Question

关注var mongoose = require('mongoose'); var studentNames = new mongoose.Schema({ firstName: {type: String, required: true}, lastName: {type: String, required: true}, telephone: {type: String}, emailAddress: {type: String}, isMember: {type: Boolean}, lessons: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}] }); var lessonNames = new mongoose.Schema({ name: {type: String}, startTime: {type: String}, endTime: {type: String}, day: {type: String} }); var locationNames = new mongoose.Schema({ name: {type: String, required: true}, address: {type: String}, lessons: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], lessonDays: [{ Monday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], Tuesday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], Wednesday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], Thursday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], Friday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], Saturday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}], Sunday: [{type: mongoose.Schema.Types.ObjectId, ref: 'lessonnames'}] }] }); mongoose.model('studentnames', studentNames); mongoose.model('lessonnames', lessonNames); mongoose.model('locationnames', locationNames);我写了哪个接受function并返回xml作为结果。

table

以下是我用来对CREATE FUNCTION FunctionTest(@ID INT,@XML_Details xml) RETURNS @RESULT TABLE ( Value1 INT, Value2 INT ) AS BEGIN DECLARE @tbl_Xml_Result Table ( Value1 INT, Value2 INT ) INSERT INTO @RESULT(Value1,Value2) SELECT l.v.value('Value2[1]','INT'), l.v.value('Value1[1]','INT') FROM @XML_Details.nodes('/Listings/listing')l(v) RETURN END上面运行的代码，但它始终返回function结果。

Empty

Answer 1

首先，尽可能避免多线功能。 Inline UDFs have better pefromance

第二个XQuery区分大小写，并且您没有'/Listings/listing'路径。

第三：您可能希望使用@ID进行过滤。

CREATE FUNCTION FunctionTest(@ID INT,@XML_Details xml)
RETURNS  TABLE
AS
RETURN(
    SELECT Value1, Value2
    FROM (SELECT  
            l.v.value('(Value2)[1]','INT') AS Value2,
            l.v.value('(Value1)[1]','INT') AS Value1
          FROM @XML_Details.nodes('/Listings/Listing')l(v)) AS sub
    WHERE Value1 = @ID
    )
GO

DECLARE @tbl_Xml_Result Table(Value1 INT,Value2 INT);
INSERT INTO @tbl_xml_Result(Value1, Value2)
values(1,2),(2,3),(3,4),(4,5),(5,6);

DECLARE @xml_Temp xml  = (SELECT * 
                          FROM @tbl_xml_Result
                          FOR XML PATH('Listing'),ROOT('Listings'));

SELECT T.Value1,
       T.Value2
FROM FunctionTest(1,@xml_Temp) T;

的 LiveDemo

Answer 2

你职能第19行的问题：

    FROM @XML_Details.nodes('/Listings/listing')l(v)

* l 正在努力 - 您需要 L 正在努力

为什么SQL函数总是返回空结果集？

2 个答案: