如何在Windows Server上与phpdbg一起使用PHP代码覆盖率？

Question

由于我仍然对PHPUnit / PHP代码覆盖率和Xdebug感到烦恼，因此我决定尝试另一种方法-使用phpdbg。

我这样做是因为票数所示。在CMD和Git Bash中尝试过，但结果相同，但失败了：

$ composer info | grep "phpunit"
phpunit/php-code-coverage           4.0.0  Library that provides collectio...
phpunit/php-file-iterator           1.4.1  FilterIterator implementation t...
phpunit/php-text-template           1.2.1  Simple template engine.
phpunit/php-timer                   1.0.8  Utility class for timing
phpunit/php-token-stream            1.4.8  Wrapper around PHP's tokenizer ...
phpunit/phpunit                     5.4.6  The PHP Unit Testing framework.
phpunit/phpunit-mock-objects        3.2.3  Mock Object library for PHPUnit

$ phpdbg -qrr ./vendor/bin/phpunit -v

dir=$(d=${0%[/\\]*}; cd "$d"; cd "../phpunit/phpunit" && pwd)

# See if we are running in Cygwin by checking for cygpath program
if command -v 'cygpath' >/dev/null 2>&1; then
        # Cygwin paths start with /cygdrive/ which will break windows PHP,
        # so we need to translate the dir path to windows format. However
        # we could be using cygwin PHP which does not require this, so we
        # test if the path to PHP starts with /cygdrive/ rather than /usr/bin
        if [[ $(which php) == /cygdrive/* ]]; then
                dir=$(cygpath -m "$dir");
        fi
fi

dir=$(echo $dir | sed 's/ /\ /g')
"${dir}/phpunit" "$@"

如何在Windows Server上使用组合PHPUnit和PHP代码覆盖率和phpdbg？

Answer 1

在Windows下，似乎其中的文件vendor/bin实际上是批处理文件，调用原始文件（而不是phpdbg可以理解的php文件）。

在这种情况下：

dir=$(d=${0%[/\\]*}; cd "$d"; cd "../phpunit/phpunit" && pwd)
                                  ^^^^^^^^^^^^^^^^^^
"${dir}/phpunit" "$@"
 ^^^^^^^^^^^^^^

即../phpunit/phpunit/phpunit（此路径相对于vendor/bin）；因此实际文件位于vendor/phpunit/phpunit/phpunit。

然后，您可以直接通过它调用它phpdbg -qrr vendor/phpunit/phpunit/phpunit。