监视磁盘分区使用情况的bash脚本 [英] bash script that monitor a disk partition's usage

问题描述

df表演

-bash-4.1# df
Filesystem           1K-blocks      Used Available Use% Mounted on
/dev/sda3            1918217320 1783986384  36791092  98% /
tmpfs                 16417312         0  16417312   0% /dev/shm
/dev/sda1               482214    148531    308784  33% /boot
/dev/sdb1            1922858352 1373513440 451669312  76% /disk2

我需要对bash脚本编写脚本，如果分区100％充满，该函数将返回1.如何才能做到这一点?我可以使用哪些命令来解析df的输出?

I need to bash script a function that returns 1 if an paritions become 100% full. how can this be done? what commands can I use to parse out the output of df?

推荐答案

不是过度拥护和笨拙的忠实拥护者，因为随着时间的推移，它确实会带来错误.

Not a huge fan of excessive greps and awks as it can really bring errors over time..

我只获取重要文件夹的信息.以下是使用stat的示例，它将为您提供文件夹中的可用BYTES，然后将其转换为MB(10 ** 6).我在RHEL6.x系统上对此进行了粗略的测试.

I would just get the information for the folders that matter. Below is a sample of using stat which will give you the available BYTES in a folder, then converts it to MB (10**6). I roughly tested this on my RHEL6.x system.

    folder_x_mb=$(($(stat -f --format="%a*%s" /folder_x)/10**6))    
    folder_y_mb=$(($(stat -f --format="%a*%s" /folder_y)/10**6))   
    folder_z_mb=$(($(stat -f --format="%a*%s" /folder_z)/10**6))     

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