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• Using EUSM to manage EUS mappings in OUD

  - by Sylvain Duloutre

  EUSM is a command line tool that can be used to manage the EUS settings starting with the 11.1 release of Oracle. In the 11.1 release the tool is not yet documented in the Oracle EUS documentation, but this is planned for a coming release. The same commands used by EUSM can be performed from the Database Console GUI or from Grid Control*. For more details, search for the document ID 1085065.1 on OTN. The examples below don't include all the EUSM options, only the options that are used by EUS. EUSM is user friendly and intuitive. Typing eusm help <option> lists the parameters to be used for any of the available options. Here are the options related to connectivity with OUD : ldap_host="gnb.fr.oracle.com" - name of the OUD server. ldap_port=1389 - nonSSL (SASL) port used for OUD connections.  ldap_user_dn="cn=directory manager" - OUD administrator nameldap_user_password="welcome1" - OUD administrator password Find below common commands: To List Enterprise roles in OUD eusm listEnterpriseRoles domain_name=<Domain> realm_dn=<realm> ldap_host=<hostname> ldap_port=<port> ldap_user_dn=<oud administrator> ldap_user_password=<oud admin password> To List Mappings eusm listMappings domain_name=<Domain> realm_dn=<realm> ldap_host=<hostname> ldap_port=<port> ldap_user_dn=<oud admin> ldap_user_password=<oud admin password> To List Enterprise Role Info eusm listEnterpriseRoleInfo enterprise_role=<rdn of enterprise role> domain_name=<Domain> realm_dn=<realm> ldap_host=<hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password=<oud admin password> To Create Enterprise Role eusm createRole enterprise_role=<rdn of the enterprise role> domain_name=<Domain> realm_dn=<realm> ldap_host=<hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password=<oud admin password> To Create User-Schema Mapping eusm createMapping database_name=<SID of target database> realm_dn="<realm>" map_type=<ENTRY/SUBTREE> map_dn="<dn of enterprise user>" schema="<name of the shared schema>" ldap_host=<oud hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password="<oud admin password>" To Create Proxy Permission eusm createProxyPerm proxy_permission=<Name of the proxypermission> domain_name=<Domain> realm_dn="<realm>" ldap_host=<hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password=<oud admin password> To Grant Proxy permission to Proxy group eusm grantProxyPerm proxy_permission=<Name of the proxy permission> domain_name=<Domain> realm_dn="<realm>" ldap_host=<hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password=<password> group_dn="<dn of the enterprise group>" To Map proxy permission to proxy user in DB eusm addTargetUser proxy_permission=<Name of the proxy permission> domain_name=<Domain> realm_dn="<realm>" ldap_host=<hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password=<oud admin password> database_name=<SID of the target database> target_user=<target database user> dbuser=<Database user with DBA privileges> dbuser_password=<database user password> dbconnect_string=<database_host>:<port>:<DBSID> Enterprise role to Global role mapping eusm addGlobalRole enterprise_role=<rdn of the enterprise role> domain_name=<Domain> realm_dn="<realm>" database_name=<SID of the target database> global_role=<name of the global role defined in the target database> dbuser=<database user> dbuser_password=<database user password> dbconnect_string=<database_host>:<port>:<DBSID> ldap_host=<oid_hostname> ldap_port=<port> ldap_user_dn="<oud admin>" ldap_user_password=<oud admin password>

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• DB function failed with error number 1 in joomla admin panel

  - by sabuj

  When i access joomla article manager or module manager then i had faced the bellow output: 500 - An error has occurred! DB function failed with error number 1 Can't create/write to file '/tmp/#sql_57c0_0.MYD' (Errcode: 17) SQL=SELECT c.*, g.name AS groupname, cc.title AS name, u.name AS editor, f.content_id AS frontpage, s.title AS section_name, v.name AS author FROM jos_content AS c LEFT JOIN jos_categories AS cc ON cc.id = c.catid LEFT JOIN jos_sections AS s ON s.id = c.sectionid LEFT JOIN jos_groups AS g ON g.id = c.access LEFT JOIN jos_users AS u ON u.id = c.checked_out LEFT JOIN jos_users AS v ON v.id = c.created_by LEFT JOIN jos_content_frontpage AS f ON f.content_id = c.id WHERE c.state != -2 ORDER BY section_name , section_name, cc.title, c.ordering LIMIT 0, 20

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• [Symfony] Admin generator and i18n

  - by David

  I have read lots of questions about i18n, but I haven't found any about performance. I have a simple backend app listing the contents of an ads table. These ads have a category, that is translated in some languages (it's defined as i18n in the Doctrine schema). So, when I add a "table_method" in my generator.yml to include de Category table, it reduces the number of queries, but there are yet some of them referencing i18n translation tables. So, if I add the category Translation table to the query, it reduces even more the queries BUT it increases the processing time considerably. Why this time penalty? Just because of the translation table? And why isn't the filter using this method to avoid so many translation queries as well? I mean, if I want to filter by category, it is making one query per category to include the translation table. Why??

