Question

我需要根据先前请求的响应创建请求。响应的格式如下。基本上，响应可以包含任意数量的组，每组可以包含任意数量的问题 示例回复

<Group>
  <Question>
    <ID>1234</ID>
    <Row>2</Row>
    <Code>1-6666</Code>
    <Text>my text</Text>
  </Question>
  <Question>
    <ID>2222</ID>
    <Row>3</Row>
    <Code>1-111</Code>
    <Text>my text</Text>
  </Question>
</Group>
<Group>
  <Question>
    <ID>4244</ID>
    <Row>0</Row>
    <Code>1-8888</Code>
    <Text>my textfgdfgd</Text>
  </Question>
</Group>

示例请求对于每个组和问题，我需要在请求中包含一些数据

<Header Stuff>
 <UpdateTargets>
   <Group>
     <Question>
       <ID>1234</ID>
       <Row>2</Row>
       <NewValue>my updated value</NewValue>
     </Question>

...对于响应中出现的每个小组和问题

我不知道该怎么做。我假设某种时髦的剧本。

Answer 1

添加一个groovy脚本testStep，在其中你可以使用XmlSlurper来读取先前响应的内容，并使用groovy.xml.MarkupBuilder从先前的数据创建新请求。请参阅以下代码，我希望它是自我解释的：

// this is your xml hardcoded for the sample, but you can
// get it using the follow line
// def xml = context.testCase.getTestStepByName('SOAP Request').getPropertyValue('response')
def xml = '''<root><Group>
  <Question>
    <ID>1234</ID>
    <Row>2</Row>
    <Code>1-6666</Code>
    <Text>my text</Text>
  </Question>
  <Question>
    <ID>2222</ID>
    <Row>3</Row>
    <Code>1-111</Code>
    <Text>my text</Text>
  </Question>
</Group>
<Group>
  <Question>
    <ID>4244</ID>
    <Row>0</Row>
    <Code>1-8888</Code>
    <Text>my textfgdfgd</Text>
  </Question>
</Group></root>
'''

// parse the response from the previous testStep
def root = new XmlSlurper().parseText(xml)

// markup to create the request
def sw = new StringWriter()
def request = new groovy.xml.MarkupBuilder(sw)
// create <UpdateTargets>
request.UpdateTargets {
    // for each <Group> in response create a group in next
    // request
    root.Group.each{ group ->
        Group {
            // for each <Question> in response create a question in next
            // request
            group.Question.each { question ->
                Question(){
                    ID(question.ID)
                    Row(question.Row)
                    NewValue('yourNewValue')
                }
            }
        }   
    }
}

log.info sw

此代码将您的新请求记录为：

<UpdateTargets>
  <Group>
    <Question>
      <ID>1234</ID>
      <Row>2</Row>
      <NewValue>yourNewValue</NewValue>
    </Question>
    <Question>
      <ID>2222</ID>
      <Row>3</Row>
      <NewValue>yourNewValue</NewValue>
    </Question>
  </Group>
  <Group>
    <Question>
      <ID>4244</ID>
      <Row>0</Row>
      <NewValue>yourNewValue</NewValue>
    </Question>
  </Group>
</UpdateTargets>

基于OP评论的编辑

如果您的请求中有静态部分，则可以包含使用groovy.xml.MarkupBuilder构建的dinamyc部分，如下所示：

... from previous script
// markup to create the request
def sw = new StringWriter()
def request = new groovy.xml.MarkupBuilder(sw)
// create <UpdateTargets>
...

// static data sample 
def staticData = 
'''<Envelope>
    <Header>someData</Header>
    <Body>
        <SomeTags>moreData</SomeTags>
    </Body>
</Envelope>'''
def newRequest = new XmlSlurper().parseText(staticData)
// dinamic part builded with markupBuilder
def dinamicPart = new XmlSlurper().parseText(sw.toString())
// append the <UpdateTargets> to your static part
newRequest.Body.appendNode(dinamicPart)
// serialize the full xml
log.info  groovy.xml.XmlUtil.serialize( newRequest )

这将输出：

<Envelope>
   <Header>someData</Header>
   <Body>
      <SomeTags>moreData</SomeTags>
      <UpdateTargets>
         <Group>
            <Question>
               <ID>1234</ID>
               <Row>2</Row>
               <NewValue>yourNewValue</NewValue>
            </Question>
            <Question>
               <ID>2222</ID>
               <Row>3</Row>
               <NewValue>yourNewValue</NewValue>
            </Question>
         </Group>
         <Group>
            <Question>
               <ID>4244</ID>
               <Row>0</Row>
               <NewValue>yourNewValue</NewValue>
            </Question>
         </Group>
      </UpdateTargets>
   </Body>
</Envelope>

希望它有所帮助，

SoapUI - 动态创建请求

1 个答案: