Question

当我单独运行$ phpunit时，它会无声地失败，这就是我试图知道问题是什么而没有运气：

$ phpunit
$ echo $?
=> 255

$ php -i | grep error

display_errors => STDOUT => STDOUT
display_startup_errors => On => On
error_append_string => no value => no value
error_log => /usr/local/var/log/php_error.log => /usr/local/var/log/php_error.log
error_prepend_string => no value => no value
error_reporting => 32767 => 32767
html_errors => Off => Off
ignore_repeated_errors => Off => Off
log_errors => On => On
log_errors_max_len => 1024 => 1024
track_errors => On => On
xmlrpc_error_number => 0 => 0
xmlrpc_errors => Off => Off
xdebug.force_display_errors => On => On
xdebug.force_error_reporting => 1 => 1
xdebug.show_error_trace => Off => Off
opcache.error_log => no value => no value


phpunit --version
PHPUnit 6.0-g969991b by Sebastian Bergmann and contributors.


php --version
PHP 7.0.10-1 (cli) ( NTS )
Copyright (c) 1997-2016 The PHP Group
Zend Engine v3.0.0, Copyright (c) 1998-2016 Zend Technologies
    with Zend OPcache v7.0.10-1, Copyright (c) 1999-2016, by Zend Technologies
    with Xdebug v2.4.0, Copyright (c) 2002-2016, by Derick Rethans

检查语法错误，返回一切正常：

for f in $(find tests/)
> do php -l $f
> done
...

作为临时解决方法，我按照以下方式运行所有测试

find . -name "*_Test.php" -type f -exec phpunit {} \;

php_error.log没有日志条目，并确保文件上没有任何内容写入：

$ truncate -s0 /usr/local/var/log/php_error.log
$ phpunit
$ tail /usr/local/var/log/php_error.log

我确实有写日志文件的写权限： -rw-rw-r-- 1 adam adam 0 Sep 2 11:49 /usr/local/var/log/php_error.log

并且安全：

$ php -r "error_log('am i logged?');"
$ tail /usr/local/var/log/php_error.log
[02-Sep-2016 13:16:37 Europe/Paris] am i logged?

当我这样做的时候 phpunit tests/* 只有根目录测试上的文件正在运行，phpunit不会递归递送

这是phpunit.xml

<?xml version="1.0" encoding="UTF-8"?>
<phpunit backupGlobals="false"
         backupStaticAttributes="false"
         colors="true"
         convertErrorsToExceptions="true"
         convertNoticesToExceptions="true"
         bootstrap="bootstrap.php"
         convertWarningsToExceptions="true"
         processIsolation="false"
         stopOnFailure="false">
    <testsuites>
        <testsuite name="Application Test Suite">
            <directory suffix=".php">./tests</directory>
        </testsuite>
    </testsuites>
</phpunit>

这是bootstrap.php文件的内容

<?php

//
// Author: Oussama Elgoumri
// Email : ktsnepyg9igfz1@gmail.com
//
// Wed Aug 31 17:13:04 WEST 2016
//


require_once __DIR__ . '/vendor/autoload.php';
require_once dirname(dirname(__FILE__)) . '/wp-load.php';
require_once ABSPATH . 'wp-admin/includes/admin.php';


use Dotenv\Dotenv;


do_action('admin_init');

(new Dotenv(__DIR__))
    ->load();

我知道没有错误，这很奇怪，echo $?在运行phpunit后返回255，因为当我运行时

psysh bootstrap.php没有错误

我会感激任何帮助，即使它是hacki，我只是想进行测试覆盖，并生成报告。谢谢

更新

作为临时workarround我使用此命令运行所有测试： find . -name "*_Test.php" -type f -exec phpunit {} \;

测试单个文件

$ phpunit tests/Upload__FN_Test.php
PHPUnit 5.5.4 by Sebastian Bergmann and contributors.

.                                                                   1 / 1 (100%)

Time: 1.27 seconds, Memory: 24.00MB

OK (1 test, 1 assertion)

测试目录中的所有文件 该目录应 NOT 包含子目录，否则将失败，如上所述，错误为255

$ phpunit tests/src/Uptobox/Workflow_PHP/
PHPUnit 5.5.4 by Sebastian Bergmann and contributors.

.....                                                               5 / 5 (100%)

Time: 7.03 seconds, Memory: 26.00MB

OK (5 tests, 6 assertions)

测试具有测试覆盖率的目录中的文件

$ phpunit tests/src/Uptobox/Workflow_PHP/ --coverage-text --whitelist src/Uptobox/Uptobox.php
PHPUnit 5.5.4 by Sebastian Bergmann and contributors.

.....                                                               5 / 5 (100%)

Time: 6.93 seconds, Memory: 28.00MB

OK (5 tests, 6 assertions)


Code Coverage Report:    
  2016-09-02 12:35:00    

 Summary:                
  Classes:  0.00% (0/1)  
  Methods: 75.00% (3/4)  
  Lines:   90.91% (10/11)

\App\Uptobox::Uptobox
  Methods:  75.00% ( 3/ 4)   Lines:  90.91% ( 10/ 11)

到目前为止我注意到只要目标目录不包含任何子目录，phpunit就会处理该目录中的所有测试文件，否则它会无声地失败。

Answer 1

您的目录后缀可能不对。

这里：您尝试将每个php文件作为测试执行：

<testsuites>
    <testsuite name="Application Test Suite">
        <directory suffix=".php">./tests</directory>
    </testsuite>
</testsuites>

但是你在这里限制它为“* _Test.php”

find . -name "*_Test.php" -type f -exec phpunit {} \;

尝试：

<testsuites>
    <testsuite name="Application Test Suite">
        <directory suffix="Test.php">./tests</directory>
    </testsuite>
</testsuites>

PHPUnit失败了

1 个答案: