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• MySQL: Query to obtain recipes using all given ingredients.

  - by John_A

  hi I have the following simplified tables: CREATE TABLE recipe(id int, name varchar(25)); CREATE TABLE ingredient(name varchar(25)); CREATE TABLE uses_ingredient(recipe_id int, name varchar(25)); I want to make a query that returns all id's of recipes that contain both Chicken and Cream. I have tried SELECT recipe_id FROM uses_ingredient INNER JOIN (SELECT * FROM ingredient WHERE name="Chicken" OR name="Cream") USING (name) GROUP BY recipe_id HAVING COUNT(recipe_id) >= (SELECT COUNT(*) FROM theme); which gives me :"ERROR 1248 (42000): Every derived table must have its own alias" and is probably wrong too. Next I tried SELECT recipe_id FROM (SELECT * FROM ingredient WHERE name="Chicken" OR name="Cream") AS t INNER JOIN uses_ingredient USING (name) GROUP BY recipe_id HAVING COUNT(recipe_id)>= (SELECT COUNT(*) FROM t); which gives "ERROR 1146 (42S02): Table 'recipedb.t' doesn't exist" I want to avoid creating temporary tables including using ENGINE=MEMORY.

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• An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a HAVING cla

  - by brz dot net

  I have to find the indentid from the status table based on below two conditions: 1. If there are more than one record having same indentid in status table and the same indentID has count1 in feasibilitystatus table then I don't want to display the record. 2. If there is only one record of indentid in status table and the same indentID has count0 in feasibilitystatus table then I don't want to display the record. Query: select distinct s.indentid from status s where s.status='true' and s.indentid not in(select case when count(s.indentid)>1 then (select indentid from feasibilitystatus group by indentid having count(indentid)>1) else (select indentid from feasibilitystatus group by indentid having count(indentid)>0) end as indentid from status) Error: An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a HAVING clause or a select list, and the column being aggregated is an outer reference.

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• LINQ. Grouping by days. How to do this easily?

  - by punkouter

  I can't seem to find any god reference on this. I have alot of data in SQL with dates. So I wanted to make a line chart to show this data over time. If I want to show it over a perioud of days then I need to group by days.. But the LOGDATE is the full date.. not the DAY.. So I have this below.. but LINQ doesnt know what 'DayOfYear' property is.. HELP var q = from x in dc.ApplicationLogs let dt = x.LogDate group x by new { dayofyear = dt.Value.DayOfYear } into g select new { iCount = g.Count(), strDate = g.Key };

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• MYSQL sum() for distinct rows

  - by makeee

  I'm looking for help using sum() in my SQL query (not posting full query since the scenario is fairly simple). I have COUNT(DISTINCT conversions.id) in my query. I use DISTINCT because I'm doing "group by" for multiple columns and this ensures the same row is not counted more than once. Now I want to add: SUM(conversions.value) as conversion_value The problem is that the "value" for each row is counted more than once (due to the multiple group bys) I basically want to do SUM(conversions.value) for each DISTINCT conversions.id. Is that possible?

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• Xcode can't find imported header files

  - by solerous

  I've used other IDE's in the past but am new to Xcode. I'm trying to bring in a bunch of C code from an open source project. I've imported them into a new Group and the .c files all show up under Implementation Files and the full list of files shows up in the Groups/Files group as well as my project directory in the finder. When I try to include or import one of these header files, code completion even works so I know Xcode is seeing them. But then when I go to build it says "no such file or directory". How can I get these files to import into my code?

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• How to get the class of an input inside a jQuery each loop?

  - by Paul Atkins

  Hi, I have function which appends inputs inside a list item when a link is clicked. I then loop through these inputs using an each loop using the code below. It is working correctly as shown, however instead of using field.name I want to use the class of the input as the array key but when i try to do this the class is shown as undefined. Here is the code I am currently using: var values = {}; $.each($('li :input').serializeArray(), function(i, field) { values[field.name] = field.value; }); Here is the code I have inside the list item once I have appended the hidden inputs using jQuery append: <li><input type="hidden" name="group" class="group" value="2"/><input type="hidden" name="condition" class="condition" value="isany"/><input type="hidden" name="value" class="value" value="1,2"/></li> I can get the name attribute fine but class is always undefined. Could anybody help with this?

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• How to retain a row which is foreign key in another table and remove other duplicate rows?

