我有一个班级:
class Node{
private Node parent;
private List<Node> children;
...
}
如何使用Apache POI将其树的项目导出为excel以获取此类文档(我只需要移动表格中的第一列):
A
B
C
D
E
F
G
答案 0 :(得分:1)
一个简单的解决方案是创建一个NodeWriter
类,它基本上将Node写入Excel电子表格:
import java.io.FileOutputStream;
import java.io.IOException;
import org.apache.poi.ss.usermodel.Cell;
import org.apache.poi.ss.usermodel.Row;
import org.apache.poi.xssf.usermodel.XSSFSheet;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
public class NodeWriter {
public void write(Node tree, String filePathName) {
XSSFWorkbook workbook = new XSSFWorkbook();
XSSFSheet sheet = workbook.createSheet("Tree");
writeHelp(0, 1, tree, sheet);
try (FileOutputStream outputStream = new FileOutputStream(filePathName)) {
workbook.write(outputStream);
workbook.close();
} catch (IOException e) {
e.printStackTrace();
}
}
private void writeHelp(int indent, int rowNum, Node tree, XSSFSheet sheet) {
if (sheet.getRow(rowNum) != null) {
writeHelp(indent, rowNum+1, tree, sheet);
} else {
Row row = sheet.createRow(rowNum);
Cell cell = row.createCell(indent);
cell.setCellValue(tree.getNodeName());
for (Node child : tree.getChildren()) {
writeHelp(indent + 1, rowNum + 1, child, sheet);
}
}
}
}
我对你的Node
课做了一些假设。此解决方案可确保您创建新行并且不会覆盖现有行(如果if
中没有writeHelp
循环那样)。
答案 1 :(得分:0)
我的解决方案 - 合并。祝你今天愉快。感谢。
看起来像这样: https://docs.oracle.com/cd/E36352_01/epm.1112/disclosure_mgmt_admin/new_files/image002.jpg
我的合并解决方案:
private int createHierarchy(Sheet sheet, Node node, int currentRowIdx, int nodeLevel) {
if(node.getParent() == null){
sheet.setColumnWidth(8, 1000);
Row row = sheet.createRow(currentRowIdx);
row.createCell(nodeLevel).setCellValue(node.getName());
row.createCell(9).setCellValue(node.getValue());
sheet.addMergedRegion(new CellRangeAddress(currentRowIdx, currentRowIdx, nodeLevel, 8));
nodeLevel++;
}
for (Node node : node.getChildren()) {
Row row = sheet.createRow(++currentRowIdx);
row.createCell(nodeLevel).setCellValue(node.getName());
row.createCell(9).setCellValue(node.getValue());
sheet.addMergedRegion(new CellRangeAddress(currentRowIdx, currentRowIdx, nodeLevel, 8));
currentRowIdx = createHierarchy(sheet, node, currentRowIdx, nodeLevel+1);
}
return currentRowIdx;
}