Mixing
xpath regular expression for xml
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I would like to read an xml using xpath regular expression.
The tags can be <PaymentCard> or <ns2:PaymentCard>.
I need to find a common regular expression to read both of the tags.
Currently I am able to read only <PaymentCard> tag using below code:
paymentCardPath = xpath.compile("//PaymentCard");
java xml xslt xpath
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
asked Jan 3 at 6:17
AmarAmar
1631312
add a comment |
I would like to read an xml using xpath regular expression.
The tags can be <PaymentCard> or <ns2:PaymentCard>.
I need to find a common regular expression to read both of the tags.
Currently I am able to read only <PaymentCard> tag using below code:
paymentCardPath = xpath.compile("//PaymentCard");
java xml xslt xpath
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
asked Jan 3 at 6:17
AmarAmar
1631312
There's no regular expression involved.
– choroba
Jan 3 at 6:19
1
Is it reallyns2.PaymentCardor actuallyns2:PaymentCard?
– Tim C
Jan 3 at 8:09
In XPath 2.0 you can use//*[matches(name(), 'PaymentCard')]or//*[ends-with(name(), 'PaymentCard')]
– Andersson
Jan 3 at 8:45
its actually ns2:PaymentCard
– Amar
Jan 3 at 10:15
add a comment |
I would like to read an xml using xpath regular expression.
The tags can be <PaymentCard> or <ns2:PaymentCard>.
I need to find a common regular expression to read both of the tags.
Currently I am able to read only <PaymentCard> tag using below code:
paymentCardPath = xpath.compile("//PaymentCard");
java xml xslt xpath
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
asked Jan 3 at 6:17
AmarAmar
1631312
I would like to read an xml using xpath regular expression.
The tags can be <PaymentCard> or <ns2:PaymentCard>.
I need to find a common regular expression to read both of the tags.
Currently I am able to read only <PaymentCard> tag using below code:
paymentCardPath = xpath.compile("//PaymentCard");
java xml xslt xpath
java xml xslt xpath
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
asked Jan 3 at 6:17
AmarAmar
1631312
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
asked Jan 3 at 6:17
AmarAmar
1631312
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
edited Jan 3 at 18:57
Daniel Haley
39.7k45481
39.7k45481
asked Jan 3 at 6:17
AmarAmar
1631312
asked Jan 3 at 6:17
AmarAmar
1631312
asked Jan 3 at 6:17
AmarAmar
1631312
1631312
There's no regular expression involved.
– choroba
Jan 3 at 6:19
1
Is it reallyns2.PaymentCardor actuallyns2:PaymentCard?
– Tim C
Jan 3 at 8:09
In XPath 2.0 you can use//*[matches(name(), 'PaymentCard')]or//*[ends-with(name(), 'PaymentCard')]
– Andersson
Jan 3 at 8:45
its actually ns2:PaymentCard
– Amar
Jan 3 at 10:15
add a comment |
There's no regular expression involved.
– choroba
Jan 3 at 6:19
1
Is it reallyns2.PaymentCardor actuallyns2:PaymentCard?
– Tim C
Jan 3 at 8:09
In XPath 2.0 you can use//*[matches(name(), 'PaymentCard')]or//*[ends-with(name(), 'PaymentCard')]
– Andersson
Jan 3 at 8:45
its actually ns2:PaymentCard
– Amar
Jan 3 at 10:15
There's no regular expression involved.
– choroba
Jan 3 at 6:19
There's no regular expression involved.
– choroba
Jan 3 at 6:19
1
1
Is it really
ns2.PaymentCard or actually ns2:PaymentCard?– Tim C
Jan 3 at 8:09
Is it really
ns2.PaymentCard or actually ns2:PaymentCard?– Tim C
Jan 3 at 8:09
In XPath 2.0 you can use
//*[matches(name(), 'PaymentCard')] or //*[ends-with(name(), 'PaymentCard')]– Andersson
Jan 3 at 8:45
In XPath 2.0 you can use
//*[matches(name(), 'PaymentCard')] or //*[ends-with(name(), 'PaymentCard')]– Andersson
Jan 3 at 8:45
its actually ns2:PaymentCard
– Amar
Jan 3 at 10:15
its actually ns2:PaymentCard
– Amar
Jan 3 at 10:15
add a comment |
2 Answers
2
active
oldest
votes
In a supplementary comment you clarified that the two names in question are actually PaymentCard and ns2:PaymentCard - that is, two names that have the same local part and different namespace URIs.
