同步四个Shell脚本以在UNIX中一个接一个地运行 [英] Synchronizing four shell scripts to run one after another in unix

问题描述

我有4个Shell脚本来生成一个文件(比如说param.txt)，该文件将由另一个工具(informatica)使用，并且随着该工具的处理完成，它将删除param.txt.

I have 4 shell script to generate a file(let's say param.txt) which is used by another tool(informatica) and as the tool is done with processing, it deletes param.txt.

这里的意图是所有四个脚本可以在不同的时间被调用，比如说12:10 am，12:13 am，12:16 am，12:17 am.第一个脚本在上午12:10运行，并创建param.txt并触发使用param.txt的informatica进程. Informatica流程又需要5到10分钟才能完成，并删除param.txt.第2个脚本在上午12:13调用并等待param.txt的不可用，并且随着informatica进程将其删除，脚本2创建了新的param.txt并再次触发相同的informatica.另外两个脚本也会发生同样的情况.

The intent here is all four scripts can get invoked at different time lets say 12:10 am, 12:13 am, 12:16 am, 12:17 am. First script runs at 12:10am and creates param.txt and trigger informatica process which uses param.txt. Informatica process takes another 5-10 minutes to complete and deletes the param.txt. The 2nd script invokes at 12:13 am and waits for unavailability of param.txt and as informatica process deletes it, script 2 creates new param.txt and triggers same informatica again. The same happen for another 2 scripts.

我正在所有4个shell脚本中使用直到和睡眠命令来检查param.txt的不可用性，如下所示:

I am using Until and sleep commands in all 4 shell script to check the unavailability of param.txt like below:

until [ ! -f "$paramfile" ]
do
    Sleep 10
done
<create param.txt file>

这里的问题是，有时当所有4个脚本开始时，第一个成功并生成param.txt(因为之前没有param.txt)和其他等待，但是当informatica进程完成并删除param.txt时，剩下的3个脚本或其中两个同时检查不可用性，其中一个会创建不可用性，但所有操作都会成功.我检查了四个脚本之间不同的睡眠间隔组合，但是这种情况几乎每次都发生.

Issue here is, sometimes when all 4 scripts begin, the first one succeeds and generates param.txt(as there was no param.txt before) and other waits but when informatica process completes and deletes param.txt, remaining 3 scripts or 2 of them checks the unavailability at same time and one of them creates it but all succeed. I have checked different combinations of sleep interval between four scripts but this situation is occurring almost every time.

推荐答案

您正在经历经典的比赛条件.要解决此问题，您需要在4个脚本之间共享一个锁"(或类似名称).

You are experiencing a classical race condition. To solve this issue, you need a shared "lock" (or similar) between your 4 scripts.

有几种方法可以实现此目的.在bash中执行此操作的一种方法是使用flock命令，并使用同意的文件名作为锁定. flock手册页具有一些类似于以下的用法示例:

There are several ways to implement this. One way to do this in bash is by using the flock command, and an agreed-upon filename to use as a lock. The flock man page has some usage examples which resemble this:

(
    flock -x 200  # try to acquire an exclusive lock on the file
    # do whatever check you want. You are guaranteed to be the only one
    # holding the lock
    if [ -f "$paramfile" ]; then
        # do something
    fi
) 200>/tmp/lock-life-for-all-scripts
# The lock is automatically released when the above block is exited

您也可以要求flock如果无法获取锁，则立即失败，或者在超时后失败(例如，打印仍在尝试获取锁"并重新启动).

You can also ask flock to fail right away if the lock can't be acquired, or to fail after a timeout (e.g. to print "still trying to acquire the lock" and restart).

根据您的用例，您还可以将锁锁定在"informatica"二进制文件上(在这种情况下，请确保使用200<来打开文件以供读取(而不是(过度)写入)

Depending on your use case, you could also put the lock on the 'informatica' binary (be sure to use 200< in that case, to open the file for reading instead of (over)writing)

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