'PSQLException: ERROR: column does not exist Position: 8

org.postgresql.util.PSQLException: ERROR: column "feedbackid" does not exist Position: 8

I am getting this error but unable to understand what the reason behind this. It shows:- **

org.postgresql.util.PSQLException: ERROR: column "feedbackid" does not exist Hint: Perhaps you meant to reference the column "feedback.feedbackId". Position: 8

**

Database table in postgres:

create table `company`.`feedback` (

feedbackId int(10) NOT NULL, feedbackActionId int(12) NOT NULL, description varchar(200) DEFAULT NULL, feedbackText varchar(2000) DEFAULT NULL, email varchar(100) NOT NULL, createdDate date NOT NULL );

database table in postgres

public Feedback getFeedbackById(int id) throws SQLException, ClassNotFoundException {
    conn = DBConnection.setDBConnection();
    String sqlQuery = "select feedbackId, feedbackActionId, description, feedbackText, email, createdDate" +
                           "from feedback " + 
                           " where feedbackId = " + id ;
    DBConnection dbConn = new DBConnection();
    ResultSet resultSet = dbConn.getResultSet(sqlQuery, conn);
    int feedbackId = resultSet.getInt("feedbackId");
    int feedbackActionId = resultSet.getInt("feedbackActionId");
    String description = resultSet.getString("description");
    String feedbackText = resultSet.getString("feedbackText");
    String email = resultSet.getString("email");
    Date createdDate = resultSet.getDate("createdDate");
    feedback = new Feedback(feedbackId, feedbackActionId, description, feedbackText, email, createdDate);

    resultSet.close();

    return feedback;
}

Thanks in advance.

javaeclipserestjerseypostgresql-9.1


Solution 1:[1]

This question might help you. Looks like Postgres is case senstive towards column names. You might need to include field name in quotes. Are PostgreSQL column names case-sensitive?

Solution 2:[2]

Please Try to use: @GeneratedValue(strategy = GenerationType.IDENTITY) in your POJO class Id Field. This thing solved my error.

Solution 3:[3]

sometimes occurs when you use double quotes instead of single quotes for strings replace where user_id="random_id" with where user_id='random_id'

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Ravi Chandra
Solution 2 Raj Shukla
Solution 3 procrastinator

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