我想在字符串中嵌入单引号 [英] I want to embed a single quote in a string

问题描述

通常，为了在字符串中插入引号，请使用 \ (反斜杠).

Usually, in order to embed a quotation mark in a string, \ (backslash) is used.

Bash脚本中的反斜杠含义不同吗?

Does a backslash have a different meaning in a Bash script?

我下面的代码不起作用:不包括引号，并且报告了以下错误:

My code below isn't working: the quotation mark wasn't included, and the following errors were reported:

recursive_merge.sh: line 7: unexpected EOF while looking for matching `''
recursive_merge.sh: line 14: syntax error: unexpected end of file

我没有解释.第14行甚至不存在.

I have no explanation. Line 14 doesn't even exist.

 #!/bin/bash
#############this file should be in the directory directly upper than p0x. sphnum.txt should also be at the same directory
for i in 02 03 04 05 06 07 09 10 11 12 13 14 15 16 17 20 21 22 23 24 25; do
 x=$(grep $i sphnum.txt |cut -c5-6)
 y=$(grep $i sphnum.txt |cut -c8-9)
 z=$(echo '\''$i'.ala.r'$x'.sph '$i'.ala.r'$y'.sph\'')
 w=$(echo $i'.ala.r'$x'r'$y'.sph')
 echo $z
 echo $w
 cd p$i/spheres.10_2_75/sph/
 /project/Biogroup/Software/Docking/MergeSpheres.pl -s $z -o $w -n 500 &
 cd ../../../
done

推荐答案

^{如 tripleee 指出的那样关于这个问题，在这种特定情况下，最佳方法是使用双引号字符串，在其中可以嵌入两个变量引用(例如， $ i )和单引号按原样" ；例如: z ='$ i.ala.r $ x.sph $ i .ala.r $ y.sph'"
此答案集中于生成/嵌入文字'字符的各种方法.字符串，从OP的误解开始.}

^{As tripleee points out in comments on the question, the best approach in this particular scenario is to use a double-quoted string, in which you can embed both variable references (e.g., $i) and single quotes as-is; e.g.: z="'$i.ala.r$x.sph $i .ala.r$y.sph'"
This answer focuses on the various approaches to producing / embedding literal ' chars. in strings, starting with the OP's misconception.}

您对'\''的使用表明您对通常用于将整体单引号字符串中的单引号嵌入"的变通方法感到困惑.>，这不是您的代码在 z = ... 行上执行的操作，因为它以'\''开始.

Your use of '\'' suggests that you're confused by the workaround that is commonly used to "embed" a single quote in an overall single-quoted string, which is not what your code does on the z=... line, because it starts with '\''.

如果我们简化您的命令，则会得到:

If we simplify your command, we get:

echo '\''$i

这是一个语法错误，因为对于Bash来说，单引号是不平衡，因为'\' 本身是被视为包含文字 \ 的完整单引号字符串，后跟 second 的 opening '单引号字符串，永远不会闭合.

which is a syntax error, because to Bash the single quotes are unbalanced, because '\' by itself is considered a complete single-quoted string containing literal \, followed by the opening ' of a second single-quoted string, which is never closed.

^{再次值得注意的是，'$ i" 是解决此特定问题的最佳解决方案:'可以原样嵌入，包括用双引号引起来的变量引用 $ i 可以保护其值免受潜在有害的}

^{Again it's worth noting that "'$i" is the best solution to this specific problem: the ' can be embedded as-is, and including variable reference $i inside the double-quoted string protects its value from potentially unwanted word-splitting and filename expansion (globbing).}

类似POSIX的外壳程序无法将单引号嵌入单引号内的字符串-甚至无法转义.因此，'\'中的 \ 仅被视为文字(参见以下解决方法).

POSIX-like shells provide NO way to embed single quotes inside a single-quoted string - not even with escaping. Hence, the \ in '\' is simply treated as a literal (see below for a workaround).

