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plurals

PO
primarykey
Id
17994119
data
AcceptedAnswerId
0
AnswerCount
0
ClosedDate
CommentCount
0
CommunityOwnedDate
CreationDate
2013-08-01T12:30:25.190
FavoriteCount
0
LastActivityDate
2013-08-01T12:36:59.313
LastEditDate
2013-08-01T12:36:59.313
LastEditorUserId
1281433
OwnerUserId
1281433
ParentId
17986551
PostTypeId
2
Score
5
ViewCount
0
LastEditorDisplayName
text
Body
In the first <a href="http://www.w3.org/TR/rdf-sparql-query/" rel="noreferrer">SPARQL Query Language for RDF</a>, this would be rather difficult, because there are not many string manipulation functions. However, in your question, you've used <a href="http://www.w3.org/TR/sparql11-query/#func-replace" rel="noreferrer"><code>replace</code></a> which appeared in <a href="http://www.w3.org/TR/sparql11-query/" rel="noreferrer">SPARQL 1.1 Query Language</a>. This is good for you because, in addition to <code>replace</code>, SPARQL 1.1 includes more string manipulation functions. One of these, <a href="http://www.w3.org/TR/sparql11-query/#func-substr" rel="noreferrer"><code>substr</code></a>, does exactly what you need. For instance, here's a query in which <code>?string</code> is bound to the string you mentioned, and <code>substr</code> is used to extract the substring you're looking for and bind it as <code>?substring</code>. <pre><code>select * where { values ?string { "OEIUFHWOIEFWNFOPQWJFHQWOIHFEB..." } bind( substr( ?string, 5, 6 ) as ?substring ) } </code></pre> The results are: <pre><code>-------------------------------------------------- | string | substring | ================================================== | "OEIUFHWOIEFWNFOPQWJFHQWOIHFEB..." | "FHWOIE" | -------------------------------------------------- </code></pre> Note that the second argument to <code>substr</code> is the starting position (where the first index is 1), and the third is the length of the substring, not the final position. You wanted a substring, <code>FHWOIE</code>, that has six characters, to the third argument is 6.
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1. POSPARQL string RANGE
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1. PTAnswer
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1. USJoshua Taylor
UserOwnerUserId
1. USJoshua Taylor
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2. VO
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