java - 有没有办法获取xml文件中的Bean ID

Question

我正在研究 Selenium 自动化。我的团队使用 xml 文件作为测试数据。 xml 文件中的测试数据将作为 beans 和属性。

每个 bean 将是一个测试用例的数据。

现在，我们使用每个 bean 创建每个 @Test 方法，如下所示:

ApplicationContext ctx = new ClassPathXmlApplicationContext("/PostPaidRegistration.xml");
        RegistrationBean user = (RegistrationBean)ctx.getBean("postpaidTC1");

现在在测试用例中我们将使用“postpaidTC1”bean 的数据。

对于另一个测试用例，我们编写另一个 @Test 方法

ApplicationContext ctx = new ClassPathXmlApplicationContext("/PostPaidRegistration.xml");
        RegistrationBean user = (RegistrationBean)ctx.getBean("postpaidTC2");

我想知道，不是为 xml 中的 bean 创建许多 @Test 方法，而是通过获取 BeanID 来使用单个 @Test 方法作为循环来运行 xml 中的所有 bean(测试数据)。

我的Xml如下:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:mongo="http://www.springframework.org/schema/data/mongo"
    xsi:schemaLocation="http://www.springframework.org/schema/beans 
    http://www.springframework.org/schema/beans/spring-beans.xsd
    http://www.springframework.org/schema/data/mongo 
    http://www.springframework.org/schema/data/mongo/spring-mongo-1.0.xsd
    http://www.springframework.org/schema/data/repository
    http://www.springframework.org/schema/data/repository/spring-repository-1.0.xsd
    http://www.springframework.org/schema/data/jpa 
   http://www.springframework.org/schema/data/jpa/spring-jpa-1.0.xsd">
<bean id="postpaidTC1" class="com.tecnotree.CLM.beans.RegistrationBean">

    <property name="service" value="GSM SERVICE" />
    <property name="subService" value="Voice"/>
    <property name="technology" value="gsm"/>
    <property name="plan" value="POSTPAID ADV REN"/>
    <property name="mCategory" value="POSTPAID NORMAL 2"/>
    <property name="mSelection" value="Automatic"/>
    <property name="firstName" value="Sarada"/>
    <property name="middleName" value="M" />
    <property name="lastName" value="Akurathi"/>
    <property name ="dob" value="19/02/1989"/>
    <property name ="prefLang" value="ENGLISH"/>
    <property name ="prefMedium" value="Email"/>
    <property name ="prefMediumValue" value="samyak@tt.com"/>
    <property name ="streetName" value="Gandhi Nagar"/>
    <property name ="buildingNumber" value="12"/>
    <property name ="docPurpose" value="Proof Identity"/>
    <property name ="docType" value="PANCARD"/>
    <property name ="docId" value="VETY3D343"/>
    <property name ="docPlaceOfIssue" value="Chennai"/>
    <property name ="billCycle" value="5th TO 4th MONTHLY BILL CYCLE"/>
</bean>
<bean id="postpaidTC2" class="com.tecnotree.CLM.beans.RegistrationBean">

    <property name="firstName" value="Suneetha"/>
    <property name="lastName" value="ss"/>
    <property name ="dob" value="19/02/1989"/>
    <property name ="prefLang" value="ENGLISH"/>
    <property name ="prefMedium" value="Email"/>
    <property name ="prefMediumValue" value="samyak@tt.com"/>
    <property name="plan" value="POSTPAID ADV REN"/>

</bean>

</beans>

Answer 1

也许你可以做到

Map<String, RegistrationBean> beanMap = ctx.getBeansOfType(RegistrationBean.class)

然后迭代 beanMap