[HIVE-18268] Hive Prepared Statement when split with double quoted in query fails

• Type: Bug
• Status: Patch Available
• Priority: Major
• Resolution: Unresolved
• Affects Version/s: 2.3.2
• Component/s: JDBC
• Target Version/s: 2.4.0, 3.0.0
• Assignee: Choi JaeHwan
• Reporter: Choi JaeHwan
• Created: 13/Dec/17 09:15
• Updated: 29/Mar/18 06:21

Description

HIVE-13625, Change sql statement split when odd number of escape characters, and add parameter counter validation, above

// prev code
    StringBuilder newSql = new StringBuilder(parts.get(0));
    for(int i=1;i<parts.size();i++){
      if(!parameters.containsKey(i)){
        throw new SQLException("Parameter #"+i+" is unset");
      }
      newSql.append(parameters.get(i));
      newSql.append(parts.get(i));
    }

// change from HIVE-13625
    int paramLoc = 1;
    while (getCharIndexFromSqlByParamLocation(sql, '?', paramLoc) > 0) {
      // check the user has set the needs parameters
      if (parameters.containsKey(paramLoc)) {
        int tt = getCharIndexFromSqlByParamLocation(newSql.toString(), '?', 1);
        newSql.deleteCharAt(tt);
        newSql.insert(tt, parameters.get(paramLoc));
      }
      paramLoc++;
    }

If the number of split SQL and the number of parameters are not matched, an SQLException is thrown

Currently, when splitting SQL, there is no processing for double quoted, and when the token ('?' ) is between double quote, SQL is split.

i think when the token between double quoted is literal, it is correct to not split.

for example, above the query;

// Some comments here
1:  String query =  " select 1 from x where qa="?" "
2:  String query = " SELECT 1 FROM `x` WHERE (trecord LIKE "ALA[d_?]%")

? is literal, then query do not split.

Issue Links

• Blocked: HIVE-13625 Hive Prepared Statement when executed with escape characters in parameter fails