赞赏良好的链接.
警告remote: warning: unable to access '/root/.config/git/attributes': Permission denied意味着什么,它带来了什么影响？
$git clone git://git.eclipse.org/gitroot/egit/egit.git
Cloning into 'egit'...
remote: warning: unable to access '/root/.config/git/attributes': Permission denied
remote: Counting objects: 57926, done.
remote: Compressing objects: 100% (11872/11872), done.
remote: Total 57926 (delta 30734), reused 56308 (delta 29136)
Receiving objects: 100% (57926/57926), 32.29 MiB | 1021 KiB/s, done.
Resolving deltas: 100% (30734/30734), done.
Checking out files: 100% (1483/1483), done.
我应该向托管人报告潜在的基础设施问题吗？
计算机:Mac OS X,版本10.8数据库:Postgres
试图将csv文件导入postgres.
pg> copy items_ordered from '/users/darchcruise/desktop/items_ordered.csv' with CSV;
ERROR: could not open file "/users/darchcruise/desktop/items_ordered.csv" for reading: Permission denied
然后我试了一下
$> chown postgres /users/darchcruise/desktop/items_ordered.csv
chown: /users/darchcruise/desktop/items_ordered.csv: Operation not permitted
最后,我试过了
$> ls -l
-rw-r--r-- 1 darchcruise staff 1016 Oct 18 21:04 items_ordered.csv
任何帮助深表感谢!
当我在程序下面运行时,我会遇到异常
java.io.InvalidClassException: Files.SerializationMain; Files.SerializationMain; no valid constructor
at java.io.ObjectStreamClass.checkDeserialize(Unknown Source)
at java.io.ObjectInputStream.readOrdinaryObject(Unknown Source)
at java.io.ObjectInputStream.readObject0(Unknown Source)
at java.io.ObjectInputStream.readObject(Unknown Source)
at Files.SerializationClass.main(SerializationClass.java:71)
Caused by: java.io.InvalidClassException: Files.SerializationMain; no valid constructor
at java.io.ObjectStreamClass.<init>(Unknown Source)
at java.io.ObjectStreamClass.lookup(Unknown Source)
at java.io.ObjectOutputStream.writeObject0(Unknown Source)
at java.io.ObjectOutputStream.writeObject(Unknown Source)
at Files.SerializationClass.main(SerializationClass.java:61)
我读到某个地方,当我们序列化任何子类时,它的基类构造函数被触发.
class Parent123
{
int age;
String name;
Parent123(int age,String name) {
System.out.println("We are in Parent123 Constructor");
this.age=age;
this.name=name;
}
}
class SerializationMain extends Parent123 implements Serializable {
int data1;
String data2;
SerializationMain(int data1,String data2)
{
super(20,"test");
this.data1=data1; …是否有更优雅的方式来写下面的内容？
try {
... // Some throwing code
return first
}
catch {
case e:ExceptionType => {} // No code to execute. Ignore error.
}
return second
我尝试insmod一个由我移植的linux内核遗留模块.出现以下错误:
> sudo insmod camac-mx.ko
insmod: error inserting 'camac-mx.ko': -1 Invalid module format
dmesg |tail -n 1
[1312783.938299] camac_mx: disagrees about version of symbol module_layout
我该如何解决？
我需要打印一个包含scala.Long的格式化字符串.java.lang.String.format()与scala.Long(编译时)和RichLong(java.util.IllegalFormatConversionException)不兼容
编译器警告在以下工作代码上弃用Integer:
val number:Long = 3243
String.format("%d", new java.lang.Long(number))
我应该改变fomatter,数据类型还是其他什么？
Main(函数main在那里)我的程序的线程保留用于非GUI任务.它调用了许多冗长的计算功能.所有已实现的GUI都在一个单独的线程中完成它们的工作.
我现在要使用Qt实现一个GUI.Qt文档说所有GUI相关的任务都应该在主线程中完成.在我的例子中,在主线程中偶尔插入QCoreApplication :: processEvents()调用几乎没用,因为它们之间有很大的延迟.
有没有办法克服Qt的这种约束？是不可能在Qt程序的主线程中做一些非GUI相关的事情？
如何按对象的字段对对象的JSONArray进行排序？
输入:
[
{ "ID": "135", "Name": "Fargo Chan" },
{ "ID": "432", "Name": "Aaron Luke" },
{ "ID": "252", "Name": "Dilip Singh" }
];
期望的输出(按"名称"字段排序):
[
{ "ID": "432", "Name": "Aaron Luke" },
{ "ID": "252", "Name": "Dilip Singh" }
{ "ID": "135", "Name": "Fargo Chan" },
];