apache-spark - Apache Spark 独立用于匿名 UID(无用户名)

Question

我正在 OpenShift 平台上启动 Apache spark 从节点。 OpenShift 在内部以匿名用户身份启动 docker 镜像(用户没有名称，只有 UID)。我收到以下异常 <i></i>

<i>

<pre><code>17/07/17 16:46:53 INFO SignalUtils: Registered signal handler for INT
12 17/07/17 16:46:55 WARN NativeCodeLoader: Unable to load native-hadoop library for your platform... using builtin-java classes where applicable
13 Exception in thread "main" java.io.IOException: failure to login
14 at org.apache.hadoop.security.UserGroupInformation.loginUserFromSubject(UserGroupInformation.java:824)
15 at org.apache.hadoop.security.UserGroupInformation.getLoginUser(UserGroupInformation.java:761)
16 at org.apache.hadoop.security.UserGroupInformation.getCurrentUser(UserGroupInformation.java:634)
17 at org.apache.spark.util.Utils$$anonfun$getCurrentUserName$1.apply(Utils.scala:2391)
18 at org.apache.spark.util.Utils$$anonfun$getCurrentUserName$1.apply(Utils.scala:2391)
19 at scala.Option.getOrElse(Option.scala:121)
20 at org.apache.spark.util.Utils$.getCurrentUserName(Utils.scala:2391)
21 at org.apache.spark.SecurityManager.<init>(SecurityManager.scala:221)
22 at org.apache.spark.deploy.worker.Worker$.startRpcEnvAndEndpoint(Worker.scala:714)
23 at org.apache.spark.deploy.worker.Worker$.main(Worker.scala:696)
24 at org.apache.spark.deploy.worker.Worker.main(Worker.scala)
25 Caused by: javax.security.auth.login.LoginException: java.lang.NullPointerException: invalid null input: name
26 at com.sun.security.auth.UnixPrincipal.<init>(UnixPrincipal.java:71)
27 at com.sun.security.auth.module.UnixLoginModule.login(UnixLoginModule.java:133)
28 at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
29 at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
30 at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
31 at java.lang.reflect.Method.invoke(Method.java:497)
32 at javax.security.auth.login.LoginContext.invoke(LoginContext.java:755)
33 at javax.security.auth.login.LoginContext.access$000(LoginContext.java:195)
34 at javax.security.auth.login.LoginContext$4.run(LoginContext.java:682)
35 at javax.security.auth.login.LoginContext$4.run(LoginContext.java:680)
36 at java.security.AccessController.doPrivileged(Native Method)
37 at javax.security.auth.login.LoginContext.invokePriv(LoginContext.java:680)
38 at javax.security.auth.login.LoginContext.login(LoginContext.java:587)
39 at org.apache.hadoop.security.UserGroupInformation.loginUserFromSubject(UserGroupInformation.java:799)
40 at org.apache.hadoop.security.UserGroupInformation.getLoginUser(UserGroupInformation.java:761)
41 at org.apache.hadoop.security.UserGroupInformation.getCurrentUser(UserGroupInformation.java:634)
42 at org.apache.spark.util.Utils$$anonfun$getCurrentUserName$1.apply(Utils.scala:2391)
43 at org.apache.spark.util.Utils$$anonfun$getCurrentUserName$1.apply(Utils.scala:2391)
44 at scala.Option.getOrElse(Option.scala:121)
45 at org.apache.spark.util.Utils$.getCurrentUserName(Utils.scala:2391)
46 at org.apache.spark.SecurityManager.<init>(SecurityManager.scala:221)
47 at org.apache.spark.deploy.worker.Worker$.startRpcEnvAndEndpoint(Worker.scala:714)
48 at org.apache.spark.deploy.worker.Worker$.main(Worker.scala:696)
49 at org.apache.spark.deploy.worker.Worker.main(Worker.scala)
50
51 at javax.security.auth.login.LoginContext.invoke(LoginContext.java:856)
52 at javax.security.auth.login.LoginContext.access$000(LoginContext.java:195)
53 at javax.security.auth.login.LoginContext$4.run(LoginContext.java:682)
54 at javax.security.auth.login.LoginContext$4.run(LoginContext.java:680)
55 at java.security.AccessController.doPrivileged(Native Method)
56 at javax.security.auth.login.LoginContext.invokePriv(LoginContext.java:680)
57 at javax.security.auth.login.LoginContext.login(LoginContext.java:587)
58 at org.apache.hadoop.security.UserGroupInformation.loginUserFromSubject(UserGroupInformation.java:799)
59 ... 10 more
</code></pre>

</i>

<i></i>我尝试在 spark-default.conf 上设置以下属性仍然没有用。

spark.eventLog.enabled             false
spark.ui.enabled                   false
spark.acls.enable                  false
spark.admin.acls                   *
spark.modify.acls                  *
spark.modify.acls.groups           *
spark.ui.view.acls.groups          *
spark.ui.enabled                   false

你能帮我解决这个问题吗？

谢谢

纳温

Answer 1

这是一种不需要 nss_wrapper 的替代方法。

默认情况下，OpenShift 容器使用匿名用户 ID 和组 ID 0(也称为“根”组)运行。首先，设置您的图像，以便 /etc/passwd 由 group-id 0 拥有，并且具有组写入权限，例如这个 Dockerfile 片段:

RUN chgrp root /etc/passwd && chmod ug+rw /etc/passwd

那么你可以在容器启动时添加如下逻辑，例如下面的脚本可以作为一个ENTRYPOINT:

#!/bin/bash

myuid=$(id -u)
mygid=$(id -g)
uidentry=$(getent passwd $myuid)

if [ -z "$uidentry" ] ; then
    # assumes /etc/passwd has root-group (gid 0) ownership
    echo "$myuid:x:$myuid:$mygid:anonymous uid:/tmp:/bin/false" >> /etc/passwd
fi

exec "$@"

这个入口点脚本会自动为匿名 uid 提供一个 passwd 文件入口，这样需要它的工具就不会失败。

有一篇关于 OpenShift 中匿名 uid 和相关主题的不错的博客文章: https://blog.openshift.com/jupyter-on-openshift-part-6-running-as-an-assigned-user-id/