转换列表<map<string, string> &gt; 到 Object 中 Java Spring 启动 ZF0B4A299C45171493AE3215D69D </map<string,>

Question

I have a problem how approach to convert List<Map<String, String>> to custom class objects to save it to JPA. If I apply the branchRepository.save(content) it does not work.

Here's my code:

-- BranchService.java --

public List<Map<String, String>> uploadEmployee(MultipartFile multip) throws Exception {

    String fileNames = multip.getOriginalFilename();

    DataFormatter formatter = new DataFormatter();

    File file = new File("./reports/" + fileNames);
    Workbook workbook = WorkbookFactory.create(file);

    FormulaEvaluator evaluator = workbook.getCreationHelper().createFormulaEvaluator();

    Sheet sheet = workbook.getSheetAt(0);

    int headerRowNum = sheet.getFirstRowNum();

    Map<Integer, String> colHeaders = new HashMap<Integer, String>();
    Row row = sheet.getRow(headerRowNum);
    for (Cell cell : row) {
        int colIdx = cell.getColumnIndex();
        String value = formatter.formatCellValue(cell, evaluator);
        colHeaders.put(colIdx, value);
    }

    List<Map<String, String>> content = new ArrayList<Map<String, String>>();
    for (int r = headerRowNum + 1; r <= sheet.getLastRowNum(); r++) {
        row = sheet.getRow(r);
        if (row == null)
            row = sheet.createRow(r);
        Map<String, String> valuesToHeaders = new HashMap<String, String>();
        for (Map.Entry<Integer, String> entry : colHeaders.entrySet()) {
            int colIdx = entry.getKey();
            Cell cell = row.getCell(colIdx);
            if (cell == null)
                cell = row.createCell(colIdx);
            String cellValue = formatter.formatCellValue(cell, evaluator);
            valuesToHeaders.put(entry.getValue(), cellValue);
        }
        content.add(valuesToHeaders);
    }

    workbook.close();

    System.out.println(content);

    return content;
}

How to convert and apply it to JPA?

Answer 1

If the Map, as element of the resulting List, contains your entity attributes mapped to its values, you can Use Gson in order to create an instance of your JPA entity:

Gson gson = new Gson();
JsonElement jsonElement = gson.toJsonTree(map);
MyEntity pojo = gson.fromJson(jsonElement, MyEntity.class);

Answer 2

Instead of producing something so generic as a List of Map s, directly return your JPA entities instead.

So turn this:

List<Map<String, String>> content = new ArrayList<Map<String, String>>();
for (int r = headerRowNum + 1; r <= sheet.getLastRowNum(); r++) {
    row = sheet.getRow(r);
    if (row == null)
        row = sheet.createRow(r);
    Map<String, String> valuesToHeaders = new HashMap<String, String>();
    for (Map.Entry<Integer, String> entry : colHeaders.entrySet()) {
        int colIdx = entry.getKey();
        Cell cell = row.getCell(colIdx);
        if (cell == null)
            cell = row.createCell(colIdx);
        String cellValue = formatter.formatCellValue(cell, evaluator);
        valuesToHeaders.put(entry.getValue(), cellValue);
    }
    content.add(valuesToHeaders);
}

Into something more like this:

List<Branch> content = new ArrayList<>();
for (int r = headerRowNum + 1; r <= sheet.getLastRowNum(); r++) {
    row = sheet.getRow(r);
    if (row == null)
        continue; //SKIP, don't bother creating empty stuff!
    Branch branch = new Branch();
    for (Map.Entry<Integer, String> entry : colHeaders.entrySet()) {
        int colIdx = entry.getKey();
        Cell cell = row.getCell(colIdx);
        if (cell != null) {
            String cellValue = formatter.formatCellValue(cell, evaluator);
            switch(entry.getValue()) {
                 case "Description": {
                      branch.setDescription(cellValue);
                      break;
                 }
                 case "name": //example with multiple headers mapping to same field
                 case "Label": {
                      branch.setLabel(cellValue);
                       break;
                 }
            }
            //alternatively use if-else block with regex matching or some other technique to map your headers to JPA entity fields
        }
    }
    content.add(branch);
}

Using switch you can map multiple different spellings or abbreviations (if your specification allows for that sort of thing), but for ultimate flexibility you could do if/else with regular expressions. It would also be possible to do away with headers altogether and make this index based ( switch on colIdx ).

Of course, you could keep your code as-is and write the above into a new convert function, eg:

List<Branch> convert(List<Map<String,String>> content) {
    return content.stream().map(this::convert).collect(Collectors.toList());
}
Branch convert(Map<String,String> props) {
    Branch branch = new Branch();
    for(String key : props.keySet()) {
        String value = props.get(key);
        switch(key) {
             case "Description": {
                  branch.setDescription(value);
                  break;
             }
             case "name": //example with multiple headers mapping to same field
             case "Label": {
                  branch.setLabel(value);
                  break;
             }
        }

    }
    return branch;
}

That might be the tidiest option (although you'll still want to remove the empty row and cell creation code).