我有以下几行代码:
%for RSA Algorithm
clc;
disp('Implementation of RSA Algorithm');
clear all;
close all;
p = input('\nEnter value of p: ');
q = input('\nEnter value of q: ');
[Pk,Phi,d,e] = init(p,q);
M = input('\nEnter message: ','s');
x=length(M);
c=0;
for j=
当我使用PDO对mysql数据库运行此查询时,它返回错误的数据类型。
<?php
$parameters = array(":1",92323);
$query = " SELECT s.Site_ID,s.Site_Url,s.Site_Name, s.Site_Approved, s.Site_Status, s.Thumbnailed ,st.Description AS Site_Status_Desc
FROM Sites s
LEFT JOIN Sites_Status st ON st.Status_ID = s.Site_Status
WHER
挑战是更改用户的AD密码。我有一个包装ldapmodify的TCL脚本来设置密码,它是有效的:
set unicodePwd [encodePw4ad $pw]
lappend text {dn: $dn}
lappend text {changetype: modify}
lappend text {replace: unicodePwd}
lappend text {unicodePwd:: $unicodePwd}
lappend text {-}
set fn /tmp/ldiff.[clock microseconds].ldif
write_file $fn [subst [j