VSCode PHPUnit调试器不会跨入或进入代码

时间:2019-05-28 05:08:17

标签: php visual-studio-code phpunit xdebug laravel-5.8

我的环境

Version: 1.34.0 (system setup)
Commit: a622c65b2c713c890fcf4fbf07cf34049d5fe758
Date: 2019-05-15T21:59:37.030Z
Electron: 3.1.8
Chrome: 66.0.3359.181
Node.js: 10.2.0
V8: 6.6.346.32
OS: Windows_NT x64 10.0.17134

PHP 7.2.9 (cli) (built: Aug 15 2018 23:29:19) ( ZTS MSVC15 (Visual C++ 2017) x86 )
Copyright (c) 1997-2018 The PHP Group
Zend Engine v3.2.0, Copyright (c) 1998-2018 Zend Technologies
    with Xdebug v2.7.1, Copyright (c) 2002-2019, by Derick Rethans

Laravel Framework 5.8.16

PHP Debug VS Code Extension
Id: felixfbecker.php-debug
Description: Debug support for PHP with XDebug
Version: 1.13.0
Publisher: Felix Becker
VS Marketplace Link: https://marketplace.visualstudio.com/items?itemName=felixfbecker.php-debug

实际行为

我启动调试器并使用以下命令运行单元测试:

vendor/bin/phpunit.bat -c D:\Code\paylivre\backend\phpunit.noreport.xml --filter .*getUserDeposits*

设置了断点,调试器在该行停止。当我按F10(单步执行)或F11(单步执行),并且测试中的哪一行都没有关系时,调试器将跳到1199行上的vendor\phpunit\phpunit\src\Framework\TestCase.php

if ($this->expectedException !== null) {

并且即使测试通过了一些断言,也存在带有空输出或'Risky'结果的测试。

我已经重新启动VSCode,重新启动了PC,并且我无法再调试此PHPUnit测试。有任何提示和/或线索吗?

0 个答案:

没有答案