此答案的其余部分显示所有生成带字面的'的方法，包括用引号括起来的字符串的内部和外部.

The rest of this answer shows all approaches to producing a literal ', both inside and outside quoted strings.

要在带引号的字符串 外 中创建单引号，只需使用 \':

To create a single quotation mark outside of a quoted string, simply use \':

$ echo I am 6\' tall.
I am 6' tall.

这仅使用 \ 引用(转义)各个'字符.但是请注意，放置在命令行中单引号或双引号字符串的上下文之外的标记应受

This quotes (escapes) the individual ' character only, using \. But note that tokens placed outside the context of a single- or double-quoted string on a command line are subject to word-splitting and filename expansion (globbing).

要在双引号字符串中使用单引号，请按原样使用(无需转义):

To use a single quote inside a double-quoted string, use it as-is (no escaping needed):

$ echo "I am 6' tall."
I am 6' tall.

如果您还希望嵌入变量引用(例如， $ i )或命令(通过命令替换， $(...))，这是最佳选择.在您的字符串中(您可以通过将 $ 转义为 \ $ 来抑制插值).

This is the best choice if you also want to embed variable references (e.g., $i) or commands (via command substitutions, $(...)) in your string (you can suppress interpolation by escaping $ as \$).

要在单引号字符串中使用单引号(其中， no 插值(扩展)是按设计执行的)，必须使用解决方法:

To use a single quote inside a single-quoted string (in which no interpolations (expansions) are performed by design), you must use a workaround:

$ echo 'I am 6'\'' tall.'
I am 6' tall.

要解决此问题，是因为单引号的字符串根本不支持嵌入的单引号 ；'\''部分仅在用单引号括起来的字符串内"有意义:

The workaround is necessitated by single-quoted strings not supporting embedded single quotes at all; the '\'' part only makes sense "inside" a single-quoted string in that:

• 前导'终止到目前为止用单引号引起来的字符串
然后，
• \'会生成一个'文字，该文字用 \ outside 引号字符串的上下文分别转义
• 结尾的'然后重新启动"单引号字符串的其余部分.

• the leading ' terminates the single-quoted string so far
• the \' then produces a ' literal individually escaped with \ outside the context of a quoted string.
• the trailing ' then "restarts" the remainder of the single-quoted string.

换句话说:虽然您不能直接嵌入单引号，但是可以将单引号的字符串分成多段，分别插入 \ -escaped '实例 outside 根据需要将单引号字符串包括在内，然后让Bash的字符串串联(自动将直接相邻的字符串连接在一起)将它们全部拼凑在一起以形成单个字符串.

In other words: While you cannot directly embed a single quote, you can break the single-quoted string into multiple pieces, insert individually \-escaped ' instances outside the single-quoted string as needed, and let Bash's string concatenation (which automatically joins directly adjacent string) piece it all back together to form a single string.

chepner 在注释中指出，您可以替代地使用

chepner points out in a comment that you can alternatively use a here-document with a quoted opening delimiter, which acts like a single-quoted string while allowing embedding of ' chars:

read -r var <<'EOF' # quoted delimiter -> like a '...' string, but ' can be embedded
I am 6' tall.
EOF

使用 未加引号的开头定界符，此处文档的作用类似于双引号字符串，就像后者一样，允许嵌入'，同时还支持扩展:

With an unquoted opening delimiter, the here-document acts like a double-quoted string, which, just like the latter, also allows embedding ', while also supporting expansions:

read -r var <<EOF # unquoted delimiter -> like a "..." string
$USER is 6' tall.
EOF

---

最后，如果不是必须保持POSIX兼容，则可以使用

---

Finally, if remaining POSIX-compliant is not a must, you can use an ANSI C-quoted string string, which allows embedding single quotes with \';
note that such strings interpret control-character escape sequences such as \n, but otherwise, like a normal single-quoted string, do not perform interpolation of variable references or command substitutions:

$ echo $'I am 6\' tall.'
I am 6' tall.

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