JDBC Select from 2 columns within the same table - java

I've been having hard times finding the appropriate syntax for a prepared statement.
This is what I currently have:
String query = "SELECT * FROM TABLE1" + "WHERE Col1="+val1+ "AND Col2="+val2;
Can you please tell me what the actual syntax is, as I keep getting an SQL syntax error?
Thanks :)

There is no space between your table name and where clause.
String query = "SELECT * FROM TABLE1 WHERE Col1="+val1+ " AND Col2="+val2;
And if Col1 and COl2 are varchar then single quotes before and after val1 and val2.
String query = "SELECT * FROM TABLE1 WHERE Col1='"+val1+ "' AND Col2='"+val2+"'";
And better using parameters instead of giving values directly. It helps prevent SQL Injection attacks

Related

Java Regex: To know the number of rows to be returned by a SQL Query

I have a SQL query and I want to know how many rows will that SQL query return. Now the problem is that I want to know the number of results beforehand which means before running the SQL query.
I would have done this easily by ResultSet.getRow() to get the total number of rows from resultset. But as per the requirement, I can get the resultset only after knowing the number of rows to be returned by that query.
I tried the below Java Regex to solve the issue:
String orgQuery = "select * from emp where id<1210 and salary>55000;"
Pattern p= Pattern.compile("(?:)from\\s+(.*)*" , Pattern.CASE_INSENSITIVE);
Matcher m= p.matcher(orgQuery);
if (m.find()) {
countQuery = "SELECT COUNT(*) as total "+ m.group(1);
System.out.println(countQuery);
}
This work perfectly file and I get the "countQuery" as:
SELECT COUNT(*) as total from emp where id<1210 and salary>55000
By this I can easily know the number of rows to be returned beforehand but the problem occurs when my query become more complex like these two:--
even more complex in case of nested queries i.e. #query2.
#query1: select * from emp where id<1210 and salary>55000 order by dept, salary desc;
#query2: select name from emp where id IN (select id from emp where id < 1210 group by salary , id order by id ASC limit 10) order by id DESC limit 10
I think the main issue is with "Order By" clause. I can remove the "Order By" clause too by below regex:
Pattern.compile("(?:)from\\s+(.*)*" , Pattern.CASE_INSENSITIVE);
But it becomes more complex in case of Nested queries.
Can any Java Regex expert help????? I am using postgres as DB.
Wrap your existing query like so:
select count(*) from (<existing query>)
With your given example:
String orgQuery = "select * from emp where id<1210 and salary>55000";
String countQuery = "select count (*) from (" + orgQuery + ')';
I know this works with Oracle. I have not used postgres, so I am not certain if there would be anything preventing this approach from working there.
I will caution on this idea of getting a count first, however, that it might be possible for the data to change between your execution of the count and the actual query.

UCASE and UPPER sql functions

I am trying to do the following query:
String query = "SELECT * FROM EMP WHERE UCASE(LAST_NAME) ";
query += "LIKE '" + lastName.toUpperCase() + "%'";
in an example of usage of an servlet to access to a database
But I am getting the error message:
Excepcion java.sql.SQLSyntaxErrorException: ORA-00904: "UCASE": invalid identifier
On the other hand, when I use the UPPER sql function, the example works but the results do not show the values of the LASTNAME column in uppercase. I do not understand what happens.
You're just comparing the upper case values, but you're selecting the actual values with select *
to get the uppercase name in your resultset you need to use UPPER in your select list, not UCASE, like this:
String query = "SELECT UPPER(LAST_NAME) AS UPPERNAME, * FROM EMP WHERE UPPER(LAST_NAME) ";
query += "LIKE '" + lastName.toUpperCase() + "%'";
What your code is doing here is building a query string named query. Once query is complete, it will be sent to the database for parsing and running.
When you are building a query to the database, you have to use the built-in database functions for the part of the query that the database is going to parse and run. So, in your example, Java is doing toUpperCase on lastName and then putting that literal into the query string that will go to the database. UPPER(LAST_NAME) is going into the query string as is, it will get passed to the database just like that and run by the database. So it needs to be a function that the database can parse and run: an Oracle function, not a Java function.
UCASE is a DB2 function & not Oracle. For Oracle, you need to use UPPER .
Second part of your question is already answered by James Z.
Having said that, I am answering because previous answers didn't pointed out SQL injection problem with the way you listed your query.
Make it a habit to always execute parametrized queries with jdbc & not by directly appending values to query string.
String query = "SELECT * FROM EMP WHERE UCASE(LAST_NAME) LIKE ? ";
Your parameter would be - lastName.toUpperCase()+"%"
SQL Injection

