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Java TestResult.errors方法代码示例

本文整理汇总了Java中junit.framework.TestResult.errors方法的典型用法代码示例。如果您正苦于以下问题:Java TestResult.errors方法的具体用法?Java TestResult.errors怎么用?Java TestResult.errors使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在junit.framework.TestResult的用法示例。


在下文中一共展示了TestResult.errors方法的2个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Java代码示例。

示例1: run

import junit.framework.TestResult; //导入方法依赖的package包/类
public void run(TestResult result) {
    if (!canRun()) {
        return;
    }
    this.main = Thread.currentThread();
    
    TestResult mine = new TestResult();
    result.startTest(this);
    super.run(mine);
    
    if (mine.errorCount() != 0) {
        Enumeration en = mine.errors();
        while(en.hasMoreElements()) {
            TestFailure f = (TestFailure)en.nextElement();
            result.addError(this, f.thrownException());
        }
        return;
    }
    if (expectedResult != (mine.failureCount() == 0)) {
        result.addFailure(this, 
            new AssertionFailedError(
                "expectedResult: " + expectedResult + "failureCount: " + mine.failureCount() + " for " + getName()
            )
        );
        return;
    }
    
    result.endTest(this);
}
 
开发者ID:apache,项目名称:incubator-netbeans,代码行数:30,代码来源:TimeOutTest.java

示例2: executeTest

import junit.framework.TestResult; //导入方法依赖的package包/类
protected void executeTest(Class test, Permission missingPermission) {
    TestSuite suite = new TestSuite();
    suite.addTestSuite(test);
    TestResult result = new TestResult();
    suite.run(result);
    if (result.wasSuccessful()) {
        if (missingPermission == null) {
            return;
        } else {
            fail("Security test expected an AccessControlException on " + missingPermission + ", but did not receive one");
        }
    } else {
        if (missingPermission == null) {
            new SecurityTestResultPrinter(System.out).print(result);
            fail("Security test was expected to run successfully, but failed (results on System.out)");
        } else {
            //There may be more than 1 failure:  iterate to ensure that they all match the missingPermission.
            boolean otherFailure = false;
            for (Enumeration e = result.errors(); e.hasMoreElements();) {
                TestFailure failure = (TestFailure) e.nextElement();
                if (failure.thrownException() instanceof AccessControlException) {
                    AccessControlException ace = (AccessControlException) failure.thrownException();
                    if (missingPermission.implies(ace.getPermission())) {
                        continue;
                    }
                }
                otherFailure = true;
                break;
            }
            if (otherFailure) {
                new SecurityTestResultPrinter(System.out).print(result);
                fail("Security test expected an AccessControlException on " + missingPermission + ", but failed for other reasons (results on System.out)");
            }
        }
    }
}
 
开发者ID:apache,项目名称:groovy,代码行数:37,代码来源:SecurityTestSupport.java


注:本文中的junit.framework.TestResult.errors方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。