I am working on a mini application and I have to retrieve and save USER_ID in a table.
It works like this; I input the email address of the user and then the statement has to retrieve USER_ID and save it into a table.
I am having problem with my SQL Like statement. Its just that I have to insert only the first part of the email address. For example, "thomas#yahoo.com". I insert only "thomas"; not remaining part.
Here are sample of the statement and resulting message I am getting in my Java console:
Code:
String ea = txt_email.getText();
String lid = "SELECT USER_ID FROM user WHERE email_address Like '"+ea+"%' ";
Error Message:
Error: java.sql.SQLException: [Microsoft][ODBC Microsoft Access Driver] Syntax error (missing operator) in query expression ''SELECT USER_ID FROM user WHERE email_address Like 'thomas%' ''.
Try using * instead of %. MS ACCESS uses * instead of % for the LIKE statement.
Hope this helps
use PreparedStatement
PreparedStatement _query = con.prepareStatement(
"SELECT USER_ID FROM user WHERE email_address Like ?");
_query.setString(1, ea + "*"); // <== use *
_query.executeQuery();
where con is your active connection object.
One advantage of using PreparedStatement is protect from SQL Injection
Related
I understand how to:
Query JSON in Postgres
How to query JSON data from postgresql in sqlalchemy
how to use clojure.java.jdbc
But I don't know how to put it together and query json with clojure.java.jdbc.
For example, I have a table user, and a field user_info
CREATE TABLE user (id SERIAL,user_info JSON);
then I found this blog to impl some protocol,and it insert success!
(insert! conn :yxt_user {:user_info {:hello [1 "test" true]}})
But I don't know how to write code to query it like this sql from jdbc/query
SELECT * FROM user WHERE data ? 'hello';
not by jdbc/execute direct write sql like
(jdbc/execute! db-spec ["UPDATE table SET col1 = NOW() WHERE id = ?" 77])
I tried to write this query
(jdbc/query conn ["SELECT * FROM user WHERE user_info ? 'hello'"])
I got this
org.postgresql.util.PSQLException: 未设定参数值 1 的内容。
Then I tried
(jdbc/query conn ["SELECT * FROM user WHERE user_info ? 'hello'" "?"])
I got this
org.postgresql.util.PSQLException: ERROR: syntax error at or near "$1"
How do I write a query to filter on a JSON column where user_info has the JSON key hello?
If you have the latest postgresql driver you can escape the ? with a double ??
This should work:
(jdbc/query conn ["SELECT * FROM user WHERE user_info ?? 'hello'"])
For some reason when my java code attempts to execute a query that is using a variable I receive a syntax error.
String query = "SELECT userName, email, address FROM users,requests"
+ "WHERE requestingUser = userID AND rideID = ?";
ps = connection.prepareStatement(query);
ps.setInt(1, rideID);
results = ps.executeQuery()
This code produces the following MySQL error:
"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 '= userID AND rideID = 34' at
line 1"
I have tried this both with and without using PreparedStatments, and I only receive an error for those queries for which I pass in a variable. Simple queries like "SELECT field FROM table" work fine. I feel like I am going insane, I appreciate any help I can get.
You are missing a space between requests and WHERE
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.
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.