I'm using Neo4j database.Neo4j does not have date data type only have timestamp data type.
I need to compare current date with existing date using cql query.
My existing date format is "8/4/2011" that is string.
Then how can I compare it.Any way to use stored procedure [date] while csv bulk data import time.
I used APOC stored procedure but I don't know how compare it.
CALL apoc.date.format(timestamp(),"ms","dd.MM.yyyy")
07.07.2016
CALL apoc.date.parse("13.01.1975 19:00","s","dd.MM.yyyy HH:mm")
158871600
I expect like this
MATCH(dst:Distributor) WHERE dst.DIST_ID = "111137401" WITH dst CALL apoc.date.parse(dst.ENTRY_DATE,'s', 'dd/MM/yyyy') YIELD d SET dst.ENTRY_DATE = d RETURN dst;
Any possibilities please help me...
RETURN datetime("2018-06-04T10:58:30.007Z").epochMillis
1528109910007
Right query is :
USING PERIODIC COMMIT LOAD CSV WITH HEADERS FROM "file:///DST.csv" AS row
CALL apoc.date.parse(toString(row.ENTRY_DATE),"ms","dd-MMM-yy") YIELD value as date CREATE (DST:Distributor {ENTRY_DATE: date })
Related
I have table ABC in which I have column Z of datatype Date. The format of the data is YYYYMMDD. Now I am looking to convert the above format to YYYY-MON-DD format. Can someone help?
You can use to_char
TO_CHAR(Z,'YYYY-MON-DD')
Depending on what the purpose of the reformatting is, you can either explicitly cast it to a VARCHAR/CHAR and define the format, or you can change your display format to however you'd like to see all dates:
ALTER SESSION SET DATE_OUTPUT_FORMAT = 'YYYY-MON-DD';
It's important to understand that if the data is in a DATE field, then it is stored as a date, and the format of the date is dependent on your viewing preferences, not how it is stored.
Since the value of the date field is stored as a number, you have to convert it to date.
ALTER SESSION SET DATE_OUTPUT_FORMAT = 'YYYY-MON-DD';
select to_date(to_char( z ), 'YYYYMMDD');
(adding this answer to summarize and resolve the question - since the clues and answers are scattered through comments)
The question stated that column Z is of type DATE, but it really seems to be a NUMBER.
Then before parsing a number like 20201017 to a date, first you need to transform it to a STRING.
Once the original number is parsed to a date, it can be represented as a new string formatted as desired.
WITH data AS (
SELECT 20201017 AS z
)
SELECT TO_CHAR(TO_DATE(TO_CHAR(z), 'YYYYMMDD'), 'YYYY-MON-DD')
FROM data;
# 2020-Oct-17
How to resolve my issue below?
1. I am pulling data from AS/400 DB2 using (iseries access odbc driver) to SSRS.
2. I want to format the column stores in integer to date format.
Sample = 20180612 below is the SQL Query.
Using below query,
SELECT CHAR(DATE(SUBSTR(DIGITS(20180612),1,4)||'-'||
SUBSTR(DIGITS(20180612),5,2)||'-'||
SUBSTR(DIGITS(20180612),7,2)),ISO) AS RESULTSDATE
Output = 2018-06-02
Question: How can I produce a below custom date format like d/m/yyyy
I have used, ISO, USA, LOCAL, JIS, EUR but no yield.
Example: 2/6/2018
If you want to do it in SQL Server, the closest you can CONVERT it to is the USA 110 (mm-dd-yyyy) in SQL Server.
If the your dataset has a field with the 2018-06-02, you can use the CDATE function is SSRS to convert it to a date and then format the text box the way you want with the FORMAT property or the FORMAT function.
Format Function:
=FORMAT(CDATE("2018-06-02"), "M/d/yyyy")
Format Property of Text Box:
/* this is as close as I can come to the desired results there is a leading zero */
select
varchar_format(
timestamp_format(char(20180612), 'YYYYMMDD')
, 'DD/MM/YYYY') as resultdate
from my8digitdatetable
/* for export IRL send out a date */
select
cast(
timestamp_format(char(20180612), 'YYYYMMDD')
as date) as resultdate
from my8digitdatetable
I got it already. I just format it to the date without formatting.
SELECT DATE(SUBSTR(DIGITS(MIN(DTAHRS.HREMCP.EMREDT)),1,4)||'-'||
SUBSTR(DIGITS(MIN(DTAHRS.HREMCP.EMREDT)),5,2)||'-'||
SUBSTR(DIGITS(MIN(DTAHRS.HREMCP.EMREDT)),7,2)) AS RESULTSDATE
After that, I put this as sub-query so that I can pass this to the parameter.