In XPath 2.0 you can find names using the local part alone using //*:PaymentCard. The XPath 1.0 equivalent is //*[local-name()='PaymentCard'].
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
1
Though strictly speaking that will also findns1:PaymentCard.
– michael.hor257k
Jan 3 at 13:27
add a comment |
Use the union operator:
//*[self::PaymentCard | self::ns2.PaymentCard]
Note that's an XPath Expression (version 1.0), not a regular expression.
answered Jan 3 at 6:21
chorobachoroba
159k14142209
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
In a supplementary comment you clarified that the two names in question are actually PaymentCard and ns2:PaymentCard - that is, two names that have the same local part and different namespace URIs.
In XPath 2.0 you can find names using the local part alone using //*:PaymentCard. The XPath 1.0 equivalent is //*[local-name()='PaymentCard'].
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
1
Though strictly speaking that will also findns1:PaymentCard.
– michael.hor257k
Jan 3 at 13:27
add a comment |
In a supplementary comment you clarified that the two names in question are actually PaymentCard and ns2:PaymentCard - that is, two names that have the same local part and different namespace URIs.
In XPath 2.0 you can find names using the local part alone using //*:PaymentCard. The XPath 1.0 equivalent is //*[local-name()='PaymentCard'].
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
1
Though strictly speaking that will also findns1:PaymentCard.
– michael.hor257k
Jan 3 at 13:27
add a comment |
In a supplementary comment you clarified that the two names in question are actually PaymentCard and ns2:PaymentCard - that is, two names that have the same local part and different namespace URIs.
In XPath 2.0 you can find names using the local part alone using //*:PaymentCard. The XPath 1.0 equivalent is //*[local-name()='PaymentCard'].
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
In a supplementary comment you clarified that the two names in question are actually PaymentCard and ns2:PaymentCard - that is, two names that have the same local part and different namespace URIs.
In XPath 2.0 you can find names using the local part alone using //*:PaymentCard. The XPath 1.0 equivalent is //*[local-name()='PaymentCard'].
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
answered Jan 3 at 10:19
Michael KayMichael Kay
112k663119
112k663119
1
Though strictly speaking that will also findns1:PaymentCard.
– michael.hor257k
Jan 3 at 13:27
add a comment |
1
Though strictly speaking that will also findns1:PaymentCard.
– michael.hor257k
Jan 3 at 13:27
1
1
Though strictly speaking that will also find
ns1:PaymentCard.– michael.hor257k
Jan 3 at 13:27
Though strictly speaking that will also find
ns1:PaymentCard.– michael.hor257k
Jan 3 at 13:27
add a comment |
Use the union operator:
//*[self::PaymentCard | self::ns2.PaymentCard]
Note that's an XPath Expression (version 1.0), not a regular expression.
answered Jan 3 at 6:21
chorobachoroba
159k14142209
add a comment |
Use the union operator:
//*[self::PaymentCard | self::ns2.PaymentCard]
Note that's an XPath Expression (version 1.0), not a regular expression.
answered Jan 3 at 6:21
chorobachoroba
159k14142209
add a comment |
Use the union operator:
//*[self::PaymentCard | self::ns2.PaymentCard]
Note that's an XPath Expression (version 1.0), not a regular expression.
answered Jan 3 at 6:21
chorobachoroba
159k14142209
Use the union operator:
//*[self::PaymentCard | self::ns2.PaymentCard]
Note that's an XPath Expression (version 1.0), not a regular expression.
answered Jan 3 at 6:21
chorobachoroba
159k14142209
answered Jan 3 at 6:21
chorobachoroba
159k14142209
answered Jan 3 at 6:21
chorobachoroba
159k14142209
answered Jan 3 at 6:21
chorobachoroba
159k14142209
159k14142209
add a comment |
add a comment |
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There's no regular expression involved.
– choroba
Jan 3 at 6:19
1
Is it really
ns2.PaymentCardor actuallyns2:PaymentCard?– Tim C
Jan 3 at 8:09
In XPath 2.0 you can use
//*[matches(name(), 'PaymentCard')]or//*[ends-with(name(), 'PaymentCard')]– Andersson
Jan 3 at 8:45
its actually ns2:PaymentCard
– Amar
Jan 3 at 10:15