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• Why tinymce in django admin outputs html tags?

  - by israkir

  I have two apps using the same tinymce textarea configurations. However, while an input in an app does output the text properly, same input in another app does not output the text properly -it outputs the <p> <li> tags etc... I have exactly same django source codes for these two different apps. As I mentioned above, two apps using the same tinymce textarea. How come this could happen? Thanks.

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• How programmable detect if a user has admin rights (can run installer)

  - by Zaky

  Hi in my code I need to detect if current user has rights to install other application (other application VC2005 redist. package)... and it should work on Vista. As I understand CheckTokenMembership or IsAdministrator do not give a complete solution. Thank you

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• How to create multiple inputs in yii + Q&A admin page

  - by user1764357

  In yii, i am developing page for Creating Question with multiple options using forms. So i have two tables Question QuestionId Question optionId Option OptionId Option To create multiple options, option field should be provided with add button so that after clicking add button, it should display new textbox control to get the new option. All these options displayed in a gridview. So can you please help me. I am very new to yii.

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• add new records using signal in django admin

  - by ganesh

  I've a model called broadcastinfo, It has fields viz.. info,userid...userid is excluded. when i add an new info, my broadcastinfo table should get the records of all userid from user table and the given message. Im trying this via signal.Any idea is highly appreciated. Thanks

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• Default http/admin port in dropwizard project

  - by mithrandir

  I have a dropwizard project and I have maintained a config.yml file at the ROOT of the project (basically at the same level as pom.xml). Here I have specified the HTTP port to be used as follows: http: port:9090 adminPort:9091 I have the following code in my TestService.java file public class TestService extends Service<TestConfiguration> { @Override public void initialize(Bootstrap<TestConfiguration> bootstrap) { bootstrap.setName("test"); } @Override public void run(TestConfiguration config, Environment env) throws Exception { // initialize some resources here.. } public static void main(String[] args) throws Exception { new TestService().run(new String[] { "server" }); } } I expect the config.yml file to be used to determine the HTTP port. However the app always seems to start with the default ports 8080 and 8081. Also note that I am running this from eclipse. Any insights as to what am I doing wrong here ?

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• Django and ImageField Question

  - by Hellnar

  Hello I have a such model: Foo (models.Model): slug = models.SlugField(unique=True) image = models.ImageField(upload_to='uploads/') I want to do two things with this: First of all, I want my image to be forced to resize to a specific width and height after the upload. I have tried this reading the documentation but seems to getting error: image = models.ImageField(upload_to='uploads/', height_field=258, width_field=425) Secondly, when adding an item via admin panel, I want my image's file name to be renamed as same as slug, if any issue arises (like if such named image already exists, add "_" to the end as it used to do. IE: My slug is i-love-you-guys , uploaded image such have i-love-you-guys.png at the end.

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• Inline editing of ManyToMany relation in Django

  - by vorpyg

  After working through the Django tutorial I'm now trying to build a very simple invoicing application. I want to add several Products to an Invoice, and to specify the quantity of each product in the Invoice form in the Django admin. Now I've to create a new Product object if I've got different quantites of the same Product. Right now my models look like this (Company and Customer models left out): class Product(models.Model): description = models.TextField() quantity = models.IntegerField() price = models.DecimalField(max_digits=10,decimal_places=2) tax = models.ForeignKey(Tax) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) invoice_no = models.IntegerField() invoice_date = models.DateField(auto_now=True) due_date = models.DateField(default=datetime.date.today() + datetime.timedelta(days=14)) I guess the quantity should be left out of the Product model, but how can I make a field for it in the Invoice model?

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• Default value for field in Django model

  - by Daniel Garcia

  Suppose I have a model: class SomeModel(models.Model): id = models.AutoField(primary_key=True) a = models.IntegerField(max_length=10) b = models.CharField(max_length=7) Currently I am using the default admin to create/edit objects of this type. How do I set the field 'a' to have the same number as id? (default=???) Other question Suppose I have a model: event_date = models.DateTimeField( null=True) year = models.IntegerField( null=True) month = models.CharField(max_length=50, null=True) day = models.IntegerField( null=True) How can i set the year, month and day fields by default to be the same as event_date field?

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• Can't set up image upload in Django

  - by culebrón

  I can't understand what's not working here: 1) settings MEDIA_ROOT = '/var/www/satel/media' MEDIA_URL = 'http://media.satel.culebron' ADMIN_MEDIA_PREFIX = '/media/' 2) models class Photo(models.Model): id = models.AutoField(primary_key=True) name = models.CharField(max_length = 200) desc = models.TextField(max_length = 1000) img = models.ImageField(upload_to = 'upload') 3) access rights: drwxr-xrwx 3 culebron culebron 4.0K 2010-04-14 21:13 media drwxr-xrwx 2 culebron culebron 4.0K 2010-04-14 19:04 upload 4) SQL: CREATE TABLE "photos_photo" ( "id" integer NOT NULL PRIMARY KEY, "name" varchar(200) NOT NULL, "desc" text NOT NULL, "img" varchar(100) NOT NULL ); 4) run Django test server as myself. 5) result: SuspiciousOperation at /admin/photos/author/add/ Attempted access to '/var/www/satel/upload/OpenStreetMap.png' denied. Not a PIL & jpeg issue, seems not to be access rights issue. But what's wrong?

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• Django Caught an exception while rendering: No module named registration

  - by Arno Smit

  I seem to have run into a bit of an issue. I am busy creating an app, and over the last few weeks setup my server to use Git, mod_wsgi to host this app. Since deploying it, everything seems to be running smoothly however, I had to go through all my files and insert the absolute url of the project to make sure it works fine. on my local machine from registration.models import UserRegistration on server from myapp.registration.models import UserRegistration Am I doing something wrong? And this has also caused an issue for me where I cannot access my django admin interface. All i get is this: Caught an exception while rendering: No module named registration Exception Value: Caught an exception while rendering: No module named registration As far as I am concerned my app has all the relevant urls, but it does not seem to work. Thank you in advance

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• Django make model field name a link

  - by Daniel Garcia

  what Im looking to do is to have a link on the name of a field of a model. So when im filling the form using the admin interface i can access some information. I know this doesn't work but shows what i want to do class A(models.Model): item_type = models.CharField(max_length=100, choices=ITEMTYPE_CHOICES, verbose_name="<a href='http://www.quackit.com/html/codes'>Item Type</a>") Other option would be to put a description next to the field. Im not even sure where to start from.

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• Alternate datasource for django model?

  - by slypete

  I'm trying to seamlessly integrate some legacy data into a django application. I would like to know if it's possible to use an alternate datasource for a django model. For example, can I contact a server to populate a list of a model? The server would not be SQL based at all. Instead it uses some proprietary tcp based protocol. Copying the data is not an option, as the legacy application will continue to be used for some time. Would a custom manager allow me to do this? This model should behave just like any other django model. It should even pluggable to the admin interface. What do you think? Thanks, Pete

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• Django loaddata throws ValidationError: [u'Enter a valid date in YYYY-MM-DD format.'] on null=true f

  - by datakid

  When I run: django-admin.py loaddata ../data/library_authors.json the error is: ... ValidationError: [u'Enter a valid date in YYYY-MM-DD format.'] The model: class Writer(models.Model): first = models.CharField(u'First Name', max_length=30) other = models.CharField(u'Other Names', max_length=30, blank=True) last = models.CharField(u'Last Name', max_length=30) dob = models.DateField(u'Date of Birth', blank=True, null=True) class Meta: abstract = True ordering = ['last'] unique_together = ("first", "last") class Author(Writer): language = models.CharField(max_length=20, choices=LANGUAGES, blank=True) class Meta: verbose_name = 'Author' verbose_name_plural = 'Authors' Note that the dob DateField has blank=True, null=True The json file has structure: [ { "pk": 1, "model": "books.author", "fields": { "dob": "", "other": "", "last": "Carey", "language": "", "first": "Peter" } }, { "pk": 3, "model": "books.author", "fields": { "dob": "", "other": "", "last": "Brown", "language": "", "first": "Carter" } } ] The backing mysql database has the relevent date field in the relevant table set to NULL as default and Null? = YES. Any ideas on what I'm doing wrong or how I can get loaddata to accept null date values?

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• PHP Framework / Library Suggestions for a Backend

  - by frbry

  I am about to start to developing backend site of a php project. Companies and site admins will login to this site and manage their data on the project. My previous admin panel experiences were full of agony and pain. So I want to make sure that I choose correct tools for my purpose. By the way, please note, I'm not looking for scaffolding. There won't be much tables in my database. Instead, there will be complex logic between entities. I want clear seperation of markup and logic code and easy-to-use and standardized user-interface. Thank you.

