嘗試取得java spring專案下resources目錄xml內容的寫法

.LoadLocalXml

gist的title是認檔名「字串排序」最小的為準，建立後將其他檔案改名或新增檔案都無效

參考資料:
https://stackoverflow.com/questions/9536737/load-a-xml-file-from-the-classpath-in-a-spring-web-app
https://zetcode.com/spring/classpathresource/

AccountDaoImpl.java

    public boolean checkLogin(String name, String password) throws Exception {
      String sql = ParseXml.getSqlByName("Account.checkLogin");

      List<Object> condition = new ArrayList<Object>();
      condition.add(name);
      condition.add(password);

      return jdbcTemplate.queryForInt(sql, condition.toArray()) != 0;
    }

ParseXml.java

package website.common;

import java.io.IOException;

import org.dom4j.Document;
import org.dom4j.DocumentException;
import org.dom4j.io.SAXReader;
import org.dom4j.Element;
import org.springframework.core.io.ClassPathResource;

public class ParseXml {

  private static Document getSQLDocument() throws DocumentException, IOException {
    SAXReader reader = new SAXReader();
    // ClassPathResource填入的是resources內的路徑
    ClassPathResource cpr = new ClassPathResource("Sql.xml");

    Document document = reader.read(cpr.getURL());
    return document;
  }

  public static String getSqlByName (String sqlName) throws Exception {
    Document document = getSQLDocument();

    Element element = document.getRootElement();

    return element.elementText(sqlName).trim();
  }
}

Sql.xml

<?xml version="1.0" encoding="UTF-8"?>
<sqls>

  <Account.checkLogin>
    SELECT count(*) FROM ACCOUNT WHERE name = ? and password = ?
  </Account.checkLogin>

</sqls>