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你好!
在使用js-sql-parser时发现一个问题:
ast = sqlParser.parse("SELECT stime, A.names, B.names FROM (SELECT stime, names FROM iaas_data.iaas_d3c0d0681cc1900) AS A LEFT JOIN (SELECT stime, names FROM iaas_data.iaas_1071f89feaa0e100) AS B ON A.stime = B.stime")
sqlParser.stringify(ast) 结果:
SELECT stime, A.names, B.names FROM (SELECT stime, names FROM iaas_data.iaas_d3c0d0681cc1900) LEFT JOIN (SELECT stime, names FROM iaas_data.iaas_1071f89feaa0e100) ON A.stime = B.stime
问题:
sqlParser.stringify(ast)后的SQL 没有 AS A 和 AS B 导致SQL错误.
Activity
[-]Bug[/-][+]LEFT JOIN bug[/+]fix: #34
Merge pull request #35 from JavaScriptor/hotfix-tablefactor-alias
albin3 commentedon Apr 2, 2020
albin3 commentedon Apr 2, 2020
@enoch300 版本1.3.0修复了,多谢
enoch300 commentedon Apr 2, 2020
不客气,很好用。后续有什么问题再给您反馈哈 ,感谢付出