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General Tech Bugs & Fixes 2 years ago
Posted on 16 Aug 2022, this text provides information on Bugs & Fixes related to General Tech. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.
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I have PHP file which has logic based on which I run either
bin/phpunit -c Application/app --testsuite $TESTSUITE
or
bin/phpunit -c Application/app --group group1 group2
when Im doing
#!/bin/bash COMMAND=$(php travis_groups.php) echo $COMMAND //bin/phpunit -c Application/app --testsuite $TESTSUITE $COMMAND
Nothing happens but when I paste the command to sh file it works.
Why I can't I execute a command which is generated by a PHP script?
Edit:
Solution
eval $COMMAND
By COMMAND=$(php travis_groups.php) you just assign the command to COMMAND variable. If you want to run the command, you need to add $COMMAND after the assignment. Simple example setup:
COMMAND=$(php travis_groups.php)
COMMAND
$COMMAND
echo.sh file containing just line echo "ls"
echo.sh
echo "ls"
run_echoed.sh file containing two lines:
run_echoed.sh
COMMAND=$(bash echo.sh) $COMMAND
When you run run_echoed.sh (either by ./run_echoed.sh if the file is executable or bash run_echoed.sh), you print the directory contents (the ls command gets executed).
./run_echoed.sh
bash run_echoed.sh
ls
Of course make sure your php travis_groups.php actually prints out the commands (just run in the terminal). Otherwise you can't tell the other script what to assign to the variable.
php travis_groups.php
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