  - by Mithril

  I have two table: A: id code 1 A1 2 A1 3 B1 4 B1 5 C1 6 C1 ===================== B: id Aid 1 1 2 4 (B doesn't contain the Aid which link to code C1) Let me explain the overall flow: I want to make each row in table A have different code(by delete duplicate),and I want to retain the Aid which I can find in table B.If Aid which not be saved in table B,I retain the id bigger one. so I can not just do something as below: DELETE FROM A WHERE id NOT IN (SELECT MAX(id) FROM A GROUP BY code, ) I can get each duplicate_code_groups by below sql statement: SELECT code FROM A GROUP BY code HAVING COUNT(*) > 1 Is there some code in sql like for (var ids in duplicate_code_groups){ for (var id in ids) { if (id in B){ return id } } return max(ids) } and put the return id into a idtable?? I just don't know how to write such code in sql. then I can do DELETE FROM A WHERE id NOT IN idtable

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• Count number of results in a View

  - by Jukebox

  I need to count how many people belong in pre-defined groups (this is easy to do in SQL using the SELECT COUNT statement). My Views query runs fine and displays the actual data in my table, but I simply need to know how many results it found. However there doesn't seem to be a COUNT option in views. I am guessing I am going to have to use some sort of views hook, and then stick the result in the table. Here's a quick example of what i'm trying to achieve: My Table ---------------------- Group A | 20 people Group B | 63 people and so on. (I've tried using the Views_Calc module, but I get errors because it is not quite stable yet.) Anybody know of an easy way to count results in Views?

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• .NET grouping forms so that pulling up the primary form shows all other forms?

  - by toasteroven

  I have an app that can open up some other forms at the user's request, and they're set to not show in the taskbar. The problem is, if one of the secondary windows becomes hidden by another app, switching to the primary window only brings that form to the forefront. Is there a good way to "group" the forms so that giving any of them focus brings the whole group to the front? I tried calling BringToFront() on each form in the primary form's Activated event, but that also gives the secondary forms focus, making it impossible to interact with the primary form.

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• Improve SQL query performance

  - by Anax

  I have three tables where I store actual person data (person), teams (team) and entries (athlete). The schema of the three tables is: In each team there might be two or more athletes. I'm trying to create a query to produce the most frequent pairs, meaning people who play in teams of two. I came up with the following query: SELECT p1.surname, p1.name, p2.surname, p2.name, COUNT(*) AS freq FROM person p1, athlete a1, person p2, athlete a2 WHERE p1.id = a1.person_id AND p2.id = a2.person_id AND a1.team_id = a2.team_id AND a1.team_id IN ( SELECT id FROM team, athlete WHERE team.id = athlete.team_id GROUP BY team.id HAVING COUNT(*) = 2 ) GROUP BY p1.id ORDER BY freq DESC Obviously this is a resource consuming query. Is there a way to improve it?

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• Transmitting complex objects using TCP

  - by Rakesh K

  I have a client server application in which I need to transmit a user defined object from Client to Server using TCP connection. My object is of the following structure: class Conversation { private string convName, convOwner; public ArrayList convUsers; public string getConvName() { return this.convName; } public string getConvOwner() { return this.convOwner; } } Please help me how to transmit this object at from client and again de-serialize it into appropriate object at server side.

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• how to design a db like Facebook where users can update their status and of the fb page as admin

  - by Harsha M V

  i am designing a database where users can update status messages of theirs and they can create pages groups like facebook fan page and post status like the admin of the page and not as a user. user(id, name..) group(id, name...) group_admin(group_id, user_id) this is my set up. Is this the way to do it. How to post under the group as an admin. will i need to make a check to every user if he is the admin or not ?

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• help with php MVC problem

  - by aprencai

  hello, I have groups on my site and the urls have the following hierarchy: / groups / {id_group} / / groups / {id_group} / news / groups / {id_group} / gallery / groups / {id_group} / events / groups / {id_group} / events / {id_event} / groups / {id_group} / events / {id_event} / news / groups / {id_group} / events / {id_event} / gallery / groups / {id_group} / events / {id_event} / news As you can see a group can have news, gallery, etc. and in turn an event that is in a group can also have news, gallery, etc. How to implement this approach in a framework without specifying any specific one?, ie I would like some guidance on what would have modules, controllers, etc. Thanks.

