Question

I have 2 tables (Item and Code) with relation one to many(one Item can have many Codes – this is implemented on db level with foreign key).

When I am trying to insert one Item record and several Code records(they are linked to that Item record) in one transaction using spring JDBC I get

com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException: Cannot add or update a child row: a foreign key constraint fails (`mrbcodes`.`code`, CONSTRAINT `code_item_fk` FOREIGN KEY (`itemId`) REFERENCES `item` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION)

Here is my insert into db code.

public void saveCode(Code code, long itemId) {
  String insertSql = "insert into code(number, value, itemId) values (:number,:value,:itemId)";
  SqlParameterSource namedParameters = new BeanPropertySqlParameterSource(code);
  log.info("Inserting values into code table; number:" + code.getNumber() +"; value:" + code.getValue() 
    + "; itemId:" + itemId);
  simpleJdbcTemplate.update(insertSql, namedParameters);
 }

 @Transactional
 public void saveCodes(Collection <Code> codes) {
  KeyHolder keyHolder = new GeneratedKeyHolder();
  insertNewItem(keyHolder);  
  Number key = keyHolder.getKey();
  Assert.notNull(key, "After item insert; returned key must not be null");
  for (Code code:codes) {
   saveCode(code, key.longValue());
  }
 }

 private KeyHolder insertNewItem(KeyHolder keyHolder) {
  String createItemSQL = "insert into item(timeUsed) values(:defaultValue)";  
  SqlParameterSource sqlParameterSource = new MapSqlParameterSource("defaultValue", defaultValue);
  log.debug("Inserting new item");
  simpleJdbcTemplate.getNamedParameterJdbcOperations().update(createItemSQL,sqlParameterSource, keyHolder);
  log.debug("Item inserted; keyHolder " + keyHolder);
  return keyHolder;
 }

Answer 1

You don’t use itemId in saveCode, therefore Code is saved without itemId.