case FLOAT:
        case DOUBLE:
            cast_double_type;
        case STRING:
             everyting_is_fine;
        case DECIMAL:
        case BOOLEAN:
            throw new UDFArgumentTypeException(0, "Unsupported yet");
        default:
            throw new UDFArgumentTypeException(0,
                    "Unsupported type");
    }
}

另一种方法，是使用PrimitiveObjectInspectorUtils.getString方法：

Object colObject = col.getPrimitiveJavaObject(deferredObjects[0].get());
String colString = PrimitiveObjectInspectorUtils.getString(colObject, key);

它只是像示例一样的伪代码。希望能帮助到你。

于 2020-04-27T07:19:47.623 回答