我们可以替换 Java 中开头和结尾有符号的字符串的一部分来识别吗?

Can we Replace a portion of String in Java which has symbol in the start and end to Identify?

我有一个字符串 SELECT *FROM USERS WHERE ID = '@userid@' AND ROLE = '@role@' 现在我已经用实际值替换了@...@ 之间的任何字符串。

预期输出SELECT *FROM USERS WHERE ID = '4' AND ROLE = 'Admin'

这个替换将从一个方法中发生,我已经写了这个逻辑

public String replaceQueryKeyWithValueFromKeyValues(String query, int reportId) {
    try {
        REPMReportDao repmReportDao = new REPMReportDao();
        int Start = 0;
        int end;
        if (query.contains("@")) {
            boolean specialSymbolFound = false;
            for (int i = 0; i < query.length(); i++) {
                if (query.charAt(i) == '@') {
                    if (!specialSymbolFound) {
                        Start = i + 1;
                        specialSymbolFound = true;
                    } else {
                        specialSymbolFound = false;
                        end = i;
                        query = query.replace(query.substring(Start - 1, end + 1), repmReportDao.getReportManagerKeyValue(query.substring(Start - 1, end + 1).replaceAll("@", ""), reportId));

                    }
                }
            }
            return query;
        } else {
            return query;
        }

    } catch (Exception e) {
        logger.log(Priority.ERROR, e.getMessage());
        return e.getMessage();
    }
}

它工作正常,但如果存在单个“@”符号而不是开始和结束,它将失败。 喜欢:

SELECT  *FROM USERS WHERE emailid = 'xyz@gmail.com' AND ROLE = '@role@'

在这里它应该替换唯一的角色“@role@”并且应该保留电子邮件原样。

预期输出 => SELECT *FROM USERS WHERE emailid = 'xyz@gmail.com' AND ROLE = 'Admin'

使用字符串替换生成数据库查询被认为是非常 不好的做法,因为您的代码容易受到 SQL 注入攻击。我无法从您提供的小代码示例中看出,但绝大多数大型 Java 项目都使用 Spring Framework, which allows you to use either JdbcTemplate or (my preference) NamedParameterJdbcTemplate。两者都允许您以安全的方式替换变量。

包含 getReportManagerKeyValue:

返回的模拟数据的完整示例
import java.util.HashMap;
import java.util.Map;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Whosebug54842971 {

    private static Map<String, String> map;

    public static void main(String[] args) {
        // preparing test data
        map = new HashMap<>();
        map.put("role", "Admin");
        map.put("userid", "666");

        // original query string
        String query = "SELECT * FROM USERS WHERE ID = '@userid@' AND emailid = 'xyz@gmail.com' AND ROLE = '@role@' ";

        // regular expression to match everything between '@ and @' with capture group
        // omitting single quotes
        Pattern p = Pattern.compile("'(@[^@]*@)'");
        Matcher m = p.matcher(query);
        while (m.find()) {
            // every match will be replaced with value from getReportManagerKeyValue
            query = query.replace(m.group(1), getReportManagerKeyValue(m.group(1).replaceAll("@", "")));
        }
        System.out.println(query);
    }

    // you won't need this function
    private static String getReportManagerKeyValue(String key) {
        System.out.println("getting key " + key);
        if (!map.containsKey(key)) {
            return "'null'";
        }
        return map.get(key);
    }
}