Question

我有以下代码：

PROJECT_TYPES="iPad iPhone"
ANT_TARGET_NAMES="ipadf ipaf"

INDEX=0

for PROJECT_TYPE in $PROJECT_TYPES; do

echo "${PROJECT_TYPE} => ${ANT_TARGET_NAMES[$INDEX]}"

let "INDEX++"
done

显示以下行：

iPad => ipadf ipaf
iPhone =>

如何更改代码以显示：

iPad => ipadf
iPhone =>  ipaf

???

提前致谢

麦克

Answer 1

正确的方法是：

INDEX=0
PROJECT_TYPES=(iPad iPhone)
ANT_TARGET_NAMES=(ipadf ipaf)

for PROJECT_TYPE in ${PROJECT_TYPES[*]} 
do 
echo "${PROJECT_TYPE} => ${ANT_TARGET_NAMES[$INDEX]}"  
let "INDEX++" 
done

Answer 2

bash 4有关联数组，所以你可以写：

declare -A targets=([iPad]=ipadf [iPhone]=ipaf)
for project_type in "${!targets[@]}"; do
  printf "%s => %s\n" "$project_type" "${targets[$project_type]}"
done

否则，按照ennuikiller的回答声明两个数组，但我会直接迭代索引

projects=(iPad iPhone)
targets=(ipadf ipaf)
for (( i=0; i < ${#projects[@]}; i++ )); do
  printf "%s => %s\n" "${projects[$i]}" "${targets[$i]}"
done

在bash中按索引访问数组项

2 个答案: