Xel*_*ian 4 linux bash wrapper gradle
我有一个具有这种结构的Gradle项目:
ls -alh
drwxr-xr-x 9 Xelian 4.0K Aug 5 22:39 .
drwxrwxr-x 3 Xelian 4.0K Aug 5 16:19 ..
-rw-rw-r-- 1 Xelian 465 Aug 5 16:19 build.gradle
drwxrwxr-x 8 Xelian 4.0K Aug 5 16:38 buildSrc
drwxrwxr-x 3 Xelian 4.0K Aug 5 16:19 gradle
drwxrwxr-x 3 Xelian 4.0K Aug 5 16:45 .gradle
-rwxrwxrwx 1 Xelian 5.0K Aug 5 17:29 gradlew
-rw-rw-r-- 1 Xelian 2.3K Aug 5 16:19 gradlew.bat
-rw-rw-r-- 1 Xelian 17K Aug 5 16:19 README.md
-rw-rw-r-- 1 Xelian 30 Aug 5 16:19 settings.gradle
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为了执行gradlew bash文件,我写道:
gradlew -v
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输出:
No command 'gradlew' found, did you mean:
Command 'gradle' from package 'gradle' (universe)
gradlew: command not found
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如果我写:
./gradlew -v
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一切正常.
./表示当前目录吗?
键入时:
command
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您shell将按$PATH顺序搜索环境变量中列出的所有目录,以查找名为" command " 的可执行文件.
键入时:
./command
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该./部分明确告诉您的shell要在当前目录中执行名为" command "的文件(.被解释为当前目录).
如果您希望能够仅使用它来运行它command,则必须将当前目录添加到您的目录中PATH,或者将可执行文件移动到您已经存在的位置.PATH