英文:
Makefile. How do I find substring for given string and use it with if condition
问题
Case 1: "abc" in "abc" # True
Case 2: "ab" in "abc" # True
Case 3: "x" in "abc" # False
if [[ "abc" == *ab* ]]; then \
echo "YES"; \
else \
echo "NO"; \
fi;
YES
似乎 findstring 只匹配整个单词，而不是搜索子字符串（对于 Case 2 来说是不正确的）。有其他更简单的方法吗？
英文:
I'm trying to check if given string contains another. It should go like:
Case 1: "abc" in "abc" # True
Case 2: "ab" in "abc" # True
Case 3: "x" in "abc" # False
if test $(findstring "ab", "abc"); then \
echo "YES"; \
else \
echo "NO "; \
fi;
# NO
It seems that findstring is matching words only but not searching substrings (it is not correct for Case 2). Is there another simple way to do it?
答案1
得分: 1
你混合了make的语法和shell的语法。一般来说，你应该在规则（recipes）中使用shell的语法。
目标: 先决条件
case "abc" in *"ab"*) echo yes;; *) echo no;; esac
如果需要更多的变化，也可以参考 https://stackoverflow.com/questions/229551/how-to-check-if-a-string-contains-a-substring-in-bash
[，也就是test命令，有点脆弱。如果你知道你的shell实际上是Bash，你可以使用[[；但你必须另外告诉make这一点（SHELL := /bin/bash），这显然会降低你的代码的可移植性。
另一个目标: 其他先决条件
if [[ "abc" = *"ab"* ]]; then echo yes; else echo no; fi
# 如果命令很简单，还可以使用一个方便的缩写：
[[ "abc" = *"ab"* ]] && echo yes || echo no
你在使用findstring时的问题是"ab"实际上不是"abc"的子字符串（它是"ab"c的子字符串）。
英文:
You are mixing make syntax and shell syntax. Generally, you want to use shell syntax in recipes.
target: prerequisites
case "abc" in *"ab"*) echo yes;; *) echo no;; esac
For further variations, see also https://stackoverflow.com/questions/229551/how-to-check-if-a-string-contains-a-substring-in-bash
The [ aka test command is somewhat brittle. If you know your shell is actually Bash, you can use [[; but then you separately have to tell make that (SHELL := /bin/bash) and this obviously makes your code less portable.
othertarget: other prereqs
if [[ "abc" = *"ab"* ]]; then echo yes; else echo no; fi
# if the commands are simple, a convenient shorthand:
[[ "abc" = *"ab"* ]] && echo yes || echo no
Your problem with findstring is that "ab" really isn't a substring of "abc" (it would be a substring of "ab"c).
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