有百分比问题的 spark sql 查询

Question

我无法获得带有 % 符号的 spark sql 查询。假设我有以下数据框

val df = sc.parallelize(Seq(("Peter",123,23.5),("John",45,45.5)))
           .toDF("Name","Age","score(%)")
df.show

所以 table 看起来像这样：

+-----+---+--------+
| Name|Age|score(%)|
+-----+---+--------+
|Peter|123|    23.5|
| John| 45|    45.5|
+-----+---+--------+

我能做到：

sqlContext.sql("SELECT Name FROM df")

显示：

+-----+
| Name|
+-----+
|Peter|
| John|
+-----+

但是当我这样做时：

sqlContext.sql("SELECT score(%) FROM df")

它抛出以下内容：（看起来是 % 引起的问题，我尝试使用 \%，但没有帮助）

java.lang.RuntimeException: [1.14] failure: ``distinct'' expected but `%' found

SELECT score(%) FROM df
             ^
  at scala.sys.package$.error(package.scala:27)
  at org.apache.spark.sql.catalyst.AbstractSparkSQLParser.parse(AbstractSparkSQLParser.scala:36)
  at org.apache.spark.sql.catalyst.DefaultParserDialect.parse(ParserDialect.scala:67)
  at org.apache.spark.sql.SQLContext$$anonfun.apply(SQLContext.scala:175)
  at org.apache.spark.sql.SQLContext$$anonfun.apply(SQLContext.scala:175)
  at org.apache.spark.sql.SparkSQLParser$$anonfun$org$apache$spark$sql$SparkSQLParser$$others.apply(SparkSQLParser.scala:115)
  at org.apache.spark.sql.SparkSQLParser$$anonfun$org$apache$spark$sql$SparkSQLParser$$others.apply(SparkSQLParser.scala:114)
  at scala.util.parsing.combinator.Parsers$Success.map(Parsers.scala:137)
  at scala.util.parsing.combinator.Parsers$Success.map(Parsers.scala:136)
  at scala.util.parsing.combinator.Parsers$Parser$$anonfun$map.apply(Parsers.scala:237)
  at scala.util.parsing.combinator.Parsers$Parser$$anonfun$map.apply(Parsers.scala:237)
  at scala.util.parsing.combinator.Parsers$$anon.apply(Parsers.scala:217)
  at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append$$anonfun$apply.apply(Parsers.scala:249)
  at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append$$anonfun$apply.apply(Parsers.scala:249)
  at scala.util.parsing.combinator.Parsers$Failure.append(Parsers.scala:197)
  at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append.apply(Parsers.scala:249)
  at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append.apply(Parsers.scala:249)
  at scala.util.parsing.combinator.Parsers$$anon.apply(Parsers.scala:217)
  at scala.util.parsing.combinator.Parsers$$anon$$anonfun$apply.apply(Parsers.scala:882)
  at scala.util.parsing.combinator.Parsers$$anon$$anonfun$apply.apply(Parsers.scala:882)
  at scala.util.DynamicVariable.withValue(DynamicVariable.scala:58)
  at scala.util.parsing.combinator.Parsers$$anon.apply(Parsers.scala:881)
  at scala.util.parsing.combinator.PackratParsers$$anon.apply(PackratParsers.scala:110)
  at org.apache.spark.sql.catalyst.AbstractSparkSQLParser.parse(AbstractSparkSQLParser.scala:34)
  at org.apache.spark.sql.SQLContext$$anonfun.apply(SQLContext.scala:172)
  at org.apache.spark.sql.SQLContext$$anonfun.apply(SQLContext.scala:172)
  at org.apache.spark.sql.execution.datasources.DDLParser.parse(DDLParser.scala:42)
  at org.apache.spark.sql.SQLContext.parseSql(SQLContext.scala:195)
  at org.apache.spark.sql.SQLContext.sql(SQLContext.scala:725)
  ... 48 elided

（当我对 spark 进行编程以使用 spark-csv 摄取大量 csv 时遇到此问题。当我尝试执行 sql SELECT 时，我运行进入这个 % 问题。我想尽可能避免修改 header...)

Answer 1

尝试使用反引号分隔列名。

sqlContext.sql("SELECT `score(%)` FROM df")

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