Php Behat无法识别已定义的步骤
我对贝哈特和貂皮有意见。运行时，系统会提示我在函数声明中添加#2 这是我的作曲家文件的版本信息Php Behat无法识别已定义的步骤,php,behat,mink,Php,Behat,Mink,我对贝哈特和貂皮有意见。运行时，系统会提示我在函数声明中添加#2 这是我的作曲家文件的版本信息 { "require": { "behat/behat": "2.5.*@stable", "behat/mink": "~1.6", "behat/mink-extension": "~1.0", "behat/mink-goutte-driver": "~1.1", "fabpot/goutte": "~1.0.4",
{
"require": {
"behat/behat": "2.5.*@stable",
"behat/mink": "~1.6",
"behat/mink-extension": "~1.0",
"behat/mink-goutte-driver": "~1.1",
"fabpot/goutte": "~1.0.4",
"behat/mink-selenium2-driver": "*"
},
"config": {
"bin-dir": "bin/"
}
}
Given the user "XXX" has logged in using the password "YYYYY"
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
public function theUserHasLoggedInUsingThePassword($arg1, $arg2)
{
...
}
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
public function theUserHasLoggedInUsingThePassword2($arg1, $arg2)
{
$this->theUserHasLoggedInUsingThePassword($arg1,$arg2);
}
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
public function theUserHasLoggedInUsingThePassword2($arg1, $arg2)
{
throw new PendingException();
}
有人看到我遗漏了什么吗？去掉第一个片段，使用第二个片段，抛出异常，因为正则表达式定义了两次。方法名并不重要。但是，第一次不起作用是没有意义的…去掉第一个代码段，使用第二个，抛出异常是因为正则表达式定义了两次。方法名并不重要。但是，它第一次不起作用是没有意义的…从头开始，然后执行以下操作，看看会发生什么： 缓存
app/console cache:clear
app/console cache:clear --env=test (I guess you have test environment!)
"require-dev": {
"behat/behat": "2.5.5",
"behat/mink-extension": "1.3.3",
"behat/mink": "1.5.0",
"behat/symfony2-extension": "1.1.2",
"behat/mink-selenium2-driver": "1.1.1",
"behat/mink-browserkit-driver": "1.1.0",
"behat/mink-goutte-driver": "1.0.9",
"guzzle/guzzle": "3.9.2"
}
When test login "hello" "word"
class FeatureContext extends MinkContext implements KernelAwareInterface
{
/**
* @When /^test login "([^"]*)" "([^"]*)"$/
*/
public function testLogin($uid, $psw)
{
echo 'Step works fine';
}
}
从零开始，执行以下操作以查看发生了什么： 缓存
app/console cache:clear
app/console cache:clear --env=test (I guess you have test environment!)
"require-dev": {
"behat/behat": "2.5.5",
"behat/mink-extension": "1.3.3",
"behat/mink": "1.5.0",
"behat/symfony2-extension": "1.1.2",
"behat/mink-selenium2-driver": "1.1.1",
"behat/mink-browserkit-driver": "1.1.0",
"behat/mink-goutte-driver": "1.0.9",
"guzzle/guzzle": "3.9.2"
}
When test login "hello" "word"
class FeatureContext extends MinkContext implements KernelAwareInterface
{
/**
* @When /^test login "([^"]*)" "([^"]*)"$/
*/
public function testLogin($uid, $psw)
{
echo 'Step works fine';
}
}
经过数小时的探索，我发现这只是一个简单的打字/剪切和粘贴问题 所以问题实际上在于正则表达式语法
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"]
*)"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"]
*)"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"]*)"$/
*/
请注意，唯一的区别是密码“（[^”]*）”$/中的*经过几个小时的摸索，我发现这是一个简单的打字/剪切和粘贴问题 所以问题实际上在于正则表达式语法
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"]
*)"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"]
*)"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"])"$/
*/
/**
* @Given /^the user "([^"]*)" has logged in using the password "([^"]*)"$/
*/
请注意，唯一的区别是密码“（[^”]*）”$/中的*是否仅将
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