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问题 git diff numbers diff --git a/numbers b/numbers index 5f5fbe7..d184fef 100644 --- a/numbers +++ b/numbers @@ -1,3 +1,3 @@ -1 +4 +3 2 -3 Number 3 is repeated but the order is changed. Any way to ignore the reorder in git or any grep solution? I want the result of only added and deleted numbers, not reordering of the same numbers Any help? 回答1: The diffing tools are usually implemented in terms of the Myers' diff algorithm. There isn't much that you can do to control the behaviour of GNU/git

问题 Currently I am writing a script that invokes a new instance of the chrome browser. I know how to call chrome to open a *.html document in a new tab. google-chrome *.html Chrome will open a new tab to show that file. How can I close the tab in terminal without closing other tabs or closing the browser window? 回答1: only linux answer: Perhaps wmctrl could be of some assistance. You could use the -c option that closes a window gracefully: wmctrl -c "tab title" The string chrome is matched against

问题 This question already has answers here : What is indirect expansion? What does ${!var*} mean? (5 answers) Closed 2 years ago . Let var be a variable and it assigned a value /home/user as below var=/home/user when using this variable, i have seen it using both of the below format, 1) cd ${var} 2) cd ${!var} what is the difference? for me second option is not working , if i echo second option returns empty. 回答1: In this case, it's indirect expansion (a) , the var variable is expanded to create

问题 This question already has answers here : What is indirect expansion? What does ${!var*} mean? (5 answers) Closed 2 years ago . Let var be a variable and it assigned a value /home/user as below var=/home/user when using this variable, i have seen it using both of the below format, 1) cd ${var} 2) cd ${!var} what is the difference? for me second option is not working , if i echo second option returns empty. 回答1: In this case, it's indirect expansion (a) , the var variable is expanded to create

问题 This question already has answers here : What is indirect expansion? What does ${!var*} mean? (5 answers) Closed 2 years ago . Let var be a variable and it assigned a value /home/user as below var=/home/user when using this variable, i have seen it using both of the below format, 1) cd ${var} 2) cd ${!var} what is the difference? for me second option is not working , if i echo second option returns empty. 回答1: In this case, it's indirect expansion (a) , the var variable is expanded to create

问题 What is current directory of shell script? I this current directory from which I called it? Or this directory where script located? 回答1: The current(initial) directory of shell script is the directory from which you have called the script. 回答2: As already mentioned, the location will be where the script was called from. If you wish to have the script reference it's install location, it's quite simple. Below is a snippet that will print the PWD and the installed directory #!/bin/bash echo

问题 I have a shell script (called test.sh) which is called from PHP. Within the script I simply have: #!/bin/bash echo $(whoami) cordova platform version ios If I call test.sh from within terminal it works fine and returns the cordova ios version. If I try to call test.sh from with PHP I get: cordova: not found I have altered apache to run under my username instead of _www but that hasnt worked. Can anyone point me in the right direction as I'm guessing it is a permissions issue? I have now

问题 I have a shell script (called test.sh) which is called from PHP. Within the script I simply have: #!/bin/bash echo $(whoami) cordova platform version ios If I call test.sh from within terminal it works fine and returns the cordova ios version. If I try to call test.sh from with PHP I get: cordova: not found I have altered apache to run under my username instead of _www but that hasnt worked. Can anyone point me in the right direction as I'm guessing it is a permissions issue? I have now

问题 I want to do something like this in a shell script main.sh. Is it possible? set -e ******some code unset -e ******some other code. set -e 回答1: @Charles Duffy on another thread answered this. not exit a bash script when one of the sub-script fails set +e undoes set -e. However, using set -e is a bad idea in general; better to use || exit on individual commands where you want a nonzero exit status to be fatal. (Skip past the parable to the exercises if you're in a hurry). – Charles Duffy 来源：

问题 I am new to shell script. I have a script that runs several test scripts using different input files. Currently it exits if any one of the input test fails. I want the test to complete running the loop and exits with all errors accumulated at last. main.sh set -e ; set -x ; for f in $files;do ./scripts/test_script.sh $f done ====================== test_script.sh : runs few stuff and exits like this. : : : exit $? ================ 回答1: set -e is what causes your script to exit immediately