[Okta Webinar] Learn how to a build a cloud-first strategyRegister Now

x
?
Solved

C# MySqlDataReader missing one result

Posted on 2014-03-13
9
Medium Priority
?
550 Views
Last Modified: 2014-03-20
I am missing a result from my query.

using MySql.Data.MySqlClient;

public List<int> getNodeIDOfServerLike(String nodeName) {
         String fetchString = "SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName";
         Hashtable tempTable = new Hashtable();
         tempTable.Add("?nodeName", nodeName);
         openConnection();
         MySqlCommand cmd = createMySQLCommand(fetchString, tempTable);
         tempTable.Clear();
         MySqlDataReader dataReader = cmd.ExecuteReader();
         List<int> resultList = new List<int>();
         Console.WriteLine("Running: " + cmd.CommandText);
         while (dataReader.Read()) {
            Console.WriteLine("Got: " + dataReader["NODE_ID"].ToString());
            if (!String.IsNullOrEmpty(dataReader["NODE_ID"].ToString())) resultList.Add(int.Parse(dataReader["NODE_ID"].ToString()));
         }
         dataReader.Close();
         closeConnection();
         return resultList;
}

Open in new window


internal MySqlCommand createMySQLCommand(String commandText, Hashtable parameters) {
         if(SqlConnection != null) {
            MySqlCommand cmd = SqlConnection.CreateCommand();
            cmd.CommandText = commandText;
            cmd.Prepare();
            foreach (DictionaryEntry entry in parameters) {
               cmd.Parameters.Add((String)entry.Key, entry.Value);
            }
            return cmd;
         }
         return null;
      }

Open in new window


Result is:
Running: SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName
Got: 189

When I run the the same query directly agains the mysql server, I get two results:
189
457

That would be correct. But when I run the code above, I only get the first result. So what am I doing wrong here?
0
Comment
Question by:itnifl
  • 5
  • 3
9 Comments
 
LVL 45

Assisted Solution

by:AndyAinscow
AndyAinscow earned 750 total points
ID: 39925996
case sensitive vs insensitive ?
0
 
LVL 2

Author Comment

by:itnifl
ID: 39926029
It doesn't look that way. It doesn't matter what case I supply the nodeName in. Also I get the other ID(457)  if I supply the whole node name for that ID.
0
 
LVL 2

Accepted Solution

by:
itnifl earned 0 total points
ID: 39926089
Actually, this is the fix:
tempTable.Add("?nodeName", nodeName + "%");

Open in new window

0
Visualize your virtual and backup environments

Create well-organized and polished visualizations of your virtual and backup environments when planning VMware vSphere, Microsoft Hyper-V or Veeam deployments. It helps you to gain better visibility and valuable business insights.

 
LVL 45

Assisted Solution

by:AndyAinscow
AndyAinscow earned 750 total points
ID: 39