Postgresql Syntax Error At Or Near Like
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here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings syntax error at or near postgres and policies of this site About Us Learn more about Stack Overflow postgresql copy table with constraints the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags postgres copy table structure Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only postgresql duplicate table takes a minute: Sign up Postgresql - CREATE TABLE SYNTAX ERROR USING LIKE - up vote 19 down vote favorite 2 i'm trying to create a temp table from a parent table: This is the code that i execute with pgAdmin III ( or by JDBC in Java ): CREATE TEMP TABLE table1_tmp LIKE table1 INCLUDING DEFAULTS; And the error
Postgresql Create Table From Another Table
i recieved is: [WARNING ] CREATE TEMP TABLE table1_tmp LIKE table1 INCLUDING DEFAULTS ERROR: syntax error at or near «LIKE» LÍNEA 1: CREATE TEMP TABLE table1_tmp LIKE table1 INCLUDING DEFAULTS ^ Reading postgresql 8.4 documentation, create tables using this, its very easy, but i don't understand where is the syntax problem. postgresql table share|improve this question edited Jan 12 '14 at 16:24 a_horse_with_no_name 187k24236312 asked Feb 2 '11 at 12:43 jzafrilla 73121133 add a comment| 2 Answers 2 active oldest votes up vote 36 down vote accepted You need to put the like in to parens like CREATE TEMP TABLE table1_tmp ( LIKE table1 INCLUDING DEFAULTS ) ; This is not obvious from the docs if you don't count parens 1:1 share|improve this answer answered Feb 2 '11 at 13:01 Heiko Rupp 18.9k104998 add a comment| up vote 20 down vote I'm not a Postgresql user but the manual say that there is ( ) around the like setence. CREATE TEMP TABLE table1_tmp (LIKE table1 INCLUDING DEFAULTS); share|improve this answer edited Feb 2 '11 at 15:32 Frank Heikens 48.8k128191 ans
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Redshift Create Table Like
Discuss the workings and policies of this site About Us Learn more insert into select postgresql about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack postgresql rename table Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, http://stackoverflow.com/questions/4874607/postgresql-create-table-syntax-error-using-like helping each other. Join them; it only takes a minute: Sign up Postgresql Syntax error at or near “ THEN ” up vote 1 down vote favorite I'm trying to create a view in Postgresql , but when I run this code appears this error: syntax error at or near " THEN " CREATE OR REPLACE VIEW VW_MONITOR_DEVICE http://stackoverflow.com/questions/35427991/postgresql-syntax-error-at-or-near-then AS SELECT P.POSIZIONE_DEVICE_ID AS MONITOR_DEVICE_ID, P.VALID AS VALID, [...] IF (VALID == FALSE THEN 'Valid' ELSE P.REASON_FOR_INVALID) AS DESCRIPTION, [...] FROM public.TA_POSIZIONI_DEVICE P JOIN ... TA_POSIZIONI_DEVICE VALID (Boolean not null) sql postgresql syntax-error create-view share|improve this question edited Feb 16 at 9:27 a_horse_with_no_name 187k24236312 asked Feb 16 at 9:00 Marco 3929 1 case when VALID is FALSE THEN 'Valid' ELSE P.REASON_FOR_INVALID end –jarlh Feb 16 at 9:01 @jarlh Don't forget the END :-) –Tim Biegeleisen Feb 16 at 9:02 1 @TimBiegeleisen, oops, thanks! Will edit. –jarlh Feb 16 at 9:03 1 There is no IF in SQL, where in the Postgres SQL reference did you find that? –a_horse_with_no_name Feb 16 at 9:03 @jarlh Thanks You! This woks :) case when VALID is FALSE THEN 'Valid' ELSE P.REASON_FOR_INVALID end DESCRIPTION, –Marco Feb 16 at 9:08 add a comment| 3 Answers 3 active oldest votes up vote 2 down vote accepted You should use CASE The SQL CASE expression is a generic conditional expression, simil
