blocks|key|1154049|text|如果将密钥和$x连接起来会发生什么？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1154050|$_.Key+%2B+$x|1154051|写主机似乎有一个语法，你可以在其中输入变量，它将它们分开:写主机(a，b，c)+-separator+"_“|1154052|也许你正在使用的就是它的速记？|1154053|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

What happens if you concatenate the key and the $x?

$_.Key + $x

Write-Host seems to have a syntax where you can input variables and it will separate them: Write-Host (a,b,c) -separator "_"

Perhaps what you're using is shorthand for that?

blocks|key|1154059|text|这不仅仅发生在哈希表的关键字上。例如，看看这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1154060|PS+>+Write-Host+$x.Length$y
3+rrr|code-block|syntax|javascript|1154061|但$x.Length$y不是有效的表达式：|offset|length|style|CODE|1154062|PS+>+$x.Length$y
Unexpected+token+'y'+in+expression+or+statement.
At+line:1+char:11
%2B+$x.Length$y+<<<<|1154063|所以必须发生的是，在将$x.Length$y传递给写主机之前，powershell将它们解释为两个单独的表达式，并且由于有两个表达式，写主机用分隔符将它们分开。|1154064|那么这里发生了什么呢？：|1154065|PS+>+Write-Host+$x$y
qqqrrr|1154066|$x$y也不是有效的表达式：|1154067|PS+>+$x$y
Unexpected+token+'y'+in+expression+or+statement.
At+line:1+char:4
%2B+$x$y+<<<<|1154068|但是，带双引号的它是一个有效的字符串：|1154069|PS+>+"$x$y"
qqqrrr|1154070|在将$x$y作为单个参数传递给Write-Host之前，Powershell必须将其转换为字符串。|1154071|然而，基本上这一切都只是猜测。对我来说，这听起来很有道理。简短的答案是，您可以自己将字符串连接在一起，以消除空格：|1154072|Write-Host+($_.Key+%2B+$x)|1154073|entityMap^0|0|0|1|B|0|0|B|B|0|0|0|0|4|0|0|0|0|2|4|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1C|8|@$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1G|8|@$I|1H|J|1I|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1J|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1L|8|@$I|1M|J|1N|K|L]]|9|@]|A|$]]|$1|W|3|X|5|D|7|1O|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1P|8|@]|9|@]|A|$]]|$1|10|3|11|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|12|3|13|5|6|7|1R|8|@$I|1S|J|1T|K|L]]|9|@]|A|$]]|$1|14|3|15|5|6|7|1U|8|@]|9|@]|A|$]]|$1|16|3|17|5|D|7|1V|8|@]|9|@]|A|$E|F]]|$1|18|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|19|$]]

It doesn't happen just for hashtable keys. Look at this for instance:

<pre><code>PS &gt; Write-Host $x.Length$y
3 rrr
</code></pre>

But <code>$x.Length$y</code> is not a valid expression:

<pre><code>PS &gt; $x.Length$y
Unexpected token 'y' in expression or statement.
At line:1 char:11
+ $x.Length$y &lt;&lt;&lt;&lt;
</code></pre>

So what must be happening is that powershell is interpreting <code>$x.Length$y</code> as two separate expressions before passing them to Write-Host, and since there are two of them, Write-Host separates them with a separator.

So what is happening here then?:

<pre><code>PS &gt; Write-Host $x$y
qqqrrr
</code></pre>

<code>$x$y</code> isn't a valid expression either:

<pre><code>PS &gt; $x$y
Unexpected token 'y' in expression or statement.
At line:1 char:4
+ $x$y &lt;&lt;&lt;&lt;
</code></pre>

However, with double quotes it is a valid string:

<pre><code>PS &gt; "$x$y"
qqqrrr
</code></pre>

Powershell must be converting <code>$x$y</code> to a string before passing it to Write-Host as a single argument.

