Search Results

Search found 14709 results on 589 pages for 'root permission'.

Page 175/589 | < Previous Page | 171 172 173 174 175 176 177 178 179 180 181 182  | Next Page >

  • How do I change the effective user of psql?

    - by gvkv
    I'm using psql to run a simple set of COPY statements contained in a file: psql -d mydb -f 'wbf_queries.data.sql' where wbf_queries.data.sql contains lines: copy <my_query> to '/home/gvkv/mydata' delimiter ',' null ''; ... but I get a permission denied error: ... ERROR: could not open file ... for writing: Permission denied I'm connecting under my user account (gvkv) which is also a superuser in PostgreSQL. Obviously, psql is running under a different (effective) user but I don't know how to change this. Can it be done within psql or do I need some unix-fu?

    Read the article

  • zend_acl: Adding extra resources dynamically and getting a param predispatch

    - by Timmeh
    First sorry about the woffle as I'm not sure how best to describe this. Basically I am not sure how I can get param in the bootstrap before the controller is loaded, but here is the long winded version... I have got an acl class storing all my default resources in. All my page/post content is a database and I want the admin the ability to choose who which role the page would become available. I know it is possible just to loop through the database table and add them all in at once, but I am concerned that this is a drain on resources. I have it working whereby my access check plugin can call a dynamic permission function, but I need to get the parameter of the current page ID and it's permission to set it before the controller is loaded. Does that make sense or am I worry over nothing and I should just get the resources of all the pages at once? thanks in advance for reading my garble!!

    Read the article

  • how do i read strings from text files?

    - by ratty
    I like to read string from text file the text file consists of information below con = new MySqlConnection("server=localhost;user id=root; password=""; database=workplantype; pooling=false;"); i like to read server name such as"localhost" here and user id such as"root" here,how can i read this.

    Read the article

  • Customising Web-logics default 404 error page

    - by Sean McDaid
    I am running weblogic 9. When I enter an incorrect URL below the application root I redirect 404 request to a customized error page. Eg. http://localhost:7001/myApp/non-existent redirects to my customised error page. Is there a way to do this for all incorrect URLs entered, not necessarily below the application root? Eg. http://localhost:7001/anything_non-existent should redirect to my customized error page, not the web-logic default one? Thanks

    Read the article

  • Why is my producer-consumer blocking?

    - by User007
    My code is here: http://pastebin.com/Fi3h0E0P Here is the output 0 Should we take order today (y or n): y Enter order number: 100 More customers (y or n): n Stop serving customers right now. Passing orders to cooker: There are total of 1 order(s) 1 Roger, waiter. I am processing order #100 The goal is waiter must take orders and then give them to the cook. The waiter has to wait cook finishes all pizza, deliver the pizza, and then take new orders. I asked how P-V work in my previous post here. I don't think it has anything to do with \n consuming? I tried all kinds of combination of wait(), but none work. Where did I make a mistake? The main part is here: //Producer process if(pid > 0) { while(1) { printf("0"); P(emptyShelf); // waiter as P finds no items on shelf; P(mutex); // has permission to use the shelf waiter_as_producer(); V(mutex); // cooker now can use the shelf V(orderOnShelf); // cooker now can pickup orders wait(); printf("2"); P(pizzaOnShelf); P(mutex); waiter_as_consumer(); V(mutex); V(emptyShelf); printf("3 "); } } if(pid == 0) { while(1) { printf("1"); P(orderOnShelf); // make sure there is an order on shelf P(mutex); //permission to work cooker_as_consumer(); // take order and put pizza on shelf printf("return from cooker"); V(mutex); //release permission printf("just released perm"); V(pizzaOnShelf); // pizza is now on shelf printf("after"); wait(); printf("4"); } } So I imagine this is the execution path: enter waiter_as_producer, then go to child process (cooker), then transfer the control back to parent, finish waiter_as_consumer, switch back to child. The two waits switch back to parent (like I said I tried all possible wait() combination...).

