当我试图在普通的java类中创建一个工作簿时它正在创建,但当我试图在servlet上运行相同的时候它会抛出异常
普通的java类
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import org.apache.poi.ss.usermodel.Workbook;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
public class Sampleclass {
public static void main(String[] args) {
FileInputStream file;
try {
file = new FileInputStream(new File("C:\\Users\\Rajasekhar\\Desktop\\EmployeeDetails.xlsx"));
Workbook book=new XSSFWorkbook(file);
System.out.println(book);
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
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Servlet的
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import …
Run Code Online (Sandbox Code Playgroud) 我的代码是
[...]
import org.apache.poi.poifs.filesystem.POIFSFileSystem;
import org.apache.poi.ss.usermodel.Workbook;
import org.apache.poi.xssf.usermodel.XSSFSheet;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
import org.apache.poi.hssf.usermodel.HSSFCell;
import org.apache.poi.hssf.usermodel.HSSFSheet;
import org.apache.poi.hssf.usermodel.HSSFWorkbook;
import org.apache.poi.hssf.usermodel.HSSFRow;
public class ExcelRead {
public static void main( String [] args ) {
try {
File excel = new File("Book1.xlsx");
FileInputStream fis = new FileInputStream(excel);
XSSFWorkbook book = new XSSFWorkbook(fis);
XSSFSheet sheet = book.getSheetAt(0);
Iterator rows = sheet.rowIterator();
while( rows.hasNext() ) {
HSSFRow row = (HSSFRow) rows.next();
System.out.println("\n");
Iterator cells = row.cellIterator();
while( cells.hasNext() ) {
HSSFCell cell = (HSSFCell) cells.next();
if(HSSFCell.CELL_TYPE_NUMERIC==cell.getCellType()) …
Run Code Online (Sandbox Code Playgroud) 我正在从UI实现搜索功能,我希望给出具有唯一约束的列名称下拉以及任何主键列(如果存在),因此用户可以使用任何这些选定的列关联数据进行搜索.我已搜索此类查询但尚未找到
就像是:
SELECT COLUMN_NAMEs FROM TABLE WHERE CONSTRAINTS UNIQUE OR PRIMARY
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有任何查询来实现这一点......
我正在尝试进行基本身份验证以调用休息服务.
在服务文档中,CURL提到的过程是这样的
curl -X POST https://secure.clientservice.com/api/transactions.json -u [TOKEN]:[KEY] -H 'Content-type: application/json' \
-d "{\"transaction_type\":\"request\",\"amount_in_cents\":\"3000\",\"email\":\"user@example.com\"}"
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我遇到了如何[TOKEN]:[KEY]
从chrome rest client 发送身份验证的问题.我有用户令牌和密钥,doc说我必须做Http Basic access authentication
在休息探险家我正在尝试如下
但这是给予未经授权的访问,因为似乎授权标题是错误的,我编码[axkKtfBAaPABCh59SA]:[S7RwBG2eZ3y8mDs8VS]这个值为64.我正在做正确的方式或我错过了什么.
我想要一个查询来计算列中的重复值,是否有任何查询,或者我在想
total records=SELECT COUNT(column_name) FROM table_name;
distinct records=SELECT COUNT(DISTINCT column_name) FROM table_name;
duplicate count=total records-distinct records;
Run Code Online (Sandbox Code Playgroud) 这里的字段是硬编码的,但我想为我的任务动态获取字段,就像我将在这个jsp中有一个列表,其中包含字段名称,例如:list=[id,name,salary,doj]
此列表可能会针对新请求进行更改.我能有一些想法吗?
fields: {
PersonId: {
key: true,
create: false,
edit: false,
list: false
},
Name: {
title: 'Author Name',
width: '40%'
},
Age: {
title: 'Age',
width: '20%'
},
Watch: {
title: 'Watch',
width: '20%',
display: function (data) {
return '';
},
RecordDate: {
title: 'Record date',
width: '30%',
type: 'date',
create: false,
edit: false
}
}
});
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