authority

Android Fileprovider: IllegalArgumentException: Failed to find configured root that contains

由 依然范特西╮ 提交于 2020-01-01 04:12:05
问题 I have a question about the android FileProvider. I want to save a pdf document and open it with a default program. I don´t want to save it in external Storage. After I´ve successfully saved the pdf to the FilesDirectory/export/temp.pdf, I´ve tried to generate an URI by using FileProvider.getUriForFile(). File path = new File(getFilesDir(), "export"); File pdf = new File(path + File.separator + "temp.pdf"); pdf.getParentFile().mkdirs(); if (!pdf.exists()) pdf.createNewFile(); Uri uri =

android duplicate provider authority

由 纵然是瞬间 提交于 2019-12-23 08:11:12
问题 I have had users complaining of a "duplicate provider authority" error when trying to install my android application. I initially thought it might be an old version interfering with the new version, but they have never installed the app before. Any suggestions as to what might be causing this? 回答1: Somebody else used the same provider authority as you. AFAIK, that's the only way this error can occur. What did you choose as your provider authority? 来源： https://stackoverflow.com/questions

Alfresco Creating a new user group that permits only creating user and groups

由 烈酒焚心 提交于 2019-12-20 03:15:32
问题 I want to create a new user group . The members in this user group can create new users and new groups. I think creating users or groups require admin permission. This user group does not have other admin permission except creating, editing and deleting user,group. Does alfresco provide this kind of group in alfresco community 4.0c? How can I implement it? How to add a new group with specified permission? I really appreciate for help. 回答1: You could add a custom role in permissionDefinitions

Laravel Roles & Permissions with Authority

由 我只是一个虾纸丫 提交于 2019-12-11 12:41:08
问题 From my research, I have found that the Authority package (https://github.com/machuga/authority-l4) is best for implementing a role/permissions based user auth system while maintaining flexibility. I am having trouble understanding exactly how to use this package. The documentation covers it's functions and configuration, but does not explain a few things. Was hoping someone could point me in the right direction. What is the purpose of the config file? To specify permissions? Are these not

Alfresco Creating a new user group that permits only creating user and groups

由 半城伤御伤魂 提交于 2019-12-02 00:05:22
I want to create a new user group . The members in this user group can create new users and new groups. I think creating users or groups require admin permission. This user group does not have other admin permission except creating, editing and deleting user,group. Does alfresco provide this kind of group in alfresco community 4.0c? How can I implement it? How to add a new group with specified permission? I really appreciate for help. You could add a custom role in permissionDefinitions.xml but I don't think that would help in this instance, since the ability to create users and groups plus

What is the semantics of the double slash following the scheme in a URI?

由 允我心安 提交于 2019-11-30 04:48:25
问题 According to http://tools.ietf.org/html/rfc3986 and http://en.wikipedia.org/wiki/Uniform_resource_identifier, a URI may or may not contain a double slash following the scheme identifier. This makes "urn:issn:1535-3613" a valid URI just as "http://stackoverflow.com". Is there a strict/formal need to include the double slash or is it optional and in any case, what is the reason/semantics? When answering, please provide a conclusive answer - Don't just report how you browser/library/... handles

Android - Having Provider authority in the app project

由 会有一股神秘感。 提交于 2019-11-27 11:58:04
An android library project contains a few providers whose authority is defined like the following in a contract class : public static final String CONTENT_AUTHORITY = "my.com.library.providers.tester"; private static final Uri BASE_CONTENT_URI = Uri.parse("content://" + CONTENT_AUTHORITY); Now there are a lot of app projects which uses this library project. The problem I am currently having is that for every app project, I need to have a separate branch in the library project for every app just for having a unique content authority. This is creating some version management problems (like

.NET - Get protocol, host, and port

由 倾然丶 夕夏残阳落幕 提交于 2019-11-26 19:27:12
Is there a simple way in .NET to quickly get the current protocol, host, and port? For example, if I'm on the following URL: http://www.mywebsite.com:80/pages/page1.aspx I need to return: http://www.mywebsite.com:80 I know I can use Request.Url.AbsoluteUri to get the complete URL, and I know I can use Request.Url.Authority to get the host and port, but I'm not sure of the best way to get the protocol without parsing out the URL string. Any suggestions? Rick The following (C#) code should do the trick Uri uri = new Uri("http://www.mywebsite.com:80/pages/page1.aspx"); string requested = uri

Android - Having Provider authority in the app project

由 佐手、 提交于 2019-11-26 15:49:32
问题 An android library project contains a few providers whose authority is defined like the following in a contract class : public static final String CONTENT_AUTHORITY = "my.com.library.providers.tester"; private static final Uri BASE_CONTENT_URI = Uri.parse("content://" + CONTENT_AUTHORITY); Now there are a lot of app projects which uses this library project. The problem I am currently having is that for every app project, I need to have a separate branch in the library project for every app

.NET - Get protocol, host, and port

由 混江龙づ霸主 提交于 2019-11-26 06:57:49
问题 Is there a simple way in .NET to quickly get the current protocol, host, and port? For example, if I\'m on the following URL: http://www.mywebsite.com:80/pages/page1.aspx I need to return: http://www.mywebsite.com:80 I know I can use Request.Url.AbsoluteUri to get the complete URL, and I know I can use Request.Url.Authority to get the host and port, but I\'m not sure of the best way to get the protocol without parsing out the URL string. Any suggestions? 回答1: The following (C#) code should do
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