I am using Jena's SPARQL engine and trying to write a query to filter on a date range as I need to find the value of a property after a fixed date.
My date property is in the following format:
Fri May 23 10:20:13 IST 2014
How do I write a SPARQL query to get other properties with dates greater than this?
With your data in that format you can't filter on a range of it without adding a custom extension function to ARQ (which is intended for advanced users) since you would need to parse and interpret the date time string.
What you should instead be doing is translating your data into the standard date time format xsd:dateTime that all SPARQL implementations are required to support. See the XML Schema Part 2: Datatypes specification for details of this format.
Your specific example date would translate as follows:
2014-05-23T10:20:13+05:30
And you must ensure that you declare it to be a typed literal of type xsd:dateTime when you use it in data and queries. For example in the readable Turtle RDF syntax:
#prefix xsd: <http://www.w3.org/2001/XMLSchema#> .
#prefix : <http://example.org> .
:subject :date "2014-05-23T10:20:13+05:30"^^xsd:dateTime .
You could then write a SPARQL query that filters by range of dates like so:
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX : <http://example.org>
SELECT *
WHERE
{
?s :date ?date .
FILTER (?date > "2014-05-23T10:20:13+05:30"^^xsd:dateTime)
}
This finds all records where ?date is after the given date
Related
I have a table with like this:
id
values
user_id
1
["8","7","6"]
5
Now I'm running a query with WHERE condition on values column:
SELECT * from table_name WHERE values = ["8","7","6"]
But MySQL returns this error:
Error Code : 1064
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '["8","7","6"]'
If you want to compare for strict equality, you want to do the comparison as JSON objects. You can do this by using JSON_EXTRACT to parse the data as JSON instead of text:
SELECT * from table_name WHERE
JSON_EXTRACT(`values`, '$') = JSON_EXTRACT('["8","7","6"]', '$');
You should be able to use this for any type of JSON as long as you want strict equality. If you want to return all rows that match the given JSON object, use JSON_CONTAINS.
For example to find all values with the string "8" in it, you'd use the following:
SELECT * from table_name WHERE JSON_CONTAINS(`values`, '"8"');
Note that this matching is not as simple as you'd expect and matches any value in the document. If your data consists of JSON arrays, this should still be adequate.
The information about your column datatype, especially values are crucial. Since the column stores a mix of numbers and non-numbers characters, we can assume that it might be stored in VARCHAR() or TEXT datatype. But since the data format looks like a JSON array, it's also a possibility that the column datatype is JSON. Now both of these datatypes have a very different query structure.
First, let's address some issues:
Whenever the cell values include other characters than numerical, it will be considered as string. Hence, using plain .. WHERE values = ["8","7","6"] without wrapping it in quotes ('), you'll get that Error Code : 1064.
VALUES is a reserved word in MySQL so if you want to stick to it as your table column names, you always need to wrap it in backticks. If not, this will also return Error Code : 1064:
.. WHERE `values` = ..
Now let's try this:
If the column datatype for values is VARCHAR() or TEXT, you just have to simply wrap the search value in single quote like:
SELECT * from table_name WHERE `values` = '["8","7","6"]';
Refer this fiddle
updated for MariaDB
If the column datatype for values is JSON, it's something like this:
SELECT * from table_name where JSON_UNQUOTE(`values`)= '["8","7","6"]'
Refer this fiddle for JSON
The JSON method I've referred to this MariaDB documentation.
P/S: According to this documentation JSON is an alias for LONGTEXT introduced for compatibility reasons with MySQL's JSON data type. In other words, when creating a table with JSON datatype in MariaDB, it will be shown as LONGTEXT but with extra definition than just plain LONGTEXT datatype. See this fiddle for more detail.
In Sqlite I want to extract the date and time portions of a DateTime field separately in a view and return them also as a datetime, not strings. I've tried Cast, Date(), datetime(), but they all return strings.
I've read the SQLite documentation and understand how there is not an actual Date data type. Yet a Table field defined as DateTime is able to be parsed as a Date by an Excel query, but calculations on that field are not. I'm trying to do all data prep in the database view.
My data has the following field taken directly from the table definition:
LastModifiedDate datetime
I want the date (without time) to have the same DateTime data type as LastModifiedDate, not Text, because I use this view in many spreadsheets. I can apply Excel Date functions and formatting to LastModifiedDate field directly as returned from the ODBC query to Excel, and want to do the same to the Date-only part. I don't want to have to put a string-to-date conversion in every spreadsheet when I know it can get the date natively from Sqlite in LastModifiedDate.
