有人可以解释这个CSS属性吗？
text-indent: -1e+7px;
谢谢.
根据原始数组中对象的属性,将大数组更改为多个子数组的最佳方法是什么？例如,我有一个对象数组(所有对象都具有相同的属性):
array = [
{:name => "Jim", :amount => "20"},
{:name => "Jim", :amount => "40"},
{:name => "Jim", :amount => "30"},
{:name => "Eddie", :amount => "7"},
{:name => "Eddie", :amount => "12"},
{:name => "Pony", :amount => "330"},
{:name => "Pony", :amount => "220"},
{:name => "Pony", :amount => "50"}
]
请注意,具有相同名称属性的所有对象在数组中都是连续的.现在我想基于name属性将对象分组为子数组.我需要的是:
result = [
[
{:name => "Jim", :amount => "20"},
{:name => "Jim", :amount => "40"},
{:name => "Jim", :amount => "30"}
],
[ …我是壳牌的绿手.请参阅我的以下代码.如何让它工作？
[root@localhost ~]# ls {*.txt,*.exe}
a.txt b.txt a.exe b.exe
[root@localhost ~]# filter={*.txt,*.exe}
[root@localhost ~]# echo $filter
{*.txt,*.exe}
[root@localhost ~]# ls $filter
ls: {*.txt,*.exe}: No such file or directory
[root@localhost ~]#
我正在构建一些服务器,需要在它们上运行一些PowerShell脚本.但是,我需要Set-ExecutionPolicy -ExecutionPolicy remotesigned -Force在每台服务器上手动运行该命令.因为我无法在脚本中设置执行策略.我想知道PowerShell保存执行策略设置的位置,并希望直接更改它.那可能吗？
谢谢.
Microsoft已公开scriptdomAPI以解析和生成TSQL.我是新手,还在玩它.我想知道如何从像这样的查询中获取跨数据库引用.
UPDATE t3
SET description = 'abc'
FROM database1.dbo.table1 t1
INNER JOIN database2.dbo.table2 t2
ON (t1.id = t2.t1_id)
LEFT OUTER JOIN database3.dbo.table3 t3
ON (t3.id = t2.t3_id)
INNER JOIN database2.dbo.table4 t4
ON (t4.id = t2.t4_id)
我想要的是一个参考列表:
database1.dbo.table1.id = database2.dbo.table2.t1_id
database3.dbo.table3.id = database2.dbo.table2.t3_id
database2.dbo.table4.id = database2.dbo.table2.t4_id
但是,对于最后一个条目database2.dbo.table4.id = database2.dbo.table2.t4_id,来自2端的两个列都来自同一个数据库database2,这不是我想要的.所以我最后要求的结果是:
database1.dbo.table1.id = database2.dbo.table2.t1_id
database3.dbo.table3.id = database2.dbo.table2.t3_id
有可能实现scriptdom吗？
假设我有下面的文字.
aaaaaaa
aaaaaaa
bbb
bbb
bbb
ccccccccccccc
ddddd
ddddd
有没有办法修改文本如下.
1 aaaaaaa
1 aaaaaaa
2 bbb
2 bbb
2 bbb
3 ccccccccccccc
4 ddddd
4 ddddd
我刚刚注意到在sqlite 3源代码文件的开头,它说:
/******************************************************************************
** This file is an amalgamation of many separate C source files from SQLite
** version 3.27.1. By combining all the individual C code files into this
** single large file, the entire code can be compiled as a single translation
** unit. This allows many compilers to do optimizations that would not be
** possible if the files were compiled separately. Performance improvements
** of 5% or more are commonly seen when SQLite is compiled as …我正在学习 Rust，以下代码来自在线书籍The Rust Programming Language。
fn main() {
let mut s = String::from("hello world");
let word = first_word(&s);
s.clear(); // error!
println!("the first word is: {}", word);
}
fn first_word(s: &String) -> &str {
let bytes = s.as_bytes();
for (i, &item) in bytes.iter().enumerate() {
if item == b' ' {
return &s[0..i];
}
}
&s[..]
}
当我运行它时，我得到这个：
C:/Users/administrator/.cargo/bin/cargo.exe run --color=always --package rust2 --bin rust2
Compiling rust2 v0.1.0 (C:\my_projects\rust2)
error[E0502]: cannot borrow `s` as mutable because it is also …我在玩 Rust 并找到了以下示例：
fn main() {
let mut x = [3, 4, 5].to_vec();
x;
println!("{:?}", x);
}
编译器告诉我
18 | let mut x = [3, 4, 5].to_vec();
| ----- move occurs because `x` has type `std::vec::Vec<i32>`, which does not implement the `Copy` trait
...
21 | x;
| - value moved here
22 | println!("{:?}", x);
| ^ value borrowed here after move
似乎x;引起的语句x移到了某个地方，之后就不能使用了。移动目的地在哪里，这里究竟发生了什么？
我四处搜索，找不到任何解释这一点的信息。也许我使用了错误的关键字。
顺便说一句，我正在使用这个版本的 Rust： rustc 1.41.0-nightly (99b89533d 2019-12-16)
Rust cursive是一个 Rust TUI 框架。它指出您可以切换后端，其中一些是 Windows 支持的。但是，我无法让它发挥作用。当我运行以下命令（从这里抓取）时，出现错误：
$ git clone https://github.com/gyscos/cursive
$ cd cursive/examples
$ cargo run -v --no-default-features --features pancurses-backend --example select
error: Package `cursive-examples v0.1.1-alpha.0 (C:\my_projects\cursive\examples)`
does not have the feature `pancurses-backend`
在 Windows 上运行这些示例的正确方法是什么？