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执行下面的查询xml auto或xml路径?
SELECT AssessmentFileGUID as '@FileId',AssessmentFileTypeId as '@FileTypeId',AssessmentFileName
FROM rat.AssessmentFile where AssessmentId=17
for xml path('File'),root('Files')
给了我下面的结果:
<Files>
<File FileId="2A23D836-612F-418E-BDE8-F182C5432A0D" FileTypeId="1">
<AssessmentFileName>File-123213.pdf</AssessmentFileName>
</File>
<File FileId="CDA853B9-C587-4365-BAF5-972F8D217BAC" FileTypeId="2">
<AssessmentFileName>File-343455.png</AssessmentFileName>
</File>
</Files>
不过,我需要它的格式如下:
<Files>
<File FileId="2A23D836-612F-418E-BDE8-F182C5432A0D" FileTypeId="1">File-123213.pdf</File>
<File FileId="CDA853B9-C587-4365-BAF5-972F8D217BAC" FileTypeId="2">File-343455.png</File>
</Files>
,让我 <文件的fileid =” 2A23D836-612F-418E-BDE8-F182C5432A0D“FileTypeId =”1“> 欢迎使用HDFC Bank NetBanking-4-12-12.pdf <文件的fileid = “CDA853B9-C587-4365-BAF5-972F8D217BAC” FileTypeId = “2”> Ipad_safari,文件上传,issue.png –
对不起比拉尔了错误的答案,我不确定一下,但是你可以试试这个 SELECT AssessmentFileGUID as'File/@ FileId',AssessmentFileTypeId as'File/@ FileTypeId',AssessmentFileName as'File/text()“ FROM rat.AssessmentFile其中AssessmentId = 17 for xml路径('文件') – Manibhadra
谢谢!但这工作对我来说: AssessmentFileName as'data() 谢谢你的尝试! :) –