我正在尝试从字符串中删除最后15个字符，我必须使用“ ksh”运行该BASH脚本。它与“ bash”一起使用时效果很好，但与“ ksh”一起使用时效果不佳。这是我的代码，
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我在做什么错了？
这是#!/bin/bash
ggate_location="'$(ps -ef|grep mgr)'"
for word in $ggate_location
do
[[ $word =~ mgr\.prm$ ]] && echo ${word::-15}
done
$(ps -ef | grep mgr)答案 0 :(得分：1)
如果性能不是很关键，那么依赖外部可执行文件的以下管道应该可以正常工作：
ps -o cmd= | grep -Ewo '[^::space::]*mgr\.prm' | cut -c -15
ps -o cmd=仅要求ps显示命令行（不包含标题），grep将行过滤为包含以mgr.\prm结尾的单词的行，{{1 }}仅返回该单词的前15个字符。
请注意，cut grep ord-regexp标志不是POSIX定义的，除非您使用的是GNU -w，否则可能不起作用。在那种情况下，我建议您将grep与grep CRE regex风味标志一起使用，并在模式末尾添加一个单词-P边界，或向{模式的结尾。