尝试打开Excel工作簿时,尝试/ Catch不会激活
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了尝试打开Excel工作簿时,尝试/ Catch不会激活相关的知识,希望对你有一定的参考价值。
使用apache-POI打开excel工作簿,如果不存在则创建一个工作簿。由于某种原因,正在打开的工作簿已损坏,导致在使用错误注释注释的行上发生错误。
不知怎的,这部分代码的try / catch似乎没有激活。任何想法,以及我如何正确处理这些错误?另外,有没有办法在我的if(file.exists() && file.length() != 0) {
条件期间检查文件的完整性?
public XSSFWorkbook OpenWB(String directory, String name) {
File file = new File(directory + "\" + name + ".xlsx");
FileInputStream fIP;
if(file.exists() && file.length() != 0) {
try {
fIP = new FileInputStream(file);
//Get the workbook instance for XLSX file
workbook = new XSSFWorkbook(fIP); //*********error occurs here**********
fIP.close();
System.out.println(name + ".xlsx file open successfully.");
return workbook;
} catch (IOException e) {
e.printStackTrace();
System.out.println("Error to open " + name + ".xlsx file, creating blank");
//Create Blank workbook
workbook = new XSSFWorkbook();
Integer i = 0;
while (file.isFile() && file.exists()) {
name = name.concat(i.toString());
file = new File(directory + "\" + name + ".xlsx");
i++;
}
return workbook;
}
} else {
System.out.println("Error to open " + name + ".xlsx file, creating blank");
//Create Blank workbook
workbook = new XSSFWorkbook();
return workbook;
}
}
答案
尝试这段代码,它会给你一个错误“打开random.xlx文件时出错,创建空白”,这意味着你的try catch正在运行。你似乎忘了初始化你的变量“工作簿”。
package stackoverflow;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
public class Solution {
public XSSFWorkbook OpenWB(String directory, String name) {
File file = new File(directory + "\" + name + ".xlsx");
FileInputStream fIP;
XSSFWorkbook workbook;
if(file.exists() && file.length() != 0) {
try {
fIP = new FileInputStream(file);
//Get the workbook instance for XLSX file
workbook = new XSSFWorkbook(fIP); //*********error occurs here**********
fIP.close();
System.out.println(name + ".xlsx file open successfully.");
return workbook;
} catch (IOException e) {
e.printStackTrace();
System.out.println("Error to open " + name + ".xlsx file, creating blank");
//Create Blank workbook
workbook = new XSSFWorkbook();
Integer i = 0;
while (file.isFile() && file.exists()) {
name = name.concat(i.toString());
file = new File(directory + "\" + name + ".xlsx");
i++;
}
return workbook;
}
} else {
System.out.println("Error to open " + name + ".xlsx file, creating blank");
//Create Blank workbook
workbook = new XSSFWorkbook();
return workbook;
}
}
public static void main(String args[]) {
Solution s = new Solution();
s.OpenWB("D://", "random.xlx");
}
}
您可以根据需要修改Solution类部件。
以上是关于尝试打开Excel工作簿时,尝试/ Catch不会激活的主要内容,如果未能解决你的问题,请参考以下文章
打开 openpyxl 保存的工作簿时 Excel 有不可读的内容