car*_*ier 2 bytearray response soapui
我正在使用soapUI测试Web服务.其中一个响应如下:
<S:Envelope xmlns:S="http://schemas.xmlsoap.org/soap/envelope/">
<S:Body>
<ns2:getInputConfigFilesResponse xmlns:ns2="http://ws.pdb.ericsson.se/">
<return>UEsDBBQACAAIAO1GNToAAAAAAAAAAAAAAAANAAAAc2NyaXB0cy9lbGxza [...] AATAAAAAAAAAAAAAAAAAAGXAAAbGRhcF9ub2RlX2NvbmZpZ3VyYXRpb24vZ2VuZXJhdGVkX2xkaWZfZmlsZXMvX210YXM0X2N4cDQ0NF9yNF9JbXNtb2JpbGUubGRpZlBLAQIUABQACAAIAO1GNTp8eBuZRAEAABMDAAAmAAAAAAAAAAAAAAAAAKJiAABsZGFwX25vZGVfY29uZmlndXJhdGlvbi9lbGxzaC1jb21tYW5kc1BLBQYAAAAABQAFAIgBAAA6ZAAAAAA=</return>
</ns2:getInputConfigFilesResponse>
</S:Body>
</S:Envelope>
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响应是一个字节数组,应该代表一个zip文件.我怎样才能验证这个?
虽然在语义上是正确的,但我相信内森提出的解决方案有点不完整.
我猜他忘了在第1步之前对Base64解码提取的文本.我还会包含一项改进,以便通过使用第三方ZIP实用程序验证临时文件不需要人工干预.
这将转化为访问临时ZIP文件并检查它是否是有效的ZIP文件.因此,实现您的要求的算法将是:
考虑到所有这些,这里是完整的Groovy脚本的样子:
import org.apache.commons.codec.binary.Base64
// Step 1: Access element Base64-encoded text content and Base64 decode it
String tempZipFilename = "temp.zip"
def textBase64 = context.expand(
'${Step#Request#//ns2:getInputConfigFilesResponse[1]/return[1]}' )
def b64 = new Base64()
def zipTextBytes = b64.decode(textBase64.getBytes())
// Step 2: Output ZIP raw text into a temporary file
def zipFile = new java.io.File(tempZipFilename)
FileOutputStream fos = new java.io.FileOutputStream(zipFile)
fos.write( zipTextBytes )
fos.flush()
fos.close()
log.info "Temporary ZIP file stored as: ${zipFile.getCanonicalPath()}"
// Step 3: Check if the temporarily created ZIP file is valid
boolean responseValid
try {
zipFile = new java.util.zip.ZipFile(tempZipFilename)
responseValid = true
log.info "Number of files in the ZIP file: ${zipFile.size()}"
} catch (java.util.zip.ZipException e) {
responseValid = false
log.error "The received response contains a bad ZIP"
}
log.info "Web service response is " + (responseValid ? "valid" : "invalid")
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如果这对你有用,请告诉我.:-)
干杯!
Shonzilla
ps我建议在这个问题上添加"ZIP"标签,以便人们可以更容易地找到一个解决方案来处理嵌入在这里的Groovy的ZIP.
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