删除精确的bash数组元素？

Question

我试图从bash数组中删除任何一个关键字的第一个外观.

ARRAY=(foo bar and any number of keywords)
keywords=(red, rednet, rd3.0)

我删除了这样的关键字:ARRAY=( ${ARRAY[@]/"$keyword"/} )然后如果"red'是第一个找到的关键字,它将从两个关键字中删除' red '并返回"foo bar net"而不是"foo bar rednet".

编辑:这是一个例子,希望这使它更清晰.

for keyword in ${ARRAY[@]}; do
      if [ "$keyword" = "red" ] || [ "$keyword" = "rd3.0" ] || [ "$keyword" = "rednet" ]; then
           # HERE IS TROUBLE
           ARRAY=( ${ARRAY[@]/"$keyword"/} )
           echo "ARRAY is now ${ARRAY[@]}"
           break
      fi
 done

如果ARRAY=(red rednet rd3.0)返回net rd3.0 而不是rednet rd3.0

如果我使用unset,:unset ${ARRAY["$keyword"]}bash抱怨rd3.0是否在数组中::syntax error: invalid arithmetic operator (error token is ".0") 什么是从数组中取消设置或删除完全匹配的安全方法？

Answer 1

使用unset带有数组值的命令index,如下所示:

#!/usr/bin/env bash
ARRAY=(foo bar any red alpha number of keywords rd3.0 and)
keywords=(red, rednet, rd3.0)

index=0
for keyword in ${ARRAY[@]}; do
      if [ "$keyword" = "red" ] || [ "$keyword" = "rd3.0" ] || [ "$keyword" = "rednet" ]; then
           # HERE IS TROUBLE
           # ARRAY=( ${ARRAY[@]/"$p"/} )
           unset ARRAY[$index]
           echo "ARRAY is now: ${ARRAY[@]}"
           break
      fi
      let index++
 done