Java and MYSQL Syntax Issue

I'm trying to insert data into my MYSQL databse. I want to insert an int into the database which I have no problem doing. However, I want to INSERT INTO (VALUES) WHERE. I get a MYSQL syntax error when I try this.
I can INSERT and SELECT WHERE as long as they are in two seperate statements. Here is my code:
String query = ("INSERT INTO `accounts` (inventory) " + "VALUES ('"
+ Inventory.inventory + "') WHERE username='" + Frame.username
+ "' and password = '" + Frame.password + "'");
Basically, an INSERT statement can not have a WHERE clause. I am thinking that you want to UPDATE a certain record, eg
UPDATE accounts
SET inventory = 'valueHere'
WHERE userName = 'userHEre' AND password = 'passHere'
The only time an INSERT statement can have a WHERE clause is when you are inserting records from the result of a SELECT statement, eg
INSERT INTO tableName (col1, ..., colN)
SELECT col1, ..., colN
FROM table2
// WHERE ..your conditions here..
As a sidenote, your current coding style is vulnerable with SQL Injection. Consider using PreparedStatement.
Basic example of a PreparedStatement
String updateString = "UPDATE accounts SET inventory = ? WHERE userName = ? AND password = ?";
PreparedStatement updateStmt = con.prepareStatement(updateString);
updateStmt.setString(1, Inventory.inventory);
updateStmt.setString(2, Frame.username);
updateStmt.setString(3, Frame.password);
updateStmt.executeUpdate();
JDBC PreparedStatement
MySQL INSERT Syntax does not support the WHERE clause so that's why you have a syntax issue. Maybe you're looking for an UPDATE :
UPDATE [LOW_PRIORITY] [IGNORE] tbl_name
SET col_name1=expr1 [, col_name2=expr2 ...]
[WHERE where_definition]
[ORDER BY ...]
[LIMIT row_count]
Not a direct answer but more of a best practice....
You should avoid doing this type of string concatenation for any sql. You vulnerable to sql injection and it does not scale well. Instead you should look at using JdbcTemplates or NamedJdbcTemplate using the opensource spring framework.
The WHERE is not applicable in INSERT INTO Syntax. You want insert a new row in the table, and you should add the username and password as well as Inventory.inventory in VALUES set.

Intermittently getting "sqlexception invalid column index" [duplicate]

I got the following error while testing some code:
SQLException: Invalid column index
What exactly does that mean?
Is there an online document explaining what all the Oracle error codes and statements?
If that's a SQLException thrown by Java, it's most likely because you are trying to get or set a value from a ResultSet, but the index you are using isn't within the range.
For example, you might be trying to get the column at index 3 from the result set, but you only have two columns being returned from the SQL query.
It sounds like you're trying to SELECT a column that doesn't exist.
Perhaps you're trying to ORDER BY a column that doesn't exist?
Any typos in your SQL statement?
Using Spring's SimpleJdbcTemplate, I got it when I tried to do this:
String sqlString = "select pwy_code from approver where university_id = '123'";
List<Map<String, Object>> rows = getSimpleJdbcTemplate().queryForList(sqlString, uniId);
I had an argument to queryForList that didn't correspond to a question mark in the SQL. The first line should have been:
String sqlString = "select pwy_code from approver where university_id = ?";
I also got this type error, problem is wrong usage of parameters to statement like, Let's say you have a query like this
SELECT * FROM EMPLOYE E WHERE E.ID = ?
and for the preparedStatement object (JDBC) if you set the parameters like
preparedStatement.setXXX(1,value);
preparedStatement.setXXX(2,value)
then it results in SQLException: Invalid column index
So, I removed that second parameter setting to prepared statement then problem solved
Just try this fix, as I faced your error:
Remove the single quotation marks around your question mark, which means, if you used your reserved parameters like ('?','?','?') you should make it look like this:
(?,?,?)
I had this problem using a prepared statement. I didn't add enough "?" for the "VALUES" My eclipse had crashed after I did add the proper amount, and lost those changes. But that didn't occur to me to be the error until I started combing through the SQL as p.campbell suggested.
I had the exact same problem when using Spring Security 3.1.0. and Oracle 11G. I was using the following query and getting the invalid column index error:
<security:jdbc-user-service data-source-ref="dataSource"
users-by-username-query="SELECT A.user_name AS username, A.password AS password FROM MB_REG_USER A where A.user_name=lower(?)"
It turns out that I needed to add: "1 as enabled" to the query:
<security:jdbc-user-service data-source-ref="dataSource" users-by-username query="SELECT A.user_name AS username, A.password AS password, 1 as enabled FROM MB_REG_USER A where A.user_name=lower(?)"
Everything worked after that. I believe this could be a bug in the Spring JDBC core package...
the final sql statement is something like:
select col_1 from table_X where col_2 = 'abcd';
i run this inside my SQL IDE and everything is ok.
Next, i try to build this statement with java:
String queryString= "select col_1 from table_X where col_2 = '?';";
PreparedStatement stmt = con.prepareStatement(queryString);
stmt.setString(1, "abcd"); //raises java.sql.SQLException: Invalid column index
Although the sql statement (the first one, ran against the database) contains quotes around string values, and also finishes with a semicolumn, the string that i pass to the PreparedStatement should not contain quotes around the wildcard character ?, nor should it finish with semicolumn.
i just removed the characters that appear on white background
"select col_1 from table_X where col_2 = ' ? ' ; ";
to obtain
"select col_1 from table_X where col_2 = ?";
(i found the solution here: https://coderanch.com/t/424689/databases/java-sql-SQLException-Invalid-column)
I had this problem in one legacy application that create prepared statement dynamically.
String firstName;
StringBuilder query =new StringBuilder("select id, name from employee where country_Code=1");
query.append("and name like '");
query.append(firstName + "' ");
query.append("and ssn=?");
PreparedStatement preparedStatement =new prepareStatement(query.toString());
when it try to set value for ssn, it was giving invalid column index error, and finally found out that it is caused by firstName having ' within; that disturb the syntax.

What does the following Oracle error mean: invalid column index

I got the following error while testing some code:
SQLException: Invalid column index
What exactly does that mean?
Is there an online document explaining what all the Oracle error codes and statements?
If that's a SQLException thrown by Java, it's most likely because you are trying to get or set a value from a ResultSet, but the index you are using isn't within the range.
For example, you might be trying to get the column at index 3 from the result set, but you only have two columns being returned from the SQL query.
It sounds like you're trying to SELECT a column that doesn't exist.
Perhaps you're trying to ORDER BY a column that doesn't exist?
Any typos in your SQL statement?
Using Spring's SimpleJdbcTemplate, I got it when I tried to do this:
String sqlString = "select pwy_code from approver where university_id = '123'";
List<Map<String, Object>> rows = getSimpleJdbcTemplate().queryForList(sqlString, uniId);
I had an argument to queryForList that didn't correspond to a question mark in the SQL. The first line should have been:
String sqlString = "select pwy_code from approver where university_id = ?";
I also got this type error, problem is wrong usage of parameters to statement like, Let's say you have a query like this
SELECT * FROM EMPLOYE E WHERE E.ID = ?
and for the preparedStatement object (JDBC) if you set the parameters like
preparedStatement.setXXX(1,value);
preparedStatement.setXXX(2,value)
then it results in SQLException: Invalid column index
So, I removed that second parameter setting to prepared statement then problem solved
Just try this fix, as I faced your error:
Remove the single quotation marks around your question mark, which means, if you used your reserved parameters like ('?','?','?') you should make it look like this:
(?,?,?)
I had this problem using a prepared statement. I didn't add enough "?" for the "VALUES" My eclipse had crashed after I did add the proper amount, and lost those changes. But that didn't occur to me to be the error until I started combing through the SQL as p.campbell suggested.
I had the exact same problem when using Spring Security 3.1.0. and Oracle 11G. I was using the following query and getting the invalid column index error:
<security:jdbc-user-service data-source-ref="dataSource"
users-by-username-query="SELECT A.user_name AS username, A.password AS password FROM MB_REG_USER A where A.user_name=lower(?)"
It turns out that I needed to add: "1 as enabled" to the query:
<security:jdbc-user-service data-source-ref="dataSource" users-by-username query="SELECT A.user_name AS username, A.password AS password, 1 as enabled FROM MB_REG_USER A where A.user_name=lower(?)"
Everything worked after that. I believe this could be a bug in the Spring JDBC core package...
the final sql statement is something like:
select col_1 from table_X where col_2 = 'abcd';
i run this inside my SQL IDE and everything is ok.
Next, i try to build this statement with java:
String queryString= "select col_1 from table_X where col_2 = '?';";
PreparedStatement stmt = con.prepareStatement(queryString);
stmt.setString(1, "abcd"); //raises java.sql.SQLException: Invalid column index
Although the sql statement (the first one, ran against the database) contains quotes around string values, and also finishes with a semicolumn, the string that i pass to the PreparedStatement should not contain quotes around the wildcard character ?, nor should it finish with semicolumn.
i just removed the characters that appear on white background
"select col_1 from table_X where col_2 = ' ? ' ; ";
to obtain
"select col_1 from table_X where col_2 = ?";
(i found the solution here: https://coderanch.com/t/424689/databases/java-sql-SQLException-Invalid-column)
I had this problem in one legacy application that create prepared statement dynamically.
String firstName;
StringBuilder query =new StringBuilder("select id, name from employee where country_Code=1");
query.append("and name like '");
query.append(firstName + "' ");
query.append("and ssn=?");
PreparedStatement preparedStatement =new prepareStatement(query.toString());
when it try to set value for ssn, it was giving invalid column index error, and finally found out that it is caused by firstName having ' within; that disturb the syntax.

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