if (searchTerm.contains(".com") && searchTerm.contains("@")) {
    System.out.println("In email check");
    psmt.setString(1, "EMAIL_ID");
    psmt.setString(2, searchTerm);
} else if (numresult==true){
    System.out.println("In number check");
    psmt.setString(1, "MOBILE_NUMBER");
    psmt.setString(2, searchTerm);
} else if (uresult == true | alphanumeic == true) {
    System.out.println("In username check");
    psmt.setString(1, "USER_NAME");
    psmt.setString(2, searchTerm);
}

rs = psmt.executeQuery();

while(rs.next()) {
    HomeVo vo = new HomeVo();
    vo.setId(rs.getInt("ID"));
    vo.setFirstName(rs.getString("FIRST_NAME"));
    vo.setLastName(rs.getString("LAST_NAME"));
    vo.setEmailId(rs.getString("EMAIL_ID"));
    vo.setNumber(rs.getString("MOBILE_NUMBER"));
    vo.setQualification(rs.getString("QUALIFICATION"));
    vo.setState(rs.getString("STATE"));
    vo.setGender(rs.getString("GENDER"));
    vo.setUserName(rs.getString("USER_NAME"));
    vo.setDob(rs.getString("DOB"));
    return vo;
}

下面是我的MySQL查询。

String FETCH_USER_BY_SEARCHTERM = "SELECT * FROM SignUpTable WHERE ? = ? ";

根据用户输入的搜索词，这个逻辑检查它是否是用户名、电子邮件或移动的号码，并将其作为输入参数，控件转到特定的correct if语句，但不会返回任何值。