8.1 / 8.2 / 8.3 / 8.4 / 9.0 PostgreSQL 8.0.26 Documentation Prev Fast Backward Chapter 31. Extending SQL https://www.postgresql.org/docs/8.0/static/xfunc-sql.html Fast Forward Next 31.4. Query Language (SQL) Functions SQL functions execute https://github.com/sequelize/sequelize/issues/4064 an arbitrary list of SQL statements, returning the result of the last query in the list. In the simple (non-set) case, the first row of the last query's result will be returned. (Bear in mind that "the first row" of a multirow result is not well-defined unless you syntax error use ORDER BY.) If the last query happens to return no rows at all, the null value will be returned. Alternatively, an SQL function may be declared to return a set, by specifying the function's return type as SETOF sometype. In this case all rows of the last query's result are returned. Further details appear below. The body of an syntax error at SQL function must be a list of SQL statements separated by semicolons. A semicolon after the last statement is optional. Unless the function is declared to return void, the last statement must be a SELECT. Any collection of commands in the SQL language can be packaged together and defined as a function. Besides SELECT queries, the commands can include data modification queries (INSERT, UPDATE, and DELETE), as well as other SQL commands. (The only exception is that you can't put BEGIN, COMMIT, ROLLBACK, or SAVEPOINT commands into a SQL function.) However, the final command must be a SELECT that returns whatever is specified as the function's return type. Alternatively, if you want to define a SQL function that performs actions but has no useful value to return, you can define it as returning void. In that case, the function body must not end with a SELECT. For example, this function removes rows with negative salaries from the emp table: CREATE FUNCTION clean_emp() RETURNS void AS ' DELETE FROM emp WHERE salary < 0; ' LANGUAGE SQL; SELECT
Sign in Pricing Blog Support Search GitHub This repository Watch 269 Star 7,806 Fork 1,577 sequelize/sequelize Code Issues 693 Pull requests 72 Projects 0 Wiki Pulse Graphs New issue Postgres $any with an array gets a sytax error #4064 Closed joelanman opened this Issue Jul 5, 2015 · 6 comments Projects None yet Labels bug postgres Milestone No milestone Assignees No one assigned 6 participants joelanman commented Jul 5, 2015 If I construct a WHERE clause with $any and an array, I get an error. var selectedAnimals = ['cow','pig']; where.title = { $iLike: { $any: selectedAnimals} }; generates this SQL: SELECT "name" FROM "Animals" AS "Animal" WHERE "Animal"."type" ILIKE ANY ARRAY['cow','pig']::VARCHAR; and this error: syntax error at or near "ARRAY" From Postgres syntax, it seems like ANY should have brackets, like this: SELECT "name" FROM "Animals" AS "Animal" WHERE "Animal"."type" ILIKE ANY (ARRAY['cow','pig']) sequelize member janmeier commented Jul 6, 2015 Hmm, from the unit tests, it looks like we should be generating with brackets. Have you tried latest master https://github.com/sequelize/sequelize/blob/bebad6e208c66f9e3dad535f8cd51204a1e3bbf2/test/unit/sql/where.test.js#L394 corbanb commented Jul 15, 2015 +1 I am getting this in 3.3.2 SELECT "id", "protocol", "password_encrypted", "provider", "identifier", "friends", "tokens", "accessToken", "createdAt", "updatedAt", "UserId" FROM "Passports" AS "Passport" WHERE "Passport"."identifier" = ANY (ARRAY['333333333333']::VARCHAR) AND "Passport"."provider" = 'facebook'; Removing ::VARCHAR allows this query to work also. corbanb commented Jul 15, 2015 @mickhansen @janmeier I updated to the bleeding edge. same bug in master. janmeier added bug postgres labels Oct 12, 2015 janmeier referenced this issue Nov 5, 2015 Closed PostgreSQL like #4813 vhmth commented Feb 24, 2016 Reproducable. error looking up relevant campaigns: { [SequelizeDatabaseError: syntax error at or near "ARRAY"] name: 'SequelizeDatabaseError', message: 'syntax error at or near "ARRAY"', parent: { [error: syntax error at or near