Bash脚本中Curl调用API失败的问题排查与修复指导

问题场景 #

有一个在Linux上运行的Bash脚本，通过curl调用API接口，接收用户输入后执行始终返回{"message": "Invalid request body"}错误，但手动执行脚本输出的curl命令却能成功调用API。

原脚本内容：

#!/bin/bash
# Constants
api_key="Thdshd760Jd"
api_url="https://execute-api.amazonaws.com/prod/v1/archive;done"

# User input
read -p "Enter projectName: " projectName
read -p "Enter siteName: " siteName
read -p "Enter emailAddress: " emailAddress
read -p "Enter the path to the file containing disk names: " disk_file

# Read disk names from the file and create a comma-separated string
disks=$(awk '{printf "\"%s\", ", $1}' "$disk_file" | sed 's/, $//')

# Create JSON payload in Bash
json_payload="'{\"projectName\": \"$projectName\", \"siteName\": \"$siteName\", \"disks\": [$disks], \"emailAddress\": \"$emailAddress\"}'"

# Make the API request using curl
curl -vv  -X POST -H "x-api-key: $api_key" -H "Content-Type: application/json" -d "$json_payload" $api_url

echo 'curl -vv -X POST -H "x-api-key: '$api_key'" -H "Content-Type: application/json" -d '$json_payload' '$api_url''

执行过程：

$ ./curl_api_call.sh
Enter projectName: my_disk001
Enter siteName: JLDC
Enter emailAddress: labtest@example.com
Enter the path to the file containing disk names: demo_file
Note: Unnecessary use of -X or --request, POST is already inferred.
{"message": "Invalid request body"}
curl -vvvvvv -X POST -H "x-api-key: Thdshd760Jd" -H "Content-Type: application/json" -d '{"projectName": "my_disk001", "siteName": "JLDC", "disks": ["my_disk001"], "emailAddress": "labtest@example.com"}' https://execute-api.amazonaws.com/prod/v1/archive

另外尝试的循环版本也报相同错误：

for i in `cat` ; do curl -vvvvvv -X POST -H "x-api-key: Thdshd760Jd" -H "Content-Type: application/json" -d '{"projectName": "$i", "siteName": "JLDC", "disks": ["$i"], "emailAddress": "labtest@example.com"}' https://execute-api.amazonaws.com/prod/v1/archive;done

错误原因 #

1. JSON payload被多余单引号包裹：脚本中json_payload="'{...}'"的写法，给整个JSON字符串套了一层单引号。curl发送的请求体实际是'{"projectName": ...}'（包含外层单引号），这不符合JSON格式规范，导致API解析失败。而手动执行时去掉了外层单引号，所以请求合法。
2. 循环版本变量未展开：循环里用单引号包裹JSON内容，$i不会被Bash解析为变量值，发送的是字面量"$i"，API自然无法识别。

修正方案 #

1. 修正原脚本 #

• 移除json_payload外层的单引号，直接用双引号包裹JSON（注意转义内部双引号）；或者用printf构建JSON，避免转义错误。
• 给api_url添加双引号，避免Bash解析;为命令分隔符。

修正后的完整脚本：

#!/bin/bash
# Constants
api_key="Thdshd760Jd"
api_url="https://execute-api.amazonaws.com/prod/v1/archive;done"

# User input
read -p "Enter projectName: " projectName
read -p "Enter siteName: " siteName
read -p "Enter emailAddress: " emailAddress
read -p "Enter the path to the file containing disk names: " disk_file

# Read disk names from the file and create a comma-separated string
disks=$(awk '{printf "\"%s\", ", $1}' "$disk_file" | sed 's/, $//')

# 用printf构建JSON，避免转义混乱
json_payload=$(printf '{"projectName": "%s", "siteName": "%s", "disks": [%s], "emailAddress": "%s"}' "$projectName" "$siteName" "$disks" "$emailAddress")

# 调用curl时给api_url加双引号，处理特殊字符
curl -vv -X POST -H "x-api-key: $api_key" -H "Content-Type: application/json" -d "$json_payload" "$api_url"

# 输出调试命令时确保引号正确
echo "curl -vv -X POST -H \"x-api-key: $api_key\" -H \"Content-Type: application/json\" -d '$json_payload' \"$api_url\""

2. 修正循环版本 #

需要让$i变量正常展开，有两种写法：

• 方法1：用双引号包裹JSON，转义内部双引号

for i in $(cat); do 
  curl -vvvvvv -X POST -H "x-api-key: Thdshd760Jd" -H "Content-Type: application/json" -d "{\"projectName\": \"$i\", \"siteName\": \"JLDC\", \"disks\": [\"$i\"], \"emailAddress\": \"labtest@example.com\"}" https://execute-api.amazonaws.com/prod/v1/archive
done

• 方法2：用单引号拼接变量（更清晰，减少转义）

for i in $(cat); do 
  curl -vvvvvv -X POST -H "x-api-key: Thdshd760Jd" -H "Content-Type: application/json" -d '{"projectName": "'"$i"'", "siteName": "JLDC", "disks": ["'"$i"'"], "emailAddress": "labtest@example.com"}' https://execute-api.amazonaws.com/prod/v1/archive
done

内容的提问来源于stack exchange，提问作者user2023