Below is my whole query, then at SSRS I can use date/time parameter control to compare with the results.
SELECT x.RESULTSDATE
FROM ( SELECT DATE(SUBSTR(DIGITS(MIN(DTAHRS.HRLVTP.LTLDTE)),1,4)||'-'||
SUBSTR(DIGITS(MIN(DTAHRS.HRLVTP.LTLDTE)),5,2)||'-'||
SUBSTR(DIGITS(MIN(DTAHRS.HRLVTP.LTLDTE)),7,2)) AS RESULTSDATE
) as x
WHERE x.RESULTSDATE >= ? AND x.RESULTSDATE <= ?
I hope it will help you too.
I have a column in my PostgreSQL database, which is in timestamp without time zone format. I would like to save the current date there using the following code:
MyStoredProc.ParamByName('date').Value := FormatDateTime('yyyy-mm-dd hh:nn:ss.zzz', Now);
The date value in my stored procedure is declared as DateTime, but setting it to PgTimeStamp does not help either. The first column is created in PostgreSQL, timestamps working correctly, but the second one, where I´m trying to save data from my code, is always showing zeros.
What am I doing wrong?
You should be able to pass directly the datetime object as parameter int his way :
MyStoredProc.ParamByName('date').Value := Now;
or :
MyStoredProc.ParamByName('date').AsDateTime := Now;
Of course the parameter of the PostgreSQL function must be timezone type.
I'm very new to sql/hive. At first, I loaded a txt file into hive using:
drop table if exists Tran_data;
create table Tran_data(tran_time string,
resort string, settled double)
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t' LINES TERMINATED BY '\n';
Load data local inpath 'C:\Users\me\Documents\transaction_data.txt' into table Tran_Data;
The variable tran_time in the txt file is like this:10-APR-2014 15:01. After loading this Tran_data table, I tried to convert tran_time to a "standard" format so that I can join this table to another table using tran_time as the join key. The date format desired is 'yyyymmdd'. I searched online resources, and found this: unix_timestamp(substr(tran_time,1,11),'dd-MMM-yyyy')
So essentially, I'm doing this: unix_timestamp('10-APR-2014','dd-MMM-yyyy'). However, the output is "NULL".
So my question is: how to convert the date format to a "standard" format, and then further convert it to 'yyyymmdd' format?
from_unixtime(unix_timestamp('20150101' ,'yyyyMMdd'), 'yyyy-MM-dd')
My current Hive Version: Hive 0.12.0-cdh5.1.5
I converted datetime in first column to date in second column using the below hive date functions. Hope this helps!
select inp_dt, from_unixtime(unix_timestamp(substr(inp_dt,0,11),'dd-MMM-yyyy')) as todateformat from table;
inp_dt todateformat
12-Mar-2015 07:24:55 2015-03-12 00:00:00
unix_timestamp function will convert given string date format to unix timestamp in seconds , but not like this format dd-mm-yyyy.
You need to write your own custom udf to convert a given string date to the format that you need as present Hive do not have any predefined functions. We have to_date function to convert a timestamp to date , remaining all unix_timestamp functions won't help your problem.
select from_unixtime(unix_timestamp('01032018' ,'MMddyyyy'), 'yyyyMMdd');
input format: mmddyyyy
01032018
output after query: yyyymmdd
20180103
To help someone in the future:
The following function should work as it worked in my case
to_date(from_unixtime(UNIX_TIMESTAMP('10-APR-2014','dd-MMM-yyyy'))
unix_timestamp('2014-05-01','dd-mmm-yyyy') will work, your input string should be in this format for hive yyyy-mm-dd or yyyy-mm-dd hh:mm:ss
Where as you are trying with '01-MAY-2014' hive won't understand it as a date string
I'm currently trying to do it that way:
// Creating date object
$date = new Zend_Date();
// Adding to it 4 weeks
$date->add('4', Zend_Date::WEEK); // it's expire day
// Getting date in integer(i guess it's unix timestamp yes?)
$date->get();
// Saving it to Mysql in field 'expire' with type Varchar
Then, when needed to get rows, that have date bigger(that haven't yet expired), than current I just add to SQL a simple statement WHERE expire >= $current_date.
Or there is better way to do it? Or how it happens usually?
I would recommend using the native MySQL DATETIME column in your table. This is how you'd retrieve the date for MySQL:
$date->get('yyyy-MM-dd HH:mm:ss');