Error Permission Denied For Language C Postgres
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Postgres Permission Denied For Sequence
about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack postgres permission denied for schema Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping postgres permission denied for large object each other. Join them; it only takes a minute: Sign up ERROR: permission denied for language c up vote 20 down vote favorite 8 When create a function something like this with a non-super user i am getting the below error: ERROR: permission denied for language c SQL state: 42501 The function created is : CREATE OR
Postgres Permission Denied For Schema Public
REPLACE FUNCTION dblink_connect (text) RETURNS text AS '$libdir/dblink','dblink_connect' LANGUAGE C STRICT; But if i wanted to give permission on langauge C to my non-super user, i am getting the below error: postgres=# grant usage on language c to caixa; ERROR: language "c" is not trusted That means, non-super user can't create function with language C? or is there any thing i am doing wrong? c postgresql share|improve this question edited Oct 7 '15 at 7:32 iharob 39.4k42851 asked Aug 10 '11 at 16:34 vchitta 65341527 Probably you don't need grant USAGE privilege to caixa role explicitly, because it's granted to PUBLIC (all roles) already. The key here is lanpltrusted property from pg_language system catalog. –Grzegorz Szpetkowski Aug 10 '11 at 17:01 add a comment| 2 Answers 2 active oldest votes up vote 27 down vote accepted That's right, according to doc: Only superusers can create functions in untrusted languages Quick check: SELECT lanpltrusted FROM pg_language WHERE lanname LIKE 'c'; lanpltrusted -------------- f (1 row) If you really want this
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Postgres Permission Denied To Create Role
skills and learn from others in the community. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the http://stackoverflow.com/questions/7014437/error-permission-denied-for-language-c top RDS crosstab: ERROR: permission denied for language c up vote 3 down vote favorite We're currently using PostgreSQL 9.2 in our production database, we want to migrate to a new RDS instance, using PostgreSQL 9.4. However, we need to have permission to create a function, but only superuser can create functions on RDS and my user can't be granted as superuser. CREATE FUNCTION crosstab50(text, OUT rownumber integer, OUT c1 text,...) RETURNS SETOF record LANGUAGE c STABLE http://dba.stackexchange.com/questions/128691/rds-crosstab-error-permission-denied-for-language-c STRICT AS '$libdir/tablefunc', 'crosstab'; But we're getting this error during our restore operation: ERROR: permission denied for language c postgresql postgresql-9.4 amazonrds pivot aws share|improve this question edited Feb 12 at 8:51 Erwin Brandstetter 53.2k584143 asked Feb 9 at 7:53 Valter Henrique 1575 How to use the basic crosstab() function: PostgreSQL Crosstab Query –Erwin Brandstetter Feb 12 at 8:49 add a comment| 1 Answer 1 active oldest votes up vote 4 down vote accepted RDS doesn't let you install arbitrary C functions. It's a restricted environment. crosstab is part of the tablefunc extension. You may be able to CREATE EXTENSION tablefunc if it's on the whitelist but you won't be able to add new variants, even if the underlying C function is already loaded and approved. share|improve this answer answered Feb 9 at 7:58 Craig Ringer 28.2k14082 I was able to run CREATE EXTENSION tablefunc. How can I use it now ? I'm new to postgres. –Valter Henrique Feb 9 at 8:06 1 You can use all the predefined crosstab variants that come with tablefunc. See \df crosstab*. You cannot create new variants with different arguments. This is a limitation of AWS RDS PostgreSQL, not a limitation of PostgreSQL its self, because Amazon want to keep their RDS systems "sealed". –Craig Ringer Feb 9 at 9:17 I see, thank you so
попытке выполнить uuid-ossp.sql (этот файл содержит набор процедур для работы с uuid). Выяснилось, что хранимые процедуры на си могут создавать только суперпользователи.Если же хочется, чтобы это было доступно не только суперпользователю, можно сделать так:1. В базе сделать язык си вызывающим доверие:psql --single-transaction -c "UPDATE pg_language SET lanpltrusted = true WHERE lanname = 'c';" -d база -U суперюзер2. Дать нужному пользователю нужные права: psql -c "GRANT USAGE ON LANGUAGE c TO юзер;" -d база -U суперюзер Автор: huh-muh на 19:16 Ярлыки: postgresql Комментариев нет: Отправить комментарий Следующее Предыдущее Главная страница Подписаться на: Комментарии к сообщению (Atom) Ярлыки .NET 1С 2gis 7z аниме вау вышивание гы-гы домен железо законы касса книги КриптоПро кулинария математика мышь оружие отдых планшет ПО принтер ссылки хранилище ссылок acrobat adb alsa Android Apache audio bash batch C C# C++ CD CGI cifs COM command line converter cron cygwin DC++ debian dhcp disk image DNS dosbox dot dvd dynamic libs eToken Excel exfat fb2 firefox flash flex flv freetds ftp gcc gif git graphviz hash HASP hdd html i2p icecast iconv IE IIS Installer iptables IPTV jabber java javascript KDE kdenlive keyboard LaTeX LILO linux LUKS mail md5 mkv mount mpg123 MS Office msi MSSQL nat netbios network ntfs ntlm OE OpenOffice openssl opera oracle Outlook partimage pdf php pidgin postgresql pptp print proxy pure-ftpd pymssql python radio RDP redo backup rsync samba skype slackware SMART smbfs SMS sniffer ssh ssl svg swf thunderbird timezone tr truecrypt UAC ubuntu unixodbc usb usb bootable uuid vi video virtualbox vk.com vlc WIA Windows windows 7 wine wsus X xfce xmpp youtube Архив блога ► 2016 (13) ► октября (1) ► сентября (1) ► августа (3) ► июня (1) ► мая (1) ► апреля (2) ► марта (1) ► января (3) ► 2015 (16) ► декабря (1) ► ноября (3) ► ок