Applescript似乎没有正确地逃避字符串.我究竟做错了什么？
例：
set abc to "funky-!@#'#\"chars" display dialog abc display dialog quoted form of abc
预期/期望的产出：
funky-!@#'#"chars 'funky-!@#\'#"chars'
实际产量：
funky-!@#'#"chars 'funky-!@#'\''#"chars'
正如您所看到的,似乎在实际输出中Applescript正在添加并转发额外的’
我会很好,结尾字符是“或”,我也可以将单引号和双引号转义 – 但看起来只有单引号实际上是转义的.
反斜杠通常不会在shell中的单引号内解释.
原文链接：https://www.f2er.com/bash/387181.htmlEnclosing characters in single quotation marks preserves the literal value of each character within the single quotation marks. A single quotation mark cannot occur within single quotation marks.
A backslash cannot be used to escape a single quotation mark in a string that is set in single quotation marks. An embedded quotation mark can be created by writing,for example: ‘a’\”b’,which yields a’b.
然而,它们由echo in sh解释,这是do shell脚本使用的shell：
do shell script "echo " & quoted form of "\\t" --> "\t"
取消设置xpg_echo使其行为类似于bash中的echo：
do shell script "shopt -u xpg_echo; echo " & quoted form of "\\t" --> "\\t"
通常使用HEREDOC重定向更简单：
do shell script "rev <<< " & quoted form of "a\\tb" --> "b\\ta"
