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How do I programmatically change file permissions?
...s a conventional format that will be more readable to many developers. For APIs that accept a FileAttribute, you can wrap the set of permissions with with PosixFilePermissions.asFileAttribute().
Set<PosixFilePermission> ownerWritable = PosixFilePermissions.fromString("rw-r--r--");
FileAttribu...
Swift - which types to use? NSString or String
... using the Swift native types unless you need to interact with objective-c APIs
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How do you get a string from a MemoryStream?
...use it, think about String.TrimStart. See: docs.microsoft.com/en-us/dotnet/api/…
– Skalli
Oct 8 '19 at 13:46
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Node.js get file extension
...owfish's answer pointed out, and not write your own. More info: nodejs.org/api/path.html#path_path_extname_p
– xentek
Feb 23 '14 at 6:20
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Python Flask, how to set content type
...swered Aug 2 '12 at 8:49
Simon SapinSimon Sapin
8,32622 gold badges2929 silver badges3939 bronze badges
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MenuItemCompat.getActionView always returns null
... is that you use the SearchView from the Support V7 package and maybe your API level is set to.....22??.
Changing your code to the following in order to fix the problem:
menu.xml
<?xml version="1.0" encoding="UTF-8" ?>
<menu xmlns:android="http://schemas.android.com/apk/res/android" >...
How to create a custom exception type in Java? [duplicate]
...le I haven't forgotten some advanced things (mostly related to good style, API design). At any rate, your post has been imminently helpful.
– jpaugh
Oct 19 '15 at 0:27
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搭建高可用mongodb集群（一）——配置mongodb - 大数据 & AI - 清泛网 - 专...
...来，开发人员学习成本会大大减少。如果再对底层的sql API做一层封装，开发基本可以感觉不到mongodb和关系型数据库的区别。同样MongoDB也是号称自己能够快速搭建一个高可用可扩展的的分布式集群，网上有很多搭建的文章，在我...
Correctly determine if date string is a valid date in that format
I'm receiving a date string from an API, and it is formatted as yyyy-mm-dd .
16 Answers
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File Upload without Form
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All answers here are still using the FormData API. It is like a "multipart/form-data" upload without a form. You can also upload the file directly as content inside the body of the POST request using xmlHttpRequest like this:
var xmlHttpRequest = new XMLHttpRequest();
...