Bash脚本无法识别函数问题排查及修复方案

问题分析 #

你碰到的delete_snapshots: command not found报错，核心原因是Bash脚本是按顺序解析执行的——原脚本里你在调用delete_snapshots函数之后才定义它，导致执行到调用语句时，Bash还没读取到这个函数的定义，自然识别不了。

你的报错信息如下：

./delete_snapshots.sh
What AWS Account:
12345678910
Delete Snapshots in account: AWS Lab
./delete_snapshots.sh: line 14: delete_snapshots: command not found

原脚本的问题点 #

原脚本中函数定义放在了脚本末尾，而调用逻辑在前面，这就导致了调用时函数还未被加载：

#!/bin/bash 
echo "What AWS Account: "; read accountnumber 
declare -a arr=("12345678910" "109876543212") 
for i in "${arr[@]}" 
do 
    if [ "$i" -eq 12345678910 ]; then 
        aws_account="AWS Lab" 
        aws_key="lab" 
        aws_file="aws_lab_snapshots.txt" 
        echo "Delete Snapshots in account: $aws_account" 
        delete_snapshots  # 这里调用函数，但函数还没定义
        break 
    elif [ "$i" -eq 109876543212 ]; then 
        aws_account="AWS Billing" 
        aws_key="bill" 
        aws_file="aws_bill_snapshots.txt" 
        echo "You are currently in account: $aws_account" 
        delete_snapshots  # 同样的问题
        break 
    else 
        echo "Unkown account" 
    fi 
done 
# 函数定义在调用之后，Bash无法提前识别
delete_snapshots(){ 
    for i in $(cat $aws_file) 
    do 
        echo "*****************************************************************" 
        echo "deleting snapshot: $i" 
        aws ec2 delete-snapshot --snapshot-id=$i --profile=lab 2>&1 | sed -e 's/^An error occurred.*when calling the DeleteSnapshot operation: //' 
        echo "*****************************************************************" 
        echo; echo; echo; echo; echo 
        sleep 5 
        echo "*****************************************************************" 
        echo "Verify that snapshot: $i is gone:" 
        aws ec2 describe-snapshots --snapshot-ids=$i --profile=lab 2>&1 | sed -e 's/^An error occurred.*when calling the DescribeSnapshots operation: //g' 
        echo "*****************************************************************" 
        echo; echo; echo; echo; echo 
    done 
} 

修复后的可运行脚本 #

把函数定义移到脚本最开头，同时做了几个关键优化让脚本更健壮：

#!/bin/bash 

delete_snapshots(){ 
    for i in $(cat $aws_file) 
    do 
        echo "*****************************************************************" 
        echo "deleting snapshot: $i" 
        aws ec2 delete-snapshot --snapshot-id=$i --profile=$aws_key 2>&1 | sed -e 's/^An error occurred.*when calling the DeleteSnapshot operation: //' 
        echo "*****************************************************************" 
        echo; echo; echo; echo; echo 
        sleep 5 
        echo "*****************************************************************" 
        echo "Verify that snapshot: $i is gone:" 
        aws ec2 describe-snapshots --snapshot-ids=$i --profile=$aws_key 2>&1 | sed -e 's/^An error occurred.*when calling the DescribeSnapshots operation: //g' 
        echo "*****************************************************************" 
        echo; echo; echo; echo; echo 
    done 
} 

echo "What AWS Account: "; read accountnumber 
declare -a arr=("123456789101" "109876543212") 
for i in "${arr[@]}" 
do 
    if [ "$i" -eq 123456789101 ]; then 
        aws_account="AWS Lab" 
        aws_key="lab" 
        aws_file="source_files/aws_lab_snapshots.txt" 
        echo "Delete Snapshots in account: $aws_account" 
        delete_snapshots 
        break 
    elif [ "$i" -eq 109876543212 ]; then 
        aws_account="AWS Billing" 
        aws_key="bill" 
        aws_file="source_files/aws_bill_snapshots.txt" 
        echo "You are currently in account: $aws_account" 
        delete_snapshots 
        break 
    else 
        echo "Unkown account" 
    fi 
done 

关键修改说明 #

• 函数定义前置：将delete_snapshots函数移到脚本最顶部，确保Bash在执行调用逻辑前已经加载了函数定义
• 动态Profile参数：把aws命令中硬编码的--profile=lab改为--profile=$aws_key，让不同账户自动使用对应的AWS配置文件
• 文件路径修正：给快照列表文件加上source_files/前缀，确保脚本能正确定位到目标文件
• 账户ID修正：修正第一个账户ID的拼写，避免出现匹配错误

内容的提问来源于stack exchange，提问作者bluethundr