Basically however that is all just guesswork. It sounds plausible to me. The short answer is that you can concatenate your strings together yourself to get rid of the space:

<pre><code>Write-Host ($_.Key + $x)
</code></pre>

blocks|key|1154093|text|您好，Powershell解析器正在使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1154094|Write-Host+$_.Key$a|code-block|syntax|javascript|1154095|并通过在“$”上拆分来将其标记为两个表达式。实际上，向管道发出两个命令：|1154096|Write-Host+$_.Key
$x|1154097|我认为这主要是因为您正在访问一个属性，并将其扩展为变量旁边的一个字符串，该变量在求值时将成为一个字符串。|1154098|你应该这样做：|1154099|PS+C:\Users\jpogran>+$table+=+@{
>>+"aaa"=1
>>+}
>>
PS+C:\Users\jpogran>+$x+=+"qqq"
PS+C:\Users\jpogran>+$y+=+"rrr"
PS+C:\Users\jpogran>+$table

Name+++++++++++++++++++++++++++Value
----+++++++++++++++++++++++++++-----
aaa++++++++++++++++++++++++++++1
PS+C:\Users\jpogran>+$table.Getenumerator()+%7C+%25{+Write-host+"$($_.key)$x"+}
aaaqqq|1154100|'$()‘告诉Powershell解析器你想要扩展该对象上的属性，并将值放入字符串中任何后面/之前的内容。|1154101|詹姆斯|1154102|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

IMHO, the Powershell parser is taking 

<pre><code>Write-Host $_.Key$a
</code></pre>

And tokenizing it into two expressions by splitting on the '$'. In effect, issuing two commands to the pipeline:

<pre><code>Write-Host $_.Key
$x
</code></pre>

I think this is mostly because you are accessing a property and expanding it into a string next to a variable that when evaluated becomes a string.

The way you should be doing this is:

<pre><code>PS C:\Users\jpogran&gt; $table = @{
&gt;&gt; "aaa"=1
&gt;&gt; }
&gt;&gt;
PS C:\Users\jpogran&gt; $x = "qqq"
PS C:\Users\jpogran&gt; $y = "rrr"
PS C:\Users\jpogran&gt; $table

Name Value
---- -----
aaa 1
PS C:\Users\jpogran&gt; $table.Getenumerator() | %{ Write-host "$($_.key)$x" }
aaaqqq
</code></pre>

The '$()' tells the Powershell parser you want that property on that object expanded, and the value put into the string next to whatever comes after/before.

james

blocks|key|1161164|text|在处理命令的参数时，如果第一个片段是可识别的表达式，则PowerShell解析器会将相邻的表达式拆分为离散的参数，就像$()的情况一样。因为PowerShell是基于对象的，所以我们试图保持对象的完整性，直到我们完全必须呈现为字符串。如果你想强制字符串扩展，就像其他发帖建议的那样，在整个参数序列两边加上双引号。请注意，具有$x之类前导字符的参数将被视为双引号字符串"+a$x“。这是因为前导字符没有被识别为有效的表达式，所以我们默认将其视为可扩展的字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1161165|-bruce|1161166|微软公司视窗PowerShell团队的布鲁斯·佩耶特|1161167|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

When processing arguments to a command, the PowerShell parser splits adjacent expressions into discrete arguments if the first fragment is a recognizable expression as is the case with $(). Because PowerShell is object-based, we try to preserve object integrity until we absolutely have to render to a string. If you want to force string expansion put double-quotes around the entire argument sequence as one of the other posters suggested. Note that an argument with a leading character like a$x will be treated as if it was a double-quoted string "a$x". This is because the leading characters are not recognized as a valid expression so we default to treating it as an expandable string.

-bruce

Bruce Payette,
Windows PowerShell Team, 
Microsoft Corp.

I don't understand why there's a difference on string concatenations in this script. 
I really don't understand where the whitespace is coming from? 
Can you tell me what's wrong?

<pre><code>$table = @{
"aaa"=1
}

$x = "qqq"
$y = "rrr"

$table.GetEnumerator() | ForEach-Object { 
Write-Host $_.Key$x #THIS PRINTS "aaa qqq"
}


Write-Host $x$y #THIS PRINTS : "qqqrrr"
</code></pre>

Why is there trailing whitespaces on hashtable keys?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我不明白为什么这个脚本中的字符串连接会有所不同。我真的不明白空格是从哪里来的？你能告诉我出什么事了吗？$table = @{"aaa"=1}$x = "qqq"$y = "rrr"$table.GetEnumerator() | ForEach-Object { Write-Host $_.Key$x         ...

问为什么哈希表的键上有尾随的空格？
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问为什么哈希表的键上有尾随的空格？EN