我应该现在就起诉委员会的人。
关键是,我在日志中收到了所有关于cphulk (蛮力检测系统)无法阻止机器人的奇怪信息,因此,我发现我的网站运行速度有点慢,足以让google再支付0美元。是的,我已经超过了10到40毫秒,当我继续用webpagetest.org测试的时候,数字的变化比以前更大,所以我很确定黑客们在玩游戏。
现在来看看错误:
我查看我的邮件日志,看看那里发生了什么,我看到了这样的行:
Feb 4 04:43:19 server dovecot: auth: Error: Cpanel::MailAuth: Failed to getpwnam for user bar
Feb
让我们从文档开始:https://man7.org/linux/man-pages/man3/getpwnam.3.html 有了这个,我编写了以下C#代码: using System;
using System.Runtime.InteropServices;
if (args.Length < 1) {
Console.Error.WriteLine("Provide user name.");
Environment.Exit(-1);
}
var name = args[0];
if (RuntimeInformation.IsOSPlat
我想要构建静态,这样我就可以在嵌入式设备上使用它作为独立的二进制文件。
screen-4.2.1# ./configure LDFLAGS="-static" && make
我收到警告:
/screen.c:933: warning: Using 'getpwnam' in statically linked applications requires at runtime the shared libraries from the glibc version used for linking
原因是通过回答的。因此,如果我理解正确的话,就不可
我有一个函数:
func (struct passwd* pw)
{
struct passwd* temp;
struct passwd* save;
temp = getpwnam("someuser");
/* since getpwnam returns a pointer to a static
* data buffer, I am copying the returned struct
* to a local struct.
*/
if(temp) {
save = malloc(sizeof *save);
if (save) {
我有代码:
print('Adding user %s...' % user)
# my own class with method
sysacts.useradd()
# check if user exist after been added
try:
pwd.getpwnam(user)
print('Done.')
except KeyError as e:
print('ERROR! %s.\nExit.' %e)
#include <pwd.h>
#include <stdio.h>
struct passwd* Getpwnam_(const char* name)
{
static struct passwd* passwd;
while((passwd=getpwent())!=NULL) /* get pw entry line by line */
{
if(strcmp(passwd->pw_name, name)==0) /* find the same n
我有一个ansible剧本,它在Linux上工作得很好,但是在macOS上失败了,出现了一个奇怪的错误。
- name: Create the watcher user member of the watchers group
user:
comment: "Read-only user for folks to inspect builds"
name: "{{ watch_user }}"
group: watchers
state: present
shell: /bin/bash
createhome: