Problem in updating a data column in spring - spring

I have a Database table called ProgramData. their i have a data column called Id and executed. id set to be as auto increment.
Table structure is like this.
What i want is according to id executed column need to be updated. following is my code segment.
public void saveDtvProgDataExecuted()
{
ProgramData programeData = new ProgramData();
String SQL = "UPDATE program_data SET executed=1 WHERE programeData.id = ?";
this.jdbcTemplate.update(SQL);
}
If i run this code this gives me error like bad SQL grammar [UPDATE program_data SET executed=1 WHERE programeData.id = ?]; nested exception is com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '?' at line 1

Problem is you’re not passing the ID value to the jdbctemplate.
You should use
this.jdbctemplate.update(SQL, id);
Where id is the id of the record you’re updating.
Please refer to the documentation for more information:
http://static.springsource.org/spring/docs/3.1.x/spring-framework-reference/htmlsingle/spring-framework-reference.html#jdbc-updates

TRY THIS statement while you are passing ? in your sql query it need to be set while execution.
String SQL = "UPDATE program_data SET executed=1 WHERE programeData.id = ?";
this.jdbcTemplate.update(SQL,new PreparedStatementCallback<Boolean>(){
#Override
public Boolean doInPreparedStatement(PreparedStatement ps)
throws SQLException, DataAccessException {
ps.setInt(1,"here you need to pass value of programeData.id);
return ps.execute();
}
});

Related

Is there a better way to execute this SQL statement in Spring using JdbcTemplate?

I have an SQL statement to select a BRAND_NAME based on an input parameter. The code goes something like this:
public ResponseEntity<List<Map<String, Object>>> getBrand(String brandName){
sql = "SELECT BRAND_NAME AS \"brandName\" FROM BRAND_E WHERE LOWER(BRAND_NAME) LIKE '%" + brandName + "%'";
return new ResponseEntity<List<Map<String, Object>>>(jdbc.queryForList(sql), HttpStatus.OK);
}
I've found out that this can probably cause SQL injection attacks, so I was wondering how to code this better.
Yes.If you directly use the request param on your sql it can lead to SQL injection attacks. We can always go with the prepared statement by adding a placeholder to where the values must be added.
Use queryForList(String sql, Object... args) or queryForList(String sql, Object[] args, Class<T> elementType)
Eg:-
String employeeId= "1";
String sql = "select id,name,address from employee where id = ?";
getJdbcTemplate(). queryForList(sql, new Object[]{employeeId}, Employee.class);

Why can JPQLs modifying queries only return void or int?

When i want to modify the database via JPQL i have to mark the query as Transactional and Modiyfing. If i do so, the return type of the method representing the query has to be either void or int(representing the number of edited rows i think). Why are only the two return types allowed? If i do a HTTP-PUT request and update the object with an own JPQL query, i would like to return the updated object again. Whats the best way to do it if the return type of the query has to be void or int? Do i have to do a seperate query/request again which selects the object after it was updated?
EDIT:
Thats how i call the query:
if (inactivityListDTO.getProjectIds().size() > 0) {
projectRepository.updateProjectsIsArchivedByProjectIds(inactivityListDTO.getProjectIds(), inactivityListDTO.getIsArchived());
}
Thats the query:
#Transactional
#Modifying
#Query("UPDATE Project project SET project.isArchived = :isArchived,
project.archivedDate = current_date " +
"WHERE project.id IN :ids")
void updateProjectsIsArchivedByProjectIds(#Param("ids") List<Long> ids, #Param("isArchived") boolean isArchived);
Because it finally boils down to execute a standard UPDATE SQL in the DB , and the UPDATE in standard SQL only returns the number of records being updated and does not return a result set.
And yes , if you need get a record 's value after update , you have to query it again. Alternatively , you should consider using a JPA way to update a record , which first query the object , then update it by changing its state . Something like below (Assume you are using spring #Transactional to manage the transactional boundary):
#Transactional
public void changeEmployeeSalary(Integer employeeId , Integer salary){
Employee employee = entityManager.find(Employee.class , employeeId);
employee.setSalary(salary);
}
In this way , you do not need to query the record again after it is updated and you also do not need to manually write a UPDATE SQL.

BatchUpdateException:ORA-00942:TABLE OR VIEW DOES NOT EXIST

This issue occurred in jdbc batch insert. I queried from an Oracle datasource, parsed the resultset and then inserted into another Oracle datasource. I have got the connect metadata and printed the current username along with url, both are invalid.
But when it went to batch update, I got the ora-00942 exception. I'm pretty sure all above works fine in database. Has anyone encountered this exception and can you give me some advice?
EDIT:
Ok, I got a table named photos for example in REMOTE_USER and I queried from it. It gave me a resultset, then I parse it after that INSERT it to LOCAL_USER.photos. I did query the LOCAL_USER.photos where I logon in from PL/SQL Developer. The interesting thing was I could do the select command but not the insert. Below is some part of code.
conn = datasource.getConnection(); // notice that it was target datasource
DatabaseMetaData connMetaData = conn.getMetaData();
String userName = connMetaData.getUserName();
resultSet = ds.getResultSet();
ResultSetMetaData metaData = resultSet.getMetaData();
int count = metaData.getColumnCount();
String insertSql = generateInsertSql(count, metaData, userName);
// this was generated through metaData , the output should be
// "insert into LOCAL_USER.photos(col1,col2) values(?,...)"
logger.error("insert clause is {}", insertSql);
ps = conn.prepareStatement(insertSql);
conn.setAutoCommit(false);
while (resultSet.next()) { // this was the original datasource
stageTotalNum++;
for (int i = 1; i <= count; i++) {
Object object = resultSet.getObject(i);
dealClobColumn(ps, i, object);
}
ps.addBatch();
if (stageTotalNum % 500L == 0L) {
ps.executeBatch(); // throws batchupdateexception.
ps.clearBatch();
conn.commit();
}
}
ps.executeBatch();
conn.commit();
It should be the blob type column which I didn't handle it the right way.
First I queried from original datasource then got the blob column of the resultset by
conn.getObject(index) . Next I insert the blob column into target datasource by conn.setObject. Of course that way wasn't working at all, so I changed to the following:
conn.setBlob(rs.getBlob(index)).
Although it worked fine in my own environemnt, but when the application ran in remote server, it kept annoying about the 'table or view does not exists'.The third version is:
conn.setBinaryStream(rs.getBlob(index).getBinaryStream());
Ok, this time it worked both my pc and remote server. Thanks to #codeLover's advice and link, it really hepled me and saved my time. Appreciated it!

Query to check if the record exists in Spring Jdbc Template

I am fairly new to spring ,I am looking to check if a certain email id exists in database or not , using Spring Jdbc Template ,I looked here but could'nt find the proper answer .I am looking something like ,SELECT count(*) from table where email=?
Any help will be appreciated.
You can do something as below if you are using jdbctemplate and new version of spring
private boolean isEmailIdExists(String email) {
String sql = "SELECT count(*) FROM table WHERE email = ?";
int count = jdbcTemplate.queryForObject(sql, new Object[] { email }, Integer.class);
return count > 0;
}
queryForObject method of jdbcTemplate accepts the sql query as the first parameter, second argument is an array of objects for the sql query place holders and the third argument is the expected return value from the sql query.
In this case we only have one place holder and hence I gave the second argument as new Object[] { email } and the result we are expecting is a count which is a Integer and hence I gave it as Integer.class
I kind of got this answer from https://www.mkyong.com/spring/jdbctemplate-queryforint-is-deprecated/
You can go through it if you are interested.
private boolean isEmailIdExists(String email) {
return jdbcTemplate.queryForObject("SELECT EXISTS(SELECT FROM table WHERE email = ?)", Boolean.class, email);
}
http://www.postgresqltutorial.com/postgresql-exists/

Use WHERE clause with instances in hql

How to select a record from a table using WHERE clause and comparing instances (patient)
public History findHistory(Patient patient) { History model=null;
Session sesion=util.HibernateUtil.getSessionFactory().getCurrentSession();
String sql="FROM History h WHERE h.patients=" + patient;
try{
sesion.beginTransaction();
model=(History) sesion.createQuery(sql).uniqueResult();
sesion.beginTransaction().commit();
}
catch(Exception e){
sesion.beginTransaction().rollback();
}
return model;
}
That throws a queryException #1562
e.queryString="FROM entidad.Historia h WHERE h.pacientes=entidad.Paciente#3ad3a221"
e.detailMessage="unexpected char: '#'"
The problem with your code is that concatenating patient like you do will just append patient.toString(), which in your case is the default implementation (i.e. classname#hashcode) and it is no use for Hibernate to find out which data to retrieve in the DB.
You need to bind the parameter, first:
String sql = "FROM History h WHERE h.patients = :patient";
Then
model = (History) sesion.createQuery(sql)
.setParameter("patient", patient)
.uniqueResult();
Edit:
SQLGrammarException: could not execute query can occurs for various reason. Try to run the generated query in SqlDeveloper (or any other tool) and see what your DB says. In your case, the last part and .=? cause the error. The cross join is Harming too. I suspect your mapping is incomplete and Hibernate can't find how to join History and Patient. Try to add something like this in History entity:
#ManyToOne
#JoinColumn(name = "patient")
private Patient patient;

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