SELECT LastModifiedDate,
date(LastModifiedDate) as Datepart,
cast(LastModifiedDate as numeric) as Date2
FROM Transactions
LastModifiedDate Datepart Date2
2019-07-28 18:22:38.9165394 2019-07-28 2019
LastModifiedDate in the above query is interpreted in Excel as a date to which date formats and date functions can be applied with no further processing required. Datepart above is returned as Text to Excel, and I can't apply date functions and formats without further pre-processing in Excel. I would like Datepart to be interpreted a date in Excel just as LastModifiedDate is.
I'm looking at the ch-werner.de sqliteodbc-0.9998. It will return an ODBC TIMESTAMP type only if the column decltype starts with timestamp or datetime. It returns ODBC TIME only for decltypes starting with time and ODBC DATE only for decltypes starting with date.
sqlite3 provides this decltype only for result table columns that are direct database column references. So if your SELECT statement has some expression that is more than a plain column reference, the decltype is lost. sqlite3 works like this at least up to version 3.39.0. It is documented.
The CAST expression converts the value of given expression to a storage classes by the determined affinity of the given declared type, but does not assign decltype to the result.
If you want to see the decltypes for query columns, you can use the sqlite3 cli and give it command .stats 2. Then it'll output the column declared types for each statement it executes.
If the decltype is found, the sqliteodbc-0.9998 will always parse string values into ODBC types. If DSN Option JDConv is enabled, it'll also parse floating point julianday values (whether provided as float or a string of a float) into ODBC types and when writing it'll write floating point into database.
If you can afford to change the schema, you can add a generated virtual column. This is cheap in storage, because data is not affected, but it costs when you query the column. This column can calculate other column into the values and decltypes you need for ODBC.
ALTER TABLE data ADD COLUMN
Datepart date AS (date(LastModifiedDate))
Then to get the Datepart, you simply query the column.
SELECT Datepart FROM data
I am attempting to get information from UK Data Communities on planning decisions from here. I have tried to run the query considering the SPARQL endpoint that they provide. It's the first time I am running a query with SPARQL so I have followed general indications in here and considered a previous thread with other data from this site.
My code looks like:
library(SPARQL)
# create the query
endpoint <- "http://opendatacommunities.org/sparql"
query <-
"PREFIX dcterms: <http://purl.org/dc/terms/>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX qb: <http://purl.org/linked-data/cube#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
SELECT *
WHERE {
?s ?p ?o
}"
# submit query
qd <- SPARQL(endpoint,query)
However I get the following error:
Error: XML content does not seem to be XML: 'Request Timeout'
I have tried to edit my query by stating explicitly the format argument in SPARQL() as xml (qd <- SPARQL(endpoint,query, format = "xml")) but I have obtained similar result. I would be grateful if someone could give some hints about what is going wrong.
Try using an endpoint of "http://opendatacommunities.org/sparql.xml".
Your endpoint is does not appear to be an endpoint but an HTML page only.
I simply want to show a date and its day of the week from a table.
The following works:
select "invDate", (select extract (dow from timestamp '2014-09-22'))
from "tblInvMaster"
But the moment I try to use the actual field like the example below, it doesn't work:
select "invDate", (select extract (dow from timestamp "invDate"))
from "tblInvMaster"
The above gives a syntax error where the field name starts in timestamp.
What is the correct method of getting this to work?
The syntax
TYPENAME 'VALUE'
e.g.
TIMESTAMP '2014-01-01'
is only valid in SQL for type literals.
If you want to cast a non-literal value you must use an explicit cast. Most likely you don't require a cast at all, and can just write:
extract(dow from "invDate")
as "invDate" should already be a timestamp or date. If it isn't, you'll need to CAST("invDate" AS timestamp).
I use sqlite3 C/C++ API to retrieve rows from a table using SELECT query. I don't see any sqlite3_column_timestamp() to retrieve a timestamp column value after sqlite3_step().. How to get timestamp values ?
SQLite does not have a special timestamp data type.
When you want to use any of SQLite's date and time functions, you have to store timestamps in one of the formats supported by them, i.e., a string like YYYY-MM-DD HH:MM:SS or HH:MM:SS, a julian date number, or a Unix timestamp number.
You can declare a table column type as DATETIME, but SQLite will just ignore that type; SQLite always allows to put values of any type in any column. Such a declaration would be useful only as documentation.
The column/value accessors will only have types corresponding to the data types they support directly (NULL, INTEGER, REAL, TEXT, BLOB).
You would use the TEXT access to get/set the column value of dates.
There are some helper functions within SQL that they provide that let you to handle them in your queries.
I am not familiar with SQLite Manager, but I would assume that it is only reporting the data type that the table was declared with.
When parsing CREATE statements, sqlite understands the intention of many well supported datatypes and automatically maps them to what is appropriate for its internal storage structure. VARCHAR would be mapped to TEXT, for instance. I assume the column was declared DATETIME and sqlite just internally mapped it to TEXT.