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• Passing session data to ModelForm inside of ModelAdmin

  - by theactiveactor

  I'm trying to initialize the form attribute for MyModelAdmin class inside an instance method, as follows: class MyModelAdmin(admin.ModelAdmin): def queryset(self, request): MyModelAdmin.form = MyModelForm(request.user) My goal is to customize the editing form of MyModelForm based on the current session. When I try this however, I keep getting an error (shown below). Is this the proper place to pass session data to ModelForm? If so, then what may be causing this error? TypeError at ... Exception Type: TypeError Exception Value: issubclass() arg 1 must be a class Exception Location: /usr/lib/pymodules/python2.6/django/forms/models.py in new, line 185

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• Overwrite queryset which builds filter sidebar

  - by cw

  Hi, I'm writing a hockey database/manager. So I have the following models: class Team(models.Model): name = models.CharField(max_length=60) class Game(models.Model): home_team = models.ForeignKey(Team,related_name='home_team') away_team = models.ForeignKey(Team,related_name='away_team') class SeasonStats(models.Model): team = models.ForeignKey(Team) Ok, so my problem is the following. There are a lot of teams, but Stats are just managed for my Club. So if I use "list_display" in the admin backend, I'd like to modify/overwrite the queryset which builds the sidebar for filtering, to just display our home teams as a filter option. Is this somehow possible in Django? I already made a custom form like this class SeasonPlayerStatsAdminForm(forms.ModelForm): team = forms.ModelChoiceField(Team.objects.filter(club__home=True)) So now just the filtering is missing. Any ideas?

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• Not quite nested inlines?

  - by Lynden Shields

  Not quite sure what to call this, it's not quite nested inlines, but is probably related. I have a 3 level hierarchy of objects, A one-to-many B one-to-many C. Therefore, every C implicitly also belongs to an A. class A(models.Model): stuff = models.CharField("Stuff", max_length=50) class B(models.Model): a = models.ForeignKey(A) class C(models.Model): b = models.ForeignKey(B) I would like all C's that belong to an A to be listed on the admin page for A in an in-line. They do not have to show which B they belong to on the same page. Is this possible or is it the same problem as nested inlines anyway? If it's possible, how do I do it? I'm using django 1.3

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• How do I access admin shares on windows 2008 r2?

  - by Jim Geurts

  I am not able to access admin file shares on a Windows 2008 R2 box, if I'm logged in as a user who is part of the Administrators group. The only way I can access those shares is if I use the built in Administrator account. How can I configure the server to allow any administrator to access the file system via admin shares? Btw, this works with Windows 2003. By admin shares, I refer to: \192.168.1.4\c$ or \192.168.1.4\e$

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• Presentation software requires admin rights to install - any way to remove this requirement?

  - by James F

  I have some wireless presentation software which I use in a meeting room in order to allow end users to hold presentations here and wirelessly have their laptop screens showing on the TV on the wall. Unfortunately the .exe file requires admin rights to install therefore requiring that either the user requests temporary admin rights beforehand, I install it using my admin account or that we use a VGA/HDMI cable. Is there a way to remove the admin right requirement from a .exe or a .msi file so that it can be installed freely by any user? We are using XP for now but will be moving to 7 soon. Thanks James

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• Utility to grant admin rights to a user in Windows XP for few hours/days?

  - by user15660

  I have two accounts on my windows xp home desktop. The default regular user is used for everything and the 2nd user which has admin rights is used only for installations. I do this to avoid malware infestations during web browsing and limited user account is guarding against online threats to a good extend but many programs refuse to run under limited rights like revo uninstaller. many installs i run from limited user by selectin "run as" from right click context menu of the .exe file. but some apps need admin rights for certain. I use "switch user" to go to admin mode and do the install/uninstall. but the admin user has none of my preferences bookmarks setup nor has my locate32 indexing done and ready for fast search Is there a utility which I can use "run as" login in administration login and use that to grant my limited user admin rights for a limited amount of period like few hours or days? Please help. I guess MS might have closed many doors of it for fear of exploitation of the API. are there any?

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• Does admin=true and root has the same privileges on AIX?

  - by Boaz Tirosh

  Does a user in /etc/security/user with the parameter admin set to true (admin = true) has the same privileges as the root user? According to IBM (full information here): admin Defines the administrative status of the user. Possible values are: true The user is an administrator. Only the root user can change the attributes of users defined as administrators. false The user is not an administrator. This is the default value. Is there another type of user, or are admin and root the same?

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• What do the "ALL"s in the line " %admin ALL=(ALL) ALL " in Ubuntu's /etc/sudoers file stand for?

  - by sri

  What does each ALL mean? I understand that the whole line indicates that the admin group members get admininstartive privileges, but would like to know more info about the position of the ALLS and if they each refer to a different set of permissions or something like that? $sudo cat /etc/sudoers ... # User privilege Information root ALL=(ALL) ALL #... %sudo ALL=(ALL) ALL # #includedir /etc/sudoers.d #Members of the admin group may gain root privileges %admin ALL=(ALL) ALL # If it matters: OS: Ubuntu : 10.4

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