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Remove useless sentence about default interpreter in docs

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AhyoungRyu 2016-07-20 13:30:26 +09:00
parent 67cb828958
commit 9f44f224b8
2 changed files with 1 additions and 2 deletions
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docs/development/writingzeppelininterpreter.md
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@ -156,7 +156,6 @@ println(a)
### 0.6.0 and later
Inside of a notebook, `%[INTERPRETER_GROUP].[INTERPRETER_NAME]` directive will call your interpreter.
Note that the first interpreter configuration in zeppelin.interpreters will be the default one.
You can omit either [INTERPRETER\_GROUP] or [INTERPRETER\_NAME]. If you omit [INTERPRETER\_NAME], then first available interpreter will be selected in the [INTERPRETER\_GROUP].
Likewise, if you skip [INTERPRETER\_GROUP], then [INTERPRETER\_NAME] will be chosen from default interpreter group.

2
docs/install/install.md
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@ -377,7 +377,7 @@ You can configure Apache Zeppelin with both **environment variables** in `conf/z
<td>org.apache.zeppelin.spark.SparkInterpreter,<br />org.apache.zeppelin.spark.PySparkInterpreter,<br />org.apache.zeppelin.spark.SparkSqlInterpreter,<br />org.apache.zeppelin.spark.DepInterpreter,<br />org.apache.zeppelin.markdown.Markdown,<br />org.apache.zeppelin.shell.ShellInterpreter,<br />
...
</td>
<td>Comma separated interpreter configurations [Class] <br /> The first interpreter will be a default value. <br /> It means only the first interpreter in this list can be available without <code>%interpreter_name</code> annotation in notebook paragraph. </td>
<td>Comma separated interpreter configurations [Class]</td>
</tr>
<tr>
<td>ZEPPELIN_INTERPRETER_DIR</td>