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• Confusing alias mySQL

  - by Taylor

  I keep getting the same number outputted for the Total Sales, Minimum Sale, Largest Sale and Average Sale. The Total Invoices is working perfectly, but I cant seem to figure out how to fix the other ones. Here's the query: SELECT SUM( b.`Number of Invoices`) AS `Total Invoices`, SUM( b.`Total Customer Purchases`) AS `Total Sales`, MIN( b.`Total Customer Purchases`) AS `Minimum Sale`, MAX( b.`Total Customer Purchases`) AS `Largest Sale`, AVG( b.`Total Customer Purchases`) AS `Average Sale` FROM (SELECT a.CUS_CODE, COUNT(a.`Number of Invoices`) AS `Number of Invoices`, SUM(a.`Invoice Total`) AS `Total Customer Purchases` FROM ( SELECT CUS_CODE, LINE.INV_NUMBER AS `Number of Invoices`, SUM(LINE.LINE_UNITS * LINE.LINE_PRICE) AS `Invoice Total` FROM `ttriggs`.`INVOICE`, `ttriggs`.`LINE` WHERE INVOICE.INV_NUMBER = LINE.INV_NUMBER GROUP BY CUS_CODE, LINE.INV_NUMBER ) a ) b GROUP BY b.CUS_CODE; Heres the database diagram https://www.dropbox.com/s/b8cy5l29jwh8lyv/1_edit.jpg Subquery generates: CUS_CODE 10011 Number of Invoices 8 Total Customer Purchases 1119.03 Any help is greatly appreciated, Thanks!

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• Does '[ab]+' equal '(a|b)+' in python re module?

  - by user1477871

  I think pat1 = '[ab]' and pat2 = 'a|b' have the same function in Python(python2.7, windows) 're' module as a regular expression pattern. But I am confused with '[ab]+' and '(a|b)+', do they have the same function, if not plz explain details. ''' Created on 2012-9-4 @author: melo ''' import re pat1 = '(a|b)+' pat2 = '[ab]+' text = '22ababbbaa33aaa44b55bb66abaa77babab88' m1 = re.search(pat1, text) m2 = re.search(pat2, text) print 'search with pat1:', m1.group() print 'search with pat2:', m2.group() m11 = re.split(pat1, text) m22 = re.split(pat2, text) print 'split with pat1:', m11 print 'split with pat2:', m22 m111 = re.findall(pat1, text) m222 = re.findall(pat2, text) print 'findall with pat1:', m111 print 'findall with pat2:', m222 output as below: search with pat1: ababbbaa search with pat2: ababbbaa split with pat1: ['22', 'a', '33', 'a', '44', 'b', '55', 'b', '66', 'a', '77', 'b', '88'] split with pat2: ['22', '33', '44', '55', '66', '77', '88'] findall with pat1: ['a', 'a', 'b', 'b', 'a', 'b'] findall with pat2: ['ababbbaa', 'aaa', 'b', 'bb', 'abaa', 'babab'] why are 'pat1' and 'pat2' different and what's their difference? what kind of strings can 'pat1' actually match?

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• Count, inner join

  - by Urosh

  I have two tables: DRIVER (Driver_Id,First name,Last name,...) PARTICIPANT IN CAR ACCIDENT (Participant_Id,Driver_Id-foreign key,responsibility-yes or no,...) Now, I need to find out which driver participated in accident where responsibility is 'YES', and how many times. I did this: Select Driver_ID, COUNT (Participant.Driver_ID)as 'Number of accidents' from Participant in car accident where responsibility='YES' group by Driver_ID order by COUNT (Participant.Driver_ID) desc But, I need to add drivers first and last name from the first table(using inner join, I suppose). I don't know how, because it is not contained in either an aggregate function or the GROUP BY clause. Please help :)

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• Update multiple values in a single statement

  - by Kluge

  I have a master / detail table and want to update some summary values in the master table against the detail table. I know I can update them like this: update MasterTbl set TotalX = (select sum(X) from DetailTbl where DetailTbl.MasterID = MasterTbl.ID) update MasterTbl set TotalY = (select sum(Y) from DetailTbl where DetailTbl.MasterID = MasterTbl.ID) update MasterTbl set TotalZ = (select sum(Z) from DetailTbl where DetailTbl.MasterID = MasterTbl.ID) But, I'd like to do it in a single statement, something like this: update MasterTbl set TotalX = sum(DetailTbl.X), TotalY = sum(DetailTbl.Y), TotalZ = sum(DetailTbl.Z) from DetailTbl where DetailTbl.MasterID = MasterTbl.ID group by MasterID but that doesn't work. I've also tried versions that omit the "group by" clause. I'm not sure whether I'm bumping up against the limits of my particular database (Advantage), or the limits of my SQL. Probably the latter. Can anyone help?

