Question

需要帮助!!!

我正在使用flume将Twitter Feed导入hdfs并将其加载到hive进行分析。

步骤如下：

hdfs中的数据：

我在avro schema文件中描述了avsc并将其放入hadoop：

 {"type":"record",
 "name":"Doc",
 "doc":"adoc",
 "fields":[{"name":"id","type":"string"},
       {"name":"user_friends_count","type":["int","null"]},
       {"name":"user_location","type":["string","null"]},
       {"name":"user_description","type":["string","null"]},
       {"name":"user_statuses_count","type":["int","null"]},
       {"name":"user_followers_count","type":["int","null"]},
       {"name":"user_name","type":["string","null"]},
       {"name":"user_screen_name","type":["string","null"]},
       {"name":"created_at","type":["string","null"]},
       {"name":"text","type":["string","null"]},
       {"name":"retweet_count","type":["boolean","null"]},
       {"name":"retweeted","type":["boolean","null"]},
       {"name":"in_reply_to_user_id","type":["long","null"]},
       {"name":"source","type":["string","null"]},
       {"name":"in_reply_to_status_id","type":["long","null"]},
       {"name":"media_url_https","type":["string","null"]},
       {"name":"expanded_url","type":["string","null"]}]}

我编写了一个.hql文件来创建一个表并在其中加载数据：

 create table tweetsavro
    row format serde
        'org.apache.hadoop.hive.serde2.avro.AvroSerDe'
    stored as inputformat
        'org.apache.hadoop.hive.ql.io.avro.AvroContainerInputFormat'
    outputformat
        'org.apache.hadoop.hive.ql.io.avro.AvroContainerOutputFormat'
    tblproperties ('avro.schema.url'='hdfs:///avro_schema/AvroSchemaFile.avsc');

    load data inpath '/test/twitter_data/FlumeData.*' overwrite into table tweetsavro;

我已成功运行.hql文件，但是当我在hive中运行select *from <tablename>命令时，它显示以下错误：

error

tweetsavro的输出是：

hive> desc tweetsavro;
OK
id                      string                                      
user_friends_count      int                                         
user_location           string                                      
user_description        string                                      
user_statuses_count     int                                         
user_followers_count    int                                         
user_name               string                                      
user_screen_name        string                                      
created_at              string                                      
text                    string                                      
retweet_count           boolean                                     
retweeted               boolean                                     
in_reply_to_user_id     bigint                                      
source                  string                                      
in_reply_to_status_id   bigint                                      
media_url_https         string                                      
expanded_url            string                                      
Time taken: 0.697 seconds, Fetched: 17 row(s)

Answer 1

我面临同样的问题。问题存在于timestamp字段（在您的情况下为“created_at”列），我试图将其作为字符串插入到我的新表中。我的假设是这个数据在我的源代码中将采用[ "null","string"]格式。我分析了从sqoop import --as-avrodatafile进程生成的源avro架构。从导入生成的avro架构具有以下时间戳列的签名 { "name" : "order_date", "type" : [ "null", "long" ], "default" : null, "columnName" : "order_date", "sqlType" : "93" },

SqlType 93代表Timestamp数据类型。所以在我的目标表Avro Schema文件中，我将数据类型更改为“long”，这解决了这个问题。我的猜测可能是你的一个列中数据类型的不匹配。

异常java.io.IOException失败：org.apache.avro.AvroTypeException：发现很长，期待在hive中联合

1 个答案: