当我得到不同的记录计数时，这个问题出现了，我认为是相同的查询，一个使用not in where约束，另一个使用左连接。not in约束中的表有一个空值(坏数据)，导致该查询返回0条记录计数。我有点理解为什么，但我需要一些帮助来充分理解这个概念。

简单地说，为什么查询A返回结果而B没有?

A: select 'true' where 3 in (1, 2, 3, null)
B: select 'true' where 3 not in (1, 2, null)

这是在SQL Server 2005上。我还发现调用set ansi_nulls off会导致B返回一个结果。

DLL文件究竟是如何工作的?它们似乎有很多，但我不知道它们是什么，也不知道它们是如何工作的。

那么，他们是怎么回事?

替代标题

Xcode构建变量列表 打印Xcode构建设置列表 Clang环境变量 Xcode环境变量的规范列表

是否有一个Xcode环境变量的规范列表，可以用于构建规则等?

我正在查看这里的strlen代码，我想知道在代码中使用的优化是否真的需要?例如，为什么像下面这样的工作不一样好或更好?

unsigned long strlen(char s[]) {
    unsigned long i;
    for (i = 0; s[i] != '\0'; i++)
        continue;
    return i;
}

更简单的代码是不是更好和/或更容易编译器优化?

strlen在链接后面的页面上的代码是这样的:

/* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc. This file is part of the GNU C Library. Written by Torbjorn Granlund (tege@sics.se), with help from Dan Sahlin (dan@sics.se); commentary by Jim Blandy (jimb@ai.mit.edu). The GNU C Library is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation; either version 2.1 of the License, or (at your option) any later version. The GNU C Library is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with the GNU C Library; if not, write to the Free Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA. */ #include <string.h> #include <stdlib.h> #undef strlen /* Return the length of the null-terminated string STR. Scan for the null terminator quickly by testing four bytes at a time. */ size_t strlen (str) const char *str; { const char *char_ptr; const unsigned long int *longword_ptr; unsigned long int longword, magic_bits, himagic, lomagic; /* Handle the first few characters by reading one character at a time. Do this until CHAR_PTR is aligned on a longword boundary. */ for (char_ptr = str; ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0; ++char_ptr) if (*char_ptr == '\0') return char_ptr - str; /* All these elucidatory comments refer to 4-byte longwords, but the theory applies equally well to 8-byte longwords. */ longword_ptr = (unsigned long int *) char_ptr; /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits the "holes." Note that there is a hole just to the left of each byte, with an extra at the end: bits: 01111110 11111110 11111110 11111111 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD The 1-bits make sure that carries propagate to the next 0-bit. The 0-bits provide holes for carries to fall into. */ magic_bits = 0x7efefeffL; himagic = 0x80808080L; lomagic = 0x01010101L; if (sizeof (longword) > 4) { /* 64-bit version of the magic. */ /* Do the shift in two steps to avoid a warning if long has 32 bits. */ magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL; himagic = ((himagic << 16) << 16) | himagic; lomagic = ((lomagic << 16) << 16) | lomagic; } if (sizeof (longword) > 8) abort (); /* Instead of the traditional loop which tests each character, we will test a longword at a time. The tricky part is testing if *any of the four* bytes in the longword in question are zero. */ for (;;) { /* We tentatively exit the loop if adding MAGIC_BITS to LONGWORD fails to change any of the hole bits of LONGWORD. 1) Is this safe? Will it catch all the zero bytes? Suppose there is a byte with all zeros. Any carry bits propagating from its left will fall into the hole at its least significant bit and stop. Since there will be no carry from its most significant bit, the LSB of the byte to the left will be unchanged, and the zero will be detected. 2) Is this worthwhile? Will it ignore everything except zero bytes? Suppose every byte of LONGWORD has a bit set somewhere. There will be a carry into bit 8. If bit 8 is set, this will carry into bit 16. If bit 8 is clear, one of bits 9-15 must be set, so there will be a carry into bit 16. Similarly, there will be a carry into bit 24. If one of bits 24-30 is set, there will be a carry into bit 31, so all of the hole bits will be changed. The one misfire occurs when bits 24-30 are clear and bit 31 is set; in this case, the hole at bit 31 is not changed. If we had access to the processor carry flag, we could close this loophole by putting the fourth hole at bit 32! So it ignores everything except 128's, when they're aligned properly. */ longword = *longword_ptr++; if ( #if 0 /* Add MAGIC_BITS to LONGWORD. */ (((longword + magic_bits) /* Set those bits that were unchanged by the addition. */ ^ ~longword) /* Look at only the hole bits. If any of the hole bits are unchanged, most likely one of the bytes was a zero. */ & ~magic_bits) #else ((longword - lomagic) & himagic) #endif != 0) { /* Which of the bytes was the zero? If none of them were, it was a misfire; continue the search. */ const char *cp = (const char *) (longword_ptr - 1); if (cp[0] == 0) return cp - str; if (cp[1] == 0) return cp - str + 1; if (cp[2] == 0) return cp - str + 2; if (cp[3] == 0) return cp - str + 3; if (sizeof (longword) > 4) { if (cp[4] == 0) return cp - str + 4; if (cp[5] == 0) return cp - str + 5; if (cp[6] == 0) return cp - str + 6; if (cp[7] == 0) return cp - str + 7; } } } } libc_hidden_builtin_def (strlen)

为什么这个版本运行得很快?

它是不是做了很多不必要的工作?

当我试图访问react组件中的存储时，@connect工作得很好。但我该如何在其他代码中访问它呢。例如:让我们说我想使用授权令牌来创建我的axios实例，可以在我的应用程序中全局使用，实现这一点的最佳方法是什么?

这是我的api。js

// tooling modules
import axios from 'axios'

// configuration
const api = axios.create()
api.defaults.baseURL = 'http://localhost:5001/api/v1'
api.defaults.headers.common['Authorization'] = 'AUTH_TOKEN' // need the token here
api.defaults.headers.post['Content-Type'] = 'application/json'

export default api

现在我想从我的商店访问一个数据点，这是什么样子，如果我试图在一个react组件中使用@connect获取它

// connect to store
@connect((store) => {
  return {
    auth: store.auth
  }
})
export default class App extends Component {
  componentWillMount() {
    // this is how I would get it in my react component
    console.log(this.props.auth.tokens.authorization_token) 
  }
  render() {...}
}

有什么见解或工作流模式吗?

有64位的Visual Studio吗?为什么不呢?

我试图遵循数字海洋这篇文章中讨论的Redis安装过程，用于WSL(Windows Sub-System for Linux)。安装的Ubuntu版本为Ubuntu 18.04。

redis安装中的一切都很好，但当我试图运行这个sudo systemctl启动redis时，我得到了这个消息。

System has not been booted with systemd as init system (PID 1). Can't operate.

你知道我该怎么做吗?

我想从第一个中间件传递一些变量到另一个中间件，我试着这样做，但有“req。有些变量是一个给定的‘未定义’”。

---

//app.js
..
app.get('/someurl/', middleware1, middleware2)
...

---

////middleware1
...
some conditions
...
res.somevariable = variable1;
next();
...

---

////middleware2
...
some conditions
...
variable = req.somevariable;
...

我正在实现以下模型存储用户相关的数据在我的表-我有2列- uid(主键)和一个元列，其中存储关于JSON格式的用户的其他数据。

 uid   | meta
--------------------------------------------------
 1     | {name:['foo'], 
       |  emailid:['foo@bar.com','bar@foo.com']}
--------------------------------------------------
 2     | {name:['sann'], 
       |  emailid:['sann@bar.com','sann@foo.com']}
--------------------------------------------------

这种方法(在性能和设计方面)是否比每个属性一列模型更好?在每个属性一列模型中，表将有许多列，如uid、name、emailid。

我喜欢第一个模型的地方是，你可以添加尽可能多的字段，没有限制。

另外，我想知道，既然我已经实现了第一个模型。我如何对它执行查询，比如，我想获取所有名称为'foo'的用户?

问:在数据库中存储用户相关数据(请记住，字段的数量是不固定的)，使用JSON还是每个字段列?另外，如果实现了第一个模型，如何查询上述数据库?我应该使用这两个模型，通过存储所有的数据，可以在一个单独的行和JSON(是不同的行)的数据查询搜索?

---

更新

由于没有太多需要执行搜索的列，使用这两种模型是否明智?每列键的数据，我需要搜索和JSON为其他人(在同一个MySQL数据库)?

我目前有一个密钥存储库，其中有一个只有我应该知道的特定密码。我现在需要将对密钥库的访问权授予其他人，因此我希望:

1)修改密码，这样我就可以和其他人分享，让他们签名 2)创建一个不同的密码，并允许他们使用该密码签名。

这可能吗?如果是，怎么做?