in query "_latin" is getting concatenated before every param
java code :
PreparedStatement prepStmt = null;
QueryString = "SELECT FROM Student WHERE username=? AND password=? ";
prepStmt=con.prepareStatement(QueryString);
prepStmt.setString(1,un);
prepStmt.setString(2,pwd);
ResultSet rs;
System.out.println(prepStmt);
rs = prepStmt.executeQuery();
error :
java.sql.SQLException: [MySQL][ODBC 5.1 Driver][mysqld-5.5.24-log]You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM Student WHERE username=_latin1'wow' AND password=_latin1'hell'' at line 1
why its happening , whats the solution
and this is table :
You are missing the column list from your query. You have to either specify explicitly the columns you want, or retrieve all via wildcard. Instead of:
SELECT FROM Student WHERE username = ? AND password = ?
You want something like:
SELECT * FROM Student WHERE username = ? AND password = ?
Note that its almost always better to explicitly define the columns you want to retrieve, rather than using the wildcard expression.
Related
I am trying to delete the record from my "student" table, the table has two column rollno and student name.
I am using a PreparedStatement but I am getting some error. I could not understand the error. The error is:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1
here is my code
String query = "delete from student where rollno = ?";
PreparedStatement pst = con.prepareStatement(query);
pst.setInt(1, 7);
pst.executeUpdate(query);
Likely you are calling the execute like this:
pst.executeUpdate(query);
This will execute the raw query, without the parameter you set before.
You want to execute the prepared query instead, so just use:
pst.executeUpdate();
I have a preparedStatement for select query in mySQL.
this is what I wrote:
String sQuery = "SELECT Password FROM test WHERE Email = ?";
st = DB.prepareStatement(sQuery);
st.setString(1, email);
ResultSet rs = st.executeQuery(sQuery);
but i'm getting an exception from the glassfish server that says:
Severe: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException:
You have an error in your SQL syntax; check the manual that corresponds
to your MySQL server version for the right syntax to use near '?' at line 1
I don't understand what is the problem.. all the samples i saw, use that syntax..
You must call st.executeQuery(), without the query as argument. The query has already been passed to the statement when it was prepared.
See http://docs.oracle.com/javase/7/docs/api/java/sql/PreparedStatement.html#executeQuery%28%29
st.executeQuery() i.e. no param to executeQuery
So I am using SQL with phpMyAdmin. Now I want to make an update to my database with a prepared statement but doind it gives me the following error:
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'autor='Lol' WHERE id=44' at line 1
And this is how my statement looks like:
command = connection.prepareStatement("UPDATE books SET name=? author=? WHERE id=?");
command.setString(1, name.getText());
command.setString(2, author.getText());
command.setInt(3, IDx);
command.execute();
Wat is wrong with the statement I have made? It should be working In my opinion.
You need to separate the updated fields in a comma like:
command = connection.prepareStatement("UPDATE books SET name=?, author=? WHERE id=?");
You forgot a comma before author
UPDATE books
SET name = ?, author = ?
WHERE id = ?
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.
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.