首页 \ 问答 \ 什么关闭AbstractSerializer中的OutputStream?(what closes OutputStream in AbstractSerializer?)

什么关闭AbstractSerializer中的OutputStream?(what closes OutputStream in AbstractSerializer?)

我有自己的类扩展org.apache.cocoon.serialization.AbstractSerializer (非常基本)

public class ExcelSerializer extends AbstractSerializer {

private static final XLogger LOG = XLoggerFactory.getXLogger(ExcelSerializer.class);

private ExcelSheetCreator excelSheetCreator;

public ExcelSerializer() {
    try {
        excelSheetCreator = new ExcelSheetCreator();
    } catch (IOException e) {
        LOG.error("ERROR", e.getMessage());
    }
}



@Override
public void startElement(String uri, String loc, String raw, Attributes a) throws SAXException {
    excelSheetCreator.startElement(uri, loc, raw, a);
}

@Override
public void characters(char[] c, int start, int len) throws SAXException {
    excelSheetCreator.characters(c, start, len);
}

@Override
public void endElement(String uri, String loc, String raw) throws SAXException {
    excelSheetCreator.endElement(uri, loc, raw);
}


@Override
public void endDocument() throws SAXException {
    excelSheetCreator.setOutputStream(this.output);
    excelSheetCreator.endDocument();

}

}

第一次尝试一切正常,我得到了预期的输出。 从第二个开始,我从ExcelSerializer调用的类会抛出IOException,因为该流已经关闭。

@Override
public void endDocument() throws SAXException {
    try {
        workbook.write(outputStream);
    } catch (IOException e) {
        System.out.println("workbook");
        System.out.println("" + e.getMessage());
    }
    workbook.dispose();
}

我当然没有关闭Outputstream,或者至少没有故意关闭。 我该怎么办才能保持开放状态?

这是我的站点地图:

 <map:serializer  name="excelSerializer" logger="sitemap.serializer.excelSerializer" src="com.acrys.excel.ExcelSerializer" 
           mime-type="application/vnd.openxmlformats-officedocument.spreadsheetml.sheet">
         </map:serializer>

I have my own class extending org.apache.cocoon.serialization.AbstractSerializer (quite basic)

public class ExcelSerializer extends AbstractSerializer {

private static final XLogger LOG = XLoggerFactory.getXLogger(ExcelSerializer.class);

private ExcelSheetCreator excelSheetCreator;

public ExcelSerializer() {
    try {
        excelSheetCreator = new ExcelSheetCreator();
    } catch (IOException e) {
        LOG.error("ERROR", e.getMessage());
    }
}



@Override
public void startElement(String uri, String loc, String raw, Attributes a) throws SAXException {
    excelSheetCreator.startElement(uri, loc, raw, a);
}

@Override
public void characters(char[] c, int start, int len) throws SAXException {
    excelSheetCreator.characters(c, start, len);
}

@Override
public void endElement(String uri, String loc, String raw) throws SAXException {
    excelSheetCreator.endElement(uri, loc, raw);
}


@Override
public void endDocument() throws SAXException {
    excelSheetCreator.setOutputStream(this.output);
    excelSheetCreator.endDocument();

}

}

For the first attempt everything is OK, I got the expected output. From the second one onwards, the class that I call from ExcelSerializer throws an IOException, because the stream has already been closed.

@Override
public void endDocument() throws SAXException {
    try {
        workbook.write(outputStream);
    } catch (IOException e) {
        System.out.println("workbook");
        System.out.println("" + e.getMessage());
    }
    workbook.dispose();
}

I certainly did not close the Outputstream or at least not knowingly. What can I do to keep it open?

Here is my sitemap:

 <map:serializer  name="excelSerializer" logger="sitemap.serializer.excelSerializer" src="com.acrys.excel.ExcelSerializer" 
           mime-type="application/vnd.openxmlformats-officedocument.spreadsheetml.sheet">
         </map:serializer>

原文:https://stackoverflow.com/questions/22357338
更新时间:2023-10-17 15:10

最满意答案

其他意见是正确的。 你不能像这样映射一个对象。 有多种方法可以迭代对象属性:

for(let prop in object) {
  console.log(object[prop]);
}

然而,在你的例子中,这些对象没有任何相似之处,所以你不会从迭代它们中获得任何好处。

相反,如果您想显示访问令牌,则需要执行以下操作:

return (
   <Text>{this.state.userdetail.token.access_token}</Text>
);

The other comments are correct. You cannot map over an object like this. There are ways you can iterate over object properties:

for(let prop in object) {
  console.log(object[prop]);
}

However, in your example the objects don't have any similarities so you would not get any benefit from iterating over them.

If instead, you wanted to display the access token, you would do the following:

return (
   <Text>{this.state.userdetail.token.access_token}</Text>
);

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