    Read the article

  • Web development - relative URLs without duplicating files

    - by eshriek
    I have a site with index.php in the root folder, images in /img , and overview.php in /content . I have a sidebar.php file that is included in both index.php and overview.php . How should I refer to /img/image.gif if I include a link in each file? The location of image.gif changes relative to the location of the file that references it. Using /img/image.gif in sidebar.php will work in index.php, but it fails for the file located at /content/overview.php. The only solution that I can see is to either include a seperate sidebar.php in each subdirectory, or include an /img directory in every sub-directory. The best suggestion that I can find is to use the <base html tag as suggested here: Change relative link paths for included content in PHP However, in the same link, SamGoody suggests that the <base tag "is no longer properly supported in Internet Explorer, since version 7." I'd like some insight on the matter before committing to a course of action. Thanks. EDIT: I am using the wrong approach below with "../" Example- root/index.php: ... <link rel="stylesheet" type="text/css" href="style.css" /> <title>title</title> </head> <body> <?php include('include/header.php'); ?> <?php include('include/menu.php'); ?> ... root/include/header.php: ... <div id="header"> <span class="fl"><img src="img/dun1.png"/></span><span class="fr"><img src="img/dun2.png"/></span> ... root/content/overview.php: ... <link rel="stylesheet" type="text/css" href="../style.css" media="screen" /> <title>Overview</title> </head> <body> <?php include('../include/header.php'); ?> <?php include('../include/menu.php'); ?> ...

    Read the article

  • Other SecurityManager implementations available?

    - by mhaller
    Is there any other implementation (e.g. in an OSS project) of a Java SecurityManager available which has more features than the one in the JDK? I'm looking for features like configurable at runtime policies updateable at runtime, read from other data sources than a security.policy file Thread-aware, e.g. different policies per Thread Higher-level policies, e.g. "Disable network functions, but allow JDBC traffic" Common predefined policies, e.g. "Allow read-access to usual system properties like file.encoding or line.separator, but disallow read-access to user.home" Monitoring and audit trace logging, e.g. "Log all file access, log all network access going NOT to knownhost.example.org" Blocking jobs "requesting" a permission until an administrator grants permission, letting the thread/job continue ... I'm pretty sure that application servers (at least the commercial ones) have their own SecurityManager implementation or at least their own policy configuration. I'm wondering if there is any free project with similar requirements.

    Read the article

  • Fabric methods exceptions

    - by baobee
    I try to make Fabric func, which checks if Apache installed: from fabric.api import * def check_apache(): try: result = local('httpd -v', capture=True) except: print "check_apache exception" But if httpd not installed I get: [root@server-local ~]$ fab check_apache Fatal error: local() encountered an error (return code 127) while executing 'ahttpd -v' Aborting. check_apache exception Done. How can I get correct exception for Fabric local() method ? So I need to get exception and continue executing without any Fabric error messages: [root@server-local ~]$ fab check_apache check_apache exception Done.

    Read the article

  • Why do I get access denied to data folder when using adb?

    - by gregm
    I connected to my live device using the adb and the following commands: C:\>adb -s HT829GZ52000 shell $ ls ls sqlite_stmt_journals cache sdcard etc system sys sbin proc logo.rle init.trout.rc init.rc init.goldfish.rc init default.prop data root dev $ cd data cd data $ ls ls opendir failed, Permission denied I was surprised to see that I have access denied. How come I can't browse around the directories using the commandline like this? How do I get root access on my phone?

    Read the article

  • Unable to add PageFunction to my project.

    - by Shimmy
    I add to my project a PageFunction and I get a dozen of the following error and the project won't compile: 'ResourceDictionary' root element is a generic type and requires a x:Class attribute to support the x:TypeArguments attribute specified on the root element tag. Basically I get an error for each DataTemplate I merge in the ResourceDictionary, has anyone encoutered this problem before? Note: I use VB.NET 3.5 on VS 2010.

    Read the article

  • Staging server .htaccess for images, css and js

    - by Gavin Hall
    As we build and demo sites on our staging server with individual root folders for each such as /CLIENTNAME, we need to keep all the css, js and internal links for these sites referencing the server root. The following works for one folder each, but not sure how to adapt to work for all folders. Currently AddHandler php5-script .php RewriteEngine On RewriteRule ^(images|css|js)\/(.*) /ONEFOLDER/$1/$2 Would like AddHandler php5-script .php RewriteEngine On RewriteRule ^(images|css|js)\/(.*) /EVERYFOLDER/$1/$2 Many thanks in advance.

    Read the article

  • Page markup maintainability

    - by Tony
    Hi where js folder is under the root. if u put this JS ref in common\SomeControl.ascx, it will work fine if SomeControl is placed on ~/SomePage.aspx because SomePage is under the website root. How to put JS ref in SomeControl and allow it to be placed at any path on the website without losing the JS ref. Thanks

    Read the article

  • Content of a web-directory

    - by Webmezha
    I've got a site with some administrative pages in it's root directory. Question: Is there any possible way for a visitor to see all the pages and/or subdirectories in the root directory of this (or any other) site? If yes, what has to be done to conceal the directory's content? Thank you!

    Read the article

  • Using XNamespace to create nicely formatted XML.

    - by Steven
    I want to create a Xml file that looks something like this: <Root xmlns:ns1="name1" xmlns:ns2="name2">     <ns1:element1 />     <ns1:element2 />     <ns2:element3 /> </Root> How can I accomplish this using XAttribute, XElement, XNamespace, and XDocument where the namespaces are dynamically added.

    Read the article

  • Selective Checkout or a View, on a project in repository

    - by Yossi Zach
    I have a bunch of interconnected projects which share the same project tree. I'm looking for a version control system which provides a possibility to checkout a subset of the project tree. If my the full project tree looks like this: Project Root |-Feature1 | |-SubFeature11 | \-SubFeature12 |-Feature2 | |-SubFeature21 | \-SubFeature22 |-file1 \-file2 I want be able to checkout only subset like this: Project Root |-Feature1 | \-SubFeature12 |-Feature2 | \-SubFeature22 |-file1 \-file2 So do you know any version control system that allows to do selective checkout or a view on a repository?

    Read the article

  • How can I move my Dynamic Data folder?

    - by ProfK
    I accidentally moved my Dynamic Data' folder into myImagesfolder. The project still compiles, but it's just not right. However, when I try to move it back to the root in Visual Studio, I get an error that the destination folder already exists. If I moveDynamic Data` back to the root outside of Visual Studio, the project no longer compiles because the compiler can't find any dynamic data files. My infancy with git prompted me to ask here before embarking on an unpleasant 2am quest.

    Read the article

  • PHP is_file returns false (incorrectly) for Windows share on Ubuntu

    - by M3Mania
    Ubuntu server, PHP 5.3, connected via Samba to Windows server share. I am using file_exists() to check for availability of file on Windows machine. It returns false, although the filepath does exist. Meanwhile, file_get_contents() on the exact same filepath works fine. I'm wondering if it's a permission issue, since I'm having trouble configuring permissions of the files on the Windows share (it says I don't have permission to change permissions on them). When I look at the permissions through Nautilus, it says the user and group are both root, with 755 rights. I'd like to change the group to www-data, but can't seem to do it.

    Read the article

  • PHP & MySQL Undefined variable problem

    - by comma
    I keep getting the following error Undefined variable: id on line 91 can some one help me correct this problem? The error is on this line. $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; MySQL database tables. CREATE TABLE tags ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill VARCHAR(255) NOT NULL, experience VARCHAR(255) NOT NULL, years VARCHAR(255) NOT NULL, PRIMARY KEY (id) ); CREATE TABLE users_skills ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill_id INT UNSIGNED NOT NULL, user_id INT UNSIGNED NOT NULL, date_created DATETIME UNSIGNED NOT NULL, PRIMARY KEY (id) ); Here is the PHP & MySQL code. if (isset($_POST['info_submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id WHERE user_id='$user_id'"); if (!$dbc) { print mysqli_error($mysqli); return; } $user_id = '5'; $skill = $_POST['skill']; $experience = $_POST['experience']; $years = $_POST['years']; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON users_skills.skill_id = learned_skills.id WHERE users_skills.user_id='$user_id'"); if (mysqli_num_rows($dbc) == 0) { if (isset($_POST['skill']) && trim($_POST['skill'])!=='') { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, years) VALUES ('" . $skill . "', '" . $experience . "', '" . $years . "')"); if (mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM learned_skills WHERE id='" . $skill . "' AND experience='" . $experience . "' AND years='" . $years . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; } }

    Read the article

  • Cannot Access Kickstart.php

    - by user282659
    Hey All Let me explain you the scenario upto now.. Been a newbie to Joomla I managed to build my own website using Joomla CMS in my computers localhost. Then after the completion of the site, I wanted to upload my Local Joomla Site to a GoDaddy Server / Hosting Account.. Thanks to Google I found the below mentioned tutorial, http://zensamarketing.com/2009/07/how-to-use-joomlapack-and-kickstart-to-upload-your-local-joomla-site-to-a-godaddy-server-hosting-account/ So far I completed up to the 7th step, But I'm stuck at the step 08. I'm using Filezilla as my FTP client, Using that I uploaded the files as per the guided on step 07 and when I try to access www.mysitename.com/kickstart.php, it is giving me the below error, Forbidden You don't have permission to access /kickstart.php on this server. Additionally, a 403 Forbidden error was encountered while trying to use an ErrorDocument to handle the request. How can I set permission... Please explain in a non technical way.. Thank you so much in advance smile

    Read the article

  • One registry key for many products not deleted on uninstall

    - by NC1
    My company has many products, we want to create a registry key Software\$(var.Manufacturer)that will have all of our products if our customers have installed more than one (which is likely) I then want to have a secondary key for each of our products which get removed on uninstall but the main one does not. I have tried to achieve this like below but my main key gets deleted so all of my other products also get deleted from the registry. I know this is trivial but I cannot find an answer. <DirectoryRef Id="TARGETDIR"> <Component Id="Registry" Guid="*" MultiInstance="yes" Permanent="yes"> <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)" ForceCreateOnInstall="yes"> <RegistryValue Type="string" Name="Default" Value="true" KeyPath="yes"/> </RegistryKey> </Component> </DirectoryRef> <DirectoryRef Id="TARGETDIR"> <Component Id="RegistryEntries" Guid="*" MultiInstance="yes" > <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)\[PRODUCTNAME]" Action="createAndRemoveOnUninstall"> <RegistryValue Type="string" Name="Installed" Value="true" KeyPath="yes"/> <RegistryValue Type="string" Name="ProductName" Value="[PRODUCTNAME]"/> </RegistryKey> </Component> </DirectoryRef> EDIT: I have got my registry keys to stay using the following code. However they only all delete wen all products are deleted, not one by one as they need to. <DirectoryRef Id="TARGETDIR"> <Component Id="Registry" Guid="FF75CA48-27DE-430E-B78F-A1DC9468D699" Permanent="yes" Shared="yes" Win64="$(var.Win64)"> <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)" ForceCreateOnInstall="yes"> <RegistryValue Type="string" Name="Default" Value="true" KeyPath="yes"/> </RegistryKey> </Component> </DirectoryRef> <DirectoryRef Id="TARGETDIR"> <Component Id="RegistryEntries" Guid="D94FA576-970F-4503-B6C6-BA6FBEF8A60A" Win64="$(var.Win64)" > <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)\[PRODUCTNAME]" ForceDeleteOnUninstall="yes"> <RegistryValue Type="string" Name="Installed" Value="true" KeyPath="yes"/> <RegistryValue Type="string" Name="ProductName" Value="[PRODUCTNAME]"/> </RegistryKey> </Component> </DirectoryRef>

    Read the article

  • Java: Make a method abstract for each extending class

    - by Martijn Courteaux
    Hi, Is there any keyword or design pattern for doing this? public abstract class Root { public abstract void foo(); } public abstract class SubClass extends Root { public void foo() { // Do something } } public class SubberClass extends SubClass { // Here is it not necessary to override foo() // So is there a way to make this necessary? // A way to obligate the developer make again the override } Thanks

    Read the article

  • How to login as another user and then log out in bash script?

    - by Neuquino
    Hi, I need to write a bash script to do something as another user and then return to the initial user... Suppose I run the following as root: #!/bin/bash USER=zaraza su - "${USER}" #do some stuff as zaraza ________ #here I should logout zaraza #continue doing things as root In the console I should write "exit", but in bash is a keyword and it exits the script... Thanks in advance,

    Read the article

  • file_operations Question, how do i know if a process that opened a file for writing has decided to c

    - by djTeller
    Hi Kernel Gurus, I'm currently writing a simple "multicaster" module. Only one process can open a proc filesystem file for writing, and the rest can open it for reading. To do so i use the inode_operation .permission callback, I check the operation and when i detect someone open a file for writing I set a flag ON. i need a way to detect if a process that opened a file for writing has decided to close the file so i can set the flag OFF, so someone else can open for writing. Currently in case someone is open for writing i save the current-pid of that process and when the .close callback is called I check if that process is the one I saved earlier. Is there a better way to do that? Without saving the pid, perhaps checking the files that the current process has opened and it's permission... Thanks!

    Read the article

< Previous Page | 171 172 173 174 175 176 177 178 179 180 181 182  | Next Page >