我在用
drop table <table_name>
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如果我重新创建具有相同模式和名称的表,我将返回旧数据.我应该从hdfs文件系统中删除表目录以完全摆脱数据吗?
JavaScript中是否有一个函数返回传递给它的类型名称的默认值?例如:
var defaultValue = built_in_getDefaultValue("number");
console.log(defaultValue); // Logs the number 0 to the console.
Run Code Online (Sandbox Code Playgroud) 我正在尝试使用jconsole监视远程进程,这是我使用的命令
jconsole -debug localhost:4080
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这是我得到的堆栈跟踪
java.rmi.ConnectIOException: error during JRMP connection establishment; nested exception is:
java.io.EOFException
at sun.rmi.transport.tcp.TCPChannel.createConnection(TCPChannel.java:286)
at sun.rmi.transport.tcp.TCPChannel.newConnection(TCPChannel.java:184)
at sun.rmi.server.UnicastRef.newCall(UnicastRef.java:322)
at sun.rmi.registry.RegistryImpl_Stub.lookup(Unknown Source)
at sun.tools.jconsole.ProxyClient.checkSslConfig(ProxyClient.java:217)
at sun.tools.jconsole.ProxyClient.<init>(ProxyClient.java:110)
at sun.tools.jconsole.ProxyClient.getProxyClient(ProxyClient.java:463)
at sun.tools.jconsole.JConsole$3.run(JConsole.java:510)
Caused by: java.io.EOFException
at java.io.DataInputStream.readByte(DataInputStream.java:250)
at sun.rmi.transport.tcp.TCPChannel.createConnection(TCPChannel.java:228)
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如果我将进程的pid提供给jconsole,它就能成功附加.
我试图使用jackson将yaml文件解析为模型
型号 -
public class TestModel {
private String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
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解析
private static ObjectMapper mapper = new ObjectMapper(new YAMLFactory());
TestModel obj = mapper.readValue(yamlFile, TestModel.class);
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YAML文件
name: somename
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例外
Caused by: com.fasterxml.jackson.core.JsonParseException: Unrecognized token 'name': was expecting 'null', 'true', 'false' or NaN
[error] at [Source: /Users/amruth.s/Documents/Analytics/hydra/reportManager/target/scala-2.9.1/classes/viz/LineGraph.yaml; line: 1, column: 6]
[error] at com.fasterxml.jackson.core.JsonParser._constructError(JsonParser.java:1369)
[error] at com.fasterxml.jackson.core.base.ParserMinimalBase._reportError(ParserMinimalBase.java:599)
[error] at com.fasterxml.jackson.core.json.UTF8StreamJsonParser._reportInvalidToken(UTF8StreamJsonParser.java:2449)
[error] at com.fasterxml.jackson.core.json.UTF8StreamJsonParser._reportInvalidToken(UTF8StreamJsonParser.java:2426)
[error] at …
Run Code Online (Sandbox Code Playgroud) 我有这样的网格面板
Ext.define('sCon.view.SalesGrid',{
extend: 'Ext.grid.GridPanel',
title: 'Sales Data',
//other params
initComponent: function(){
this.callParent(arguments);
}
});
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在点击事件上,我想更改此面板的标题.我在控制器内的代码看起来像这样.
Ext.define('sCon.controller.SalesGridController', {
extend: 'Ext.app.Controller',
views: ['SalesGrid'],
// other handlers
changeTitle: function(newTitle){
this.getView('SalesGrid').setTitle('title_changed');
}
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目前它说它没有setTitle()的功能.但是文档说网格面板有setTitle()函数.我也尝试使用'title'变量更改标题
changeTitle: function(newTitle){
this.getView('SalesGrid').title = 'title_changed';
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既不起作用..请帮忙.
我想从我的 yaml 文件中加载 POJO 列表。我可以使用蛇 yaml 来做到这一点吗?
我的 yaml 文件看起来像这样——
- pty1:val1
pty2:val2
- pty1:val1
pty2:val2
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还有我的 pojo -
class pojo {
String pty1;
String pty2;
}
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Snakeyaml 文档确实说要使用 loadAs() ,但这会加载单个元素。我可以使用自动类型绑定加载此类元素的列表吗?
谢谢!