curl 命令负载中字符串的脚本连接

Question

我curl用来测试用户帐户创建 API 如下：

curl -s -X POST "https://$APISERVER/users" \
-H 'Content-Type: application/json' \
-d '{ \
"username": "'$NEWUSERNAME'", \
"firstName": "'$NEWUSERFIRSTNAME'", \
"lastName": "'$NEWUSERLASTNAME'", \
"displayName": "'$NEWUSERDISPLAYNAME'", \
"password": "'$NEWUSERPASSWORD'" \
}'

并且变量是通过命令行参数提供的：

APISERVER=http://localhost:8080
NEWUSERNAME=$1
NEWUSERPASSWORD=$2
NEWUSERFIRSTNAME=$3
NEWUSERLASTNAME=$4

# Calculated variable
NEWUSERDISPLAYNAME="${NEWUSERFIRSTNAME} ${NEWUSERLASTNAME}"

脚本调用示例如下：./test-new-user.sh jdoe Hello123 John Doe，产生以下变量值：

NEWUSERNAME=jdoe
NEWUSERPASSWORD=Hello123
NEWUSERFIRSTNAME=John
NEWUSERLASTNAME=Doe

（我打算NEWUSERDISPLAYNAME设置为“John Doe”）

但是我从服务器返回一个异常，因为curl命令中的有效负载似乎被切断、不完整或格式错误。

JSON parse error: Unexpected end-of-input in VALUE_STRING\n at [Source: 
java.io.PushbackInputStream@2eda6052; line: 1, column: 293]; nested 
exception is com.fasterxml.jackson.databind.JsonMappingException: 
Unexpected end-of-input in VALUE_STRING\n at [Source: 
java.io.PushbackInputStream@2eda6052; line: 1, column: 293]\n at 
[Source: java.io.PushbackInputStream@2eda6052; line: 1, column: 142] 
(through reference chain: 
com.mycompany.api.pojos.NewUser[\"displayName\"])"

如果我displayName在上面的 curl 命令中硬编码值（如下所示），则用户创建请求会通过并完美运行。

"displayName": "John Doe", \

我怀疑这与中的空间displayName以及我如何插入displayNameusing的值有关"'$NEWUSERDISPLAYNAME'"。是否有一种安全的方法可以在curl命令的 POST 请求负载中执行变量替换？

Answer 1

您需要引用 shell 变量：

curl -s -X POST "https://$APISERVER/users" \
-H 'Content-Type: application/json' \
-d '{ \
"username": "'"$NEWUSERNAME"'", \
"firstName": "'"$NEWUSERFIRSTNAME"'", \
"lastName": "'"$NEWUSERLASTNAME"'", \
"displayName": "'"$NEWUSERDISPLAYNAME"'", \
"password": "'"$NEWUSERPASSWORD"'" \
}'

为了避免过度引用，试试这个printf：

printf -v json -- '{ "username": "%s", "firstName": "%s", "lastName": "%s", "displayName": "%s", "password": "%s" }' \
"$NEWUSERNAME" "$NEWUSERFIRSTNAME" "$NEWUSERLASTNAME" "$NEWUSERDISPLAYNAME" "$NEWUSERPASSWORD"

curl -s -X POST "https://$APISERVER/users" \
    -H 'Content-Type: application/json' \
    -d "$json"