【发布时间】:2020-07-26 19:29:02
【问题描述】:
我有一个文件名、文件路径和文件大小的列表,我想使用 spark SQL 将这些详细信息插入到我的配置单元表中。
var fs1 = FileSystem.get(sparksession.sparkContext.hadoopConfiguration)
var file_path = fs1.listStatus(new Path("path")).filter(_.isFile).map(_.getPath).toList
var new_files = fs1.listStatus(new Path("path")).filter(_.isFile).map(_.getPath.getName).toList
var file_size = fs1.listStatus(new Path("path")).filter(_.isFile).map(_.getLen).toList
var file_data = file_path zip new_files zip file_size
for ((filedetail, size) <- file_size){
var filepath = filedetail._1
var filesize: Long = size
var filename = filedetail._2
var df = spark.sql(s"insert into mytable(file_path,filename,file_size) select '${file_path}' as file_path,'${new_files}' as filename,'${file_size}' as file_size")
df.write.insertInto("dbname.tablename")
}
它正在生成这个插入查询:
insert into mytable(file_path,filename,file_size) select 'List(path/filename.txt,path/filename4.txt,path/filename5.txt)' as file_path,'List(filename.txt, filename4.txt, filename5.txt)' as filename,'List(19, 19, 19)' as file_size;
我收到一个错误:
不匹配的输入 'file_path' 期望 {'(', 'SELECT', 'FROM', 'VALUES', 'TABLE', 'INSERT', 'MAP', 'REDUCE'}(第 1 行,第 34 行)
我想插入表格格式的数据
file_path filename file_size
path/filename.txt filename.txt 19
path/filename4.txt filename4.txt 19
path/filename5.txt filename5.txt 19
有人可以建议我如何插入上面的数据吗?
有没有办法再次将此查询拆分为 3 个不同的插入配置单元语句。
insert into mytable(file_path,filename,file_size) select 'path/filename.txt' as file_path,'filename.txt' as filename,'19' as file_size;
insert into mytable(file_path,filename,file_size) select 'path/filename3.txt' as file_path,'filename3.txt' as filename,'19' as file_size;
insert into mytable(file_path,filename,file_size) select 'path/filename4.txt' as file_path,'filename4.txt' as filename,'19' as file_size;
【问题讨论】:
标签: scala apache-spark hive apache-spark-sql