Hiérarchisation d'un XML via transformation XSL (2)

**flophie** · 15/11/2020, 13h50

Bonjour,
Suite à mon thread précédent, j'ai essayé d'appliquer la solution proposée par tsuji mais je n'arrive pas à l'étendre à des hiérarchisations complémentaires.
Voici donc un exemple plus complet de ce que je souhaite faire :

Transformer ce XML "plat" :

Code :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
<Employees>
	<Employee> <ID>1</ID> <WeekID>1</WeekID> <DayID>1</DayID> <Hours>5</Hours> </Employee>
	<Employee> <ID>1</ID> <WeekID>1</WeekID> <DayID>1</DayID> <Hours>4</Hours> </Employee>
	<Employee> <ID>1</ID> <WeekID>1</WeekID> <DayID>2</DayID> <Hours>7</Hours> </Employee>
	<Employee> <ID>1</ID> <WeekID>2</WeekID> <DayID>1</DayID> <Hours>5</Hours> </Employee>
	<Employee> <ID>1</ID> <WeekID>2</WeekID> <DayID>3</DayID> <Hours>8</Hours> </Employee>
	<Employee> <ID>2</ID> <WeekID>1</WeekID> <DayID>1</DayID> <Hours>5</Hours> </Employee>
	<Employee> <ID>2</ID> <WeekID>2</WeekID> <DayID>4</DayID> <Hours>4</Hours> </Employee>
</Employees>

En :

Code :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
<Employees>
	<Employee>
		<ID>1</ID>
		<Weeks>
			<Week>
				<WeekID>1</WeekID>
				<Days>
					<Day>
						<DayID>1</DayID>
						<Hours>5</Hours>
						<Hours>4</Hours>
					</Day>
					<Day>
						<DayID>2</DayID>
						<Hours>7</Hours>
					</Day>
				</Days>
			</Week>
			<Week>
				<WeekID>2</WeekID>
				<Days>
					<Day>
						<DayID>1</DayID>
						<Hours>5</Hours>
					</Day>
					<Day>
						<DayID>3</DayID>
						<Hours>8</Hours>
					</Day>
				</Days>
			</Week>
		</Weeks>
	</Employee>
	<Employee>
		<ID>2</ID>
		<Weeks>
			<Week>
				<WeekID>1</WeekID>
				<Days>
					<Day>
						<DayID>1</DayID>
						<Hours>5</Hours>
					</Day>
				</Days>
			</Week>
			<Week>
				<WeekID>2</WeekID>
				<Days>
					<Day>
						<DayID>4</DayID>
						<Hours>4</Hours>
					</Day>
				</Days>
			</Week>
		</Weeks>
	</Employee>
</Employees>

Quelle serait la transformation XSL à appliquer ? (petite précision qui a son importance : je ne peux pas appliquer le XSLT 2.0, il me faut trouver une solution en XSLT 1.0...)

Merci d'avance,
Flophie

**flophie** · 15/11/2020, 17h09

J'ai trouvé une solution très moche :

Code :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
 
<xsl:template match="/Employees">
<Employees>
<xsl:for-each select="Employee">
<xsl:variable name="EmployeeID" select="ID" />
<xsl:if test="not(preceding-sibling::Employee[ID=$EmployeeID])">
<Employee>
<ID><xsl:value-of select="$EmployeeID"/></ID>
<Weeks>
 
<xsl:for-each select="../Employee[ID=$EmployeeID]">
<xsl:variable name="WeekID" select="WeekID" />
<xsl:if test="not(preceding-sibling::Employee[ID=$EmployeeID and WeekID=$WeekID])">
<Week>
<WeekID><xsl:value-of select="$WeekID"/>
</WeekID>
<Days>
 
<xsl:for-each select="../Employee[ID=$EmployeeID and WeekID=$WeekID]">
<xsl:variable name="DayID" select="DayID" />
<xsl:if test="not(preceding-sibling::Employee[ID=$EmployeeID and WeekID=$WeekID and DayID=$DayID])">
<Day>
<DayID><xsl:value-of select="$DayID"/></DayID>
 
<xsl:for-each select="../Employee[ID=$EmployeeID and WeekID=$WeekID and DayID=$DayID]">
<xsl:variable name="Hours" select="Hours" />
<xsl:if test="not(preceding-sibling::Employee[ID=$EmployeeID and WeekID=$WeekID and DayID=$DayID and Hours=$Hours])">
<Hours><xsl:value-of select="Hours"/></Hours>
</xsl:if>
</xsl:for-each>
 
</Day>
</xsl:if>
</xsl:for-each>
 
</Days>
</Week>
</xsl:if> 
</xsl:for-each>
 
</Weeks>
</Employee>
</xsl:if>         
</xsl:for-each>
</Employees>
</xsl:template>
 
</xsl:stylesheet>

Si quelqu'un connait la méthode Muenchian et peut m'indiquer comment y transposer ce XSLT, je suis preneur

**tsuji** · 17/11/2020, 20h52

Si vous vouley ...

Code xslt1 :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="utf-8" omit-xml-declaration="yes" indent="yes" />
<xsl:param name="sep" select="'|'" />
 
<xsl:key name="id-g" match="Employee" use="ID" />
<xsl:key name="id-wid-g" match="Employee" use="concat(ID, $sep, WeekID)" />
<xsl:key name="id-wid-did-g" match="Employee" use="concat(ID, $sep, WeekID, $sep, DayID)" />
 
<xsl:template match="Employees">
    <xsl:copy>
        <xsl:apply-templates select="Employee" />
    </xsl:copy>
</xsl:template>
 
<xsl:template match="Employee[generate-id()=generate-id(key('id-g',ID)[1])]">
    <xsl:variable name="id" select="ID" />
    <xsl:copy>
        <ID><xsl:value-of select="$id" /></ID>
        <Weeks>
            <xsl:for-each select="/Employees/Employee[ID=$id]/WeekID">
                <xsl:apply-templates select="." />
            </xsl:for-each>
        </Weeks>
    </xsl:copy>
</xsl:template>
 
<xsl:template match="WeekID[generate-id(parent::Employee)=generate-id(key('id-wid-g',concat(preceding-sibling::ID,$sep,.))[1])]">
    <xsl:variable name="id" select="preceding-sibling::ID" />
    <xsl:variable name="wid" select="." />
    <Week>
        <WeekID><xsl:value-of select="$wid" /></WeekID>
        <Days>
            <xsl:for-each select="/Employees/Employee[ID=$id and WeekID=$wid]/DayID">
                <xsl:apply-templates select="." />
            </xsl:for-each>
        </Days>
    </Week>
</xsl:template>
 
<xsl:template match="DayID[generate-id(parent::Employee)=generate-id(key('id-wid-did-g',concat(preceding-sibling::ID,$sep,preceding-sibling::WeekID,$sep,.))[1])]">
    <xsl:variable name="id" select="preceding-sibling::ID" />
    <xsl:variable name="wid" select="preceding-sibling::WeekID" />
    <xsl:variable name="did" select="." />
    <Day>
        <DayID><xsl:value-of select="$did" /></DayID>
        <xsl:for-each select="/Employees/Employee[ID=$id and WeekID=$wid and DayID=$did]/Hours">
            <Hours><xsl:value-of select="." /></Hours>
        </xsl:for-each>
    </Day>
</xsl:template>
 
<xsl:template match="Employee" />
<xsl:template match="WeekID" />
<xsl:template match="DayID" />
 
</xsl:stylesheet>

Hiérarchisation d'un XML via transformation XSL (2)

XSL/XSLT/XPATH XML

Discussions similaires

Partager

Partager