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• CakePHP ACL use case(s)

  - by Jonathan

  I have got a simple web app in development, i want to establish a couple of user groups; Admin, Doctors & Patients. Each group would have their access restricted to particular controller actions rather than individual content. So for example, Doctors can view patient records (index & view actions), but cannot delete them. Usually i would create a groups model, and assign the various users to a group. And filter in the beforeFilter() method to determine if the user has access. But if ACL can do the job, why right the code, right? Thanks

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• How to get an item value of json using C#?

  - by user3487837

  How to get an item value of json using C#? json: [{ ID: '6512', fd: [{ titie: 'Graph-01', type: 'graph', views: { graph: { show: true, state: { group: 'DivisionName', series: ['FieldWeight', 'FactoryWeight', 'Variance'], graphType: 'lines-and-points' } } } }, { titie: 'Graph-02', type: 'Graph', views: { graph: { show: true, state: { group: 'DivisionName', series: ['FieldWeight', 'FactoryWeight', 'Variance'], graphType: 'lines-and-points' } } } }] }, { ID: '6506', fd: [{ titie: 'Map-01', type: 'map', views: { map: { show: true, state: { kpiField: 'P_BudgetAmount', kpiSlabs: [{ id: 'P_BudgetAmount', hues: ['#0fff03', '#eb0707'], scales: '10' }] } } } }] }] Above mentioned one is json, Here titie value will be get in a list please help me... my code is: string dashletsConfigPath = Url.Content("~/Content/Dashlets/Dashlets.json"); string jArray = System.IO.File.ReadAllText(Server.MapPath(dashletsConfigPath)) List<string> lists = new List<string>(); JArray list = JArray.Parse(jArray); var ll = list.Select(j => j["dashlets"]).ToList();

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• Caluculating sum of activity

  - by Maddy

  I have a table which is with following kind of information activity cost order date other information 10 1 100 -- 20 2 100 10 1 100 30 4 100 40 4 100 20 2 100 40 4 100 20 2 100 10 1 101 10 1 101 20 1 101 My requirement is to get sum of all activities over a work order ex: for order 100 1+2+4+4=11 1(for activity 10) 2(for activity 20) 4 (for activity 30) etc. i tried with group by, its taking lot time for calculation. There are 1lakh plus records in warehouse. is there any possibility in efficient way. SELECT SUM(MIN(cost)) FROM COST_WAREHOUSE a WHERE order = 100 GROUP BY (order, ACTIVITY)

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• trying to work through a list in sections

  - by user1714887

  I have a list of lists sorted by the second value of the list (the groups). I now need to iterate through this to work on each "group" at a time. the data is [name, group, data1, data2, data3, data4]. I wasn't sure if I need a while or some other sort of loop, or maybe groupby but I've never used that. any help would be appreciated. for i in range (int(max_group)): x1 = [] x2 = [] x3 = [] x4 = [] if data[i][1] == i+1: x1.append(data[2]) x2.append(data[3]) x3.append(data[4]) x4.append(data[5]) print x1 print 'next' # these are just to test where we're at

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• How to setup linux permissions for the WWW folder?

  - by Xeoncross

  Updated Summery The /var/www directory is owned by root:root which means that no one can use it and it's entirely useless. Since we all want a web server that actually works (and no-one should be logging in as "root"), then we need to fix this. Only two entities need access. PHP/Perl/Ruby/Python all need access to the folders and files since they create many of them (i.e. /uploads/). These scripting languages should be running under nginx or apache (or even some other thing like FastCGI for PHP). The developers How do they get access? I know that someone, somewhere has done this before. With however-many billions of websites out there you would think that there would be more information on this topic. I know that 777 is full read/write/execute permission for owner/group/other. So this doesn't seem to be needed as it leaves random users full permissions. What permissions are need to be used on /var/www so that... Source control like git or svn Users in a group like "websites" (or even added to "www-data") Servers like apache or lighthttpd And PHP/Perl/Ruby can all read, create, and run files (and directories) there? If I'm correct, Ruby and PHP scripts are not "executed" directly - but passed to an interpreter. So there is no need for execute permission on files in /var/www...? Therefore, it seems like the correct permission would be chmod -R 1660 which would make all files shareable by these four entities all files non-executable by mistake block everyone else from the directory entirely set the permission mode to "sticky" for all future files Is this correct? Update: I just realized that files and directories might need different permissions - I was talking about files above so i'm not sure what the directory permissions would need to be. Update 2: The folder structure of /var/www changes drastically as one of the four entities above are always adding (and sometimes removing) folders and sub folders many levels deep. They also create and remove files that the other 3 entities might need read/write access to. Therefore, the permissions need to do the four things above for both files and directories. Since non of them should need execute permission (see question about ruby/php above) I would assume that rw-rw-r-- permission would be all that is needed and completely safe since these four entities are run by trusted personal (see #2) and all other users on the system only have read access. Update 3: This is for personal development machines and private company servers. No random "web customers" like a shared host. Update 4: This article by slicehost seems to be the best at explaining what is needed to setup permissions for your www folder. However, I'm not sure what user or group apache/nginx with PHP OR svn/git run as and how to change them. Update 5: I have (I think) finally found a way to get this all to work (answer below). However, I don't know if this is the correct and SECURE way to do this. Therefore I have started a bounty. The person that has the best method of securing and managing the www directory wins.

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• ssh authentication nfs

  - by user40135

  Hi all I would like to do ssh from machine "ub0" to another machine "ub1" without using passwords. I setup using nfs on "ub0" but still I am asked to insert a password. Here is my scenario: * machine ub0 and ub1 have the same user "mpiu", with same pwd, same userid, and same group id * the 2 servers are sharing a folder that is the HOME directory for "mpiu" * I did a chmod 700 on the .ssh * I created a key using ssh-keygene -t dsa * I did "cat id_dsa.pub authorized_keys". On this last file I tried also chmod 600 and chmod 640 * off course I can guarantee that on machine ub1 the user "shared_user" can see the same fodler that wes mounted with no problem. Below the content of my .ssh folder Code: authorized_keys id_dsa id_dsa.pub known_hosts After all of this calling wathever function "ssh ub1 hostname" I am requested my password. Do you know what I can try? I also UNcommented in the ssh_config file for both machines this line IdentityFile ~/.ssh/id_dsa I also tried ssh -i $HOME/.ssh/id_dsa mpiu@ub1 Below the ssh -vv Code: OpenSSH_5.1p1 Debian-3ubuntu1, OpenSSL 0.9.8g 19 Oct 2007 OpenSSH_5.1p1 Debian-3ubuntu1, OpenSSL 0.9.8g 19 Oct 2007 debug1: Reading configuration data /etc/ssh/ssh_config debug1: Applying options for * debug2: ssh_connect: needpriv 0 debug1: Connecting to ub1 [192.168.2.9] port 22. debug1: Connection established. debug2: key_type_from_name: unknown key type '-----BEGIN' debug2: key_type_from_name: unknown key type '-----END' debug1: identity file /mirror/mpiu/.ssh/id_dsa type 2 debug1: Checking blacklist file /usr/share/ssh/blacklist.DSA-1024 debug1: Checking blacklist file /etc/ssh/blacklist.DSA-1024 debug1: Remote protocol version 2.0, remote software version lshd-2.0.4 lsh - a GNU ssh debug1: no match: lshd-2.0.4 lsh - a GNU ssh debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.1p1 Debian-3ubuntu1 debug2: fd 3 setting O_NONBLOCK debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug2: kex_parse_kexinit: diffie-hellman-group-exchange-sha256,diffie-hellman-group-exchange-sha1,diffie-hellman-group14-sha1,diffie-hellman-group1-sha1 debug2: kex_parse_kexinit: ssh-rsa,ssh-dss debug2: kex_parse_kexinit: aes128-cbc,3des-cbc,blowfish-cbc,cast128-cbc,arcfour128,arcfour256,arcfour,aes192-cbc,aes256-cbc,[email protected],aes128-ctr,aes192-ctr,aes256-ctr debug2: kex_parse_kexinit: aes128-cbc,3des-cbc,blowfish-cbc,cast128-cbc,arcfour128,arcfour256,arcfour,aes192-cbc,aes256-cbc,[email protected],aes128-ctr,aes192-ctr,aes256-ctr debug2: kex_parse_kexinit: hmac-md5,hmac-sha1,[email protected],hmac-ripemd160,[email protected],hmac-sha1-96,hmac-md5-96 debug2: kex_parse_kexinit: hmac-md5,hmac-sha1,[email protected],hmac-ripemd160,[email protected],hmac-sha1-96,hmac-md5-96 debug2: kex_parse_kexinit: none,[email protected],zlib debug2: kex_parse_kexinit: none,[email protected],zlib debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: first_kex_follows 0 debug2: kex_parse_kexinit: reserved 0 debug2: kex_parse_kexinit: diffie-hellman-group14-sha1,diffie-hellman-group1-sha1 debug2: kex_parse_kexinit: ssh-rsa,spki-sign-rsa debug2: kex_parse_kexinit: aes256-cbc,3des-cbc,blowfish-cbc,arcfour debug2: kex_parse_kexinit: aes256-cbc,3des-cbc,blowfish-cbc,arcfour debug2: kex_parse_kexinit: hmac-sha1,hmac-md5 debug2: kex_parse_kexinit: hmac-sha1,hmac-md5 debug2: kex_parse_kexinit: none,zlib debug2: kex_parse_kexinit: none,zlib debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: debug2: kex_parse_kexinit: first_kex_follows 0 debug2: kex_parse_kexinit: reserved 0 debug2: mac_setup: found hmac-md5 debug1: kex: server-client 3des-cbc hmac-md5 none debug2: mac_setup: found hmac-md5 debug1: kex: client-server 3des-cbc hmac-md5 none debug2: dh_gen_key: priv key bits set: 183/384 debug2: bits set: 1028/2048 debug1: sending SSH2_MSG_KEXDH_INIT debug1: expecting SSH2_MSG_KEXDH_REPLY debug1: Host 'ub1' is known and matches the RSA host key. debug1: Found key in /mirror/mpiu/.ssh/known_hosts:1 debug2: bits set: 1039/2048 debug1: ssh_rsa_verify: signature correct debug2: kex_derive_keys debug2: set_newkeys: mode 1 debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug2: set_newkeys: mode 0 debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /mirror/mpiu/.ssh/id_dsa (0xb874b098) debug1: Authentications that can continue: password,publickey debug1: Next authentication method: publickey debug1: Offering public key: /mirror/mpiu/.ssh/id_dsa debug2: we sent a publickey packet, wait for reply debug1: Authentications that can continue: password,publickey debug2: we did not send a packet, disable method debug1: Next authentication method: password mpiu@ub1's password: I hangs here!

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• Why is IIS Anonymous authentication being used with administrative UNC drive access?

  - by Mark Lindell

  My account is local administrator on my machine. If I try to browse to a non-existent drive letter on my own box using a UNC path name: \mymachine\x$ my account would get locked out. I would also get the following warning (Event ID 100, Type “Warning”) 5 times under the “System” group in Event Viewer on my box: The server was unable to logon the Windows NT account 'ourdomain\myaccount' due to the following error: Logon failure: unknown user name or bad password. I would also get the following warning 3 times: The server was unable to logon the Windows NT account 'ourdomain\myaccount' due to the following error: The referenced account is currently locked out and may not be logged on to. On the domain controller, Event ID 680 of type “Failure Audit” would appear 4 times under the “Security” group in Event Viewer: Logon attempt by: MICROSOFT_AUTHENTICATION_PACKAGE_V1_0 Logon account: myaccount Followed by Event ID 644: User Account Locked Out: Target Account Name: myaccount Target Account ID: OURDOMAIN\myaccount Caller Machine Name: MYMACHINE Caller User Name: STAN$ Caller Domain: OURDOMAIN Caller Logon ID: (0x0,0x3E7) Followed by another 4 errors having Event ID 680. Strangely, every time I tried to browse to the UNC path I would be prompted for a user name and password, the above errors would be written to the log, and my account would be locked out. When I hit “Cancel” in response to the user name/password prompt, the following message box would display: Windows cannot find \mymachine\x$. Check the spelling and try again, or try searching for the item by clicking the Start button and then clicking Search. I checked with others in the group using XP and they only got the above message box when browsing to a “bad” drive letter on their box. No one else was prompted for a user name/password and then locked out. So, every time I tried to browse to the “bad” drive letter, behind the scenes XP was trying to login 8 times using bad credentials (or, at least a bad password as the login was correct), causing my account to get locked out on the 4th try. Interestingly, If I tried browsing to a “good” drive such as “c$” it would work fine. As a test, I tried logging on to my box as a different login and browsing the “bad” UNC path. Strangely, my “ourdomain\myaccount” account was getting locked out – not the one I was logged in as! I was totally confused as to why the credentials for the other login were being passed. After much Googling, I found a link referring to some IIS settings I was vaguely familiar with from the past but could not see how they would affect this issue. It was related to the IIS directory security setting “Anonymous access and authentication control” located under: Control Panel/Administrative Tools/Computer Management/Services and Applications/Internet Information Services/Web Sites/Default Web Site/Properties/Directory Security/Anonymous access and authentication control/Edit/Password I found no indication while scouring the Internet that this property was related to my UNC problem. But, I did notice that this property was set to my domain user name and password. And, my password did age recently but I had not reset the password accordingly for this property. Sure enough, keying in the new password corrected the problem. I was no longer prompted for a user name/password when browsing the UNC path and the account lock-outs ceased. Now, a couple of questions: Why would an IIS setting affect the browsing of a UNC path on a local box? Why had I not encountered this problem before? My password has aged several times and I’ve never encountered this problem. And, I can’t remember the last time I updated the “Anonymous access” IIS password it’s been so long. I’ve run the script after a password reset before and never had my account locked-out due to the UNC problem (the script accesses UNC paths as a normal part of its processing). Windows Update did install “Cumulative Security Update for Internet Explorer 7 for Windows XP (KB972260)” on my box on 7/29/2009. I wonder if this is responsible.

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• How to setup linux permissions the WWW folder?

  - by Xeoncross

  Updated Summery The /var/www directory is owned by root:root which means that no one can use it and it's entirely useless. Since we all want a web server that actually works (and no-one should be logging in as "root"), then we need to fix this. Only two entities need access. PHP/Perl/Ruby/Python all need access to the folders and files since they create many of them (i.e. /uploads/). These scripting languages should be running under nginx or apache (or even some other thing like FastCGI for PHP). The developers How do they get access? I know that someone, somewhere has done this before. With however-many billions of websites out there you would think that there would be more information on this topic. I know that 777 is full read/write/execute permission for owner/group/other. So this doesn't seem to be needed as it leaves random users full permissions. What permissions are need to be used on /var/www so that... Source control like git or svn Users in a group like "websites" (or even added to "www-data") Servers like apache or lighthttpd And PHP/Perl/Ruby can all read, create, and run files (and directories) there? If I'm correct, Ruby and PHP scripts are not "executed" directly - but passed to an interpreter. So there is no need for execute permission on files in /var/www...? Therefore, it seems like the correct permission would be chmod -R 1660 which would make all files shareable by these four entities all files non-executable by mistake block everyone else from the directory entirely set the permission mode to "sticky" for all future files Is this correct? Update: I just realized that files and directories might need different permissions - I was talking about files above so i'm not sure what the directory permissions would need to be. Update 2: The folder structure of /var/www changes drastically as one of the four entities above are always adding (and sometimes removing) folders and sub folders many levels deep. They also create and remove files that the other 3 entities might need read/write access to. Therefore, the permissions need to do the four things above for both files and directories. Since non of them should need execute permission (see question about ruby/php above) I would assume that rw-rw-r-- permission would be all that is needed and completely safe since these four entities are run by trusted personal (see #2) and all other users on the system only have read access. Update 3: This is for personal development machines and private company servers. No random "web customers" like a shared host. Update 4: This article by slicehost seems to be the best at explaining what is needed to setup permissions for your www folder. However, I'm not sure what user or group apache/nginx with PHP OR svn/git run as and how to change them. Update 5: I have (I think) finally found a way to get this all to work (answer below). However, I don't know if this is the correct and SECURE way to do this. Therefore I have started a bounty. The person that has the best method of securing and managing the www directory wins.

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