Combining the use of preceding- and following-sibling in the same XPath query up to first instance
Following the post here Combining the use of preceding and following sibling in the same xpath query I am wanting to do the same except I want to get nodes up to the first occurrence of p2.
So the modified example data is
<a>
<b property="p1">zyx</b>
<b>wvu</b>
<b>tsr</b>
<b>dcv</b>
<b property="p2">qpo</b>
<b>qcs</b>
<b property="p2">wee</b>
<b>tbg</b>
<b>rty</b>
<b property="p2">qwe</b>
<b>jkl</b>
</a>
I have tried
/a/b[preceding-sibling::b/@property='p1' and following-sibling::b[1]/@property='p2']
which gives me dcv, qcs, and rty.
Then I tried
/a/b[preceding-sibling::b/@property='p1' and (following-sibling::b/@property='p2')[1]]
which gives me wvu, tsr, dcv, qpo, qcs, wee, tbg, and rty.
What I actually need is from p1 to the first p2 or wvu, tsr, and dcv. I have tried almost every combination of [1] but have failed and would like some help on the construct.
xpath siblings find-occurrences
edited Nov 15 '18 at 23:01
zx485
15k133048
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
add a comment |
Following the post here Combining the use of preceding and following sibling in the same xpath query I am wanting to do the same except I want to get nodes up to the first occurrence of p2.
So the modified example data is
<a>
<b property="p1">zyx</b>
<b>wvu</b>
<b>tsr</b>
<b>dcv</b>
<b property="p2">qpo</b>
<b>qcs</b>
<b property="p2">wee</b>
<b>tbg</b>
<b>rty</b>
<b property="p2">qwe</b>
<b>jkl</b>
</a>
I have tried
/a/b[preceding-sibling::b/@property='p1' and following-sibling::b[1]/@property='p2']
which gives me dcv, qcs, and rty.
Then I tried
/a/b[preceding-sibling::b/@property='p1' and (following-sibling::b/@property='p2')[1]]
which gives me wvu, tsr, dcv, qpo, qcs, wee, tbg, and rty.
What I actually need is from p1 to the first p2 or wvu, tsr, and dcv. I have tried almost every combination of [1] but have failed and would like some help on the construct.
xpath siblings find-occurrences
edited Nov 15 '18 at 23:01
zx485
15k133048
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
add a comment |
Following the post here Combining the use of preceding and following sibling in the same xpath query I am wanting to do the same except I want to get nodes up to the first occurrence of p2.
So the modified example data is
<a>
<b property="p1">zyx</b>
<b>wvu</b>
<b>tsr</b>
<b>dcv</b>
<b property="p2">qpo</b>
<b>qcs</b>
<b property="p2">wee</b>
<b>tbg</b>
<b>rty</b>
<b property="p2">qwe</b>
<b>jkl</b>
</a>
I have tried
/a/b[preceding-sibling::b/@property='p1' and following-sibling::b[1]/@property='p2']
which gives me dcv, qcs, and rty.
Then I tried
/a/b[preceding-sibling::b/@property='p1' and (following-sibling::b/@property='p2')[1]]
which gives me wvu, tsr, dcv, qpo, qcs, wee, tbg, and rty.
What I actually need is from p1 to the first p2 or wvu, tsr, and dcv. I have tried almost every combination of [1] but have failed and would like some help on the construct.
xpath siblings find-occurrences
edited Nov 15 '18 at 23:01
zx485
15k133048
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
Following the post here Combining the use of preceding and following sibling in the same xpath query I am wanting to do the same except I want to get nodes up to the first occurrence of p2.
So the modified example data is
<a>
<b property="p1">zyx</b>
<b>wvu</b>
<b>tsr</b>
<b>dcv</b>
<b property="p2">qpo</b>
<b>qcs</b>
<b property="p2">wee</b>
<b>tbg</b>
<b>rty</b>
<b property="p2">qwe</b>
<b>jkl</b>
</a>
I have tried
/a/b[preceding-sibling::b/@property='p1' and following-sibling::b[1]/@property='p2']
which gives me dcv, qcs, and rty.
Then I tried
/a/b[preceding-sibling::b/@property='p1' and (following-sibling::b/@property='p2')[1]]
which gives me wvu, tsr, dcv, qpo, qcs, wee, tbg, and rty.
What I actually need is from p1 to the first p2 or wvu, tsr, and dcv. I have tried almost every combination of [1] but have failed and would like some help on the construct.
xpath siblings find-occurrences
xpath siblings find-occurrences
edited Nov 15 '18 at 23:01
zx485
15k133048
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
edited Nov 15 '18 at 23:01
zx485
15k133048
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
edited Nov 15 '18 at 23:01
zx485
15k133048
edited Nov 15 '18 at 23:01
zx485
15k133048
edited Nov 15 '18 at 23:01
zx485
15k133048
15k133048
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
asked Nov 15 '18 at 22:59
sinDizzysinDizzy
59951845
59951845
add a comment |
add a comment |
1 Answer
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One way to achieve this is by using the following XPath-1.0 expression:
/a/b[preceding-sibling::b/@property='p1' and not(preceding-sibling::b/@property='p2' or @property='p2')]
It checks if a given b element has a preceding-sibling named p1 and is not a preceding-sibling of p2 or p2 itself. That's all.
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
One way to achieve this is by using the following XPath-1.0 expression:
/a/b[preceding-sibling::b/@property='p1' and not(preceding-sibling::b/@property='p2' or @property='p2')]
It checks if a given b element has a preceding-sibling named p1 and is not a preceding-sibling of p2 or p2 itself. That's all.
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
add a comment |
One way to achieve this is by using the following XPath-1.0 expression:
/a/b[preceding-sibling::b/@property='p1' and not(preceding-sibling::b/@property='p2' or @property='p2')]
It checks if a given b element has a preceding-sibling named p1 and is not a preceding-sibling of p2 or p2 itself. That's all.
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
add a comment |
One way to achieve this is by using the following XPath-1.0 expression:
/a/b[preceding-sibling::b/@property='p1' and not(preceding-sibling::b/@property='p2' or @property='p2')]
It checks if a given b element has a preceding-sibling named p1 and is not a preceding-sibling of p2 or p2 itself. That's all.
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
One way to achieve this is by using the following XPath-1.0 expression:
/a/b[preceding-sibling::b/@property='p1' and not(preceding-sibling::b/@property='p2' or @property='p2')]
It checks if a given b element has a preceding-sibling named p1 and is not a preceding-sibling of p2 or p2 itself. That's all.
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
answered Nov 15 '18 at 23:18
zx485zx485
15k133048
15k133048
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
add a comment |
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
Very clever answer. Being relatively new to xpath i didn't know how to put it together. Many thanks.
– sinDizzy
Nov 15 '18 at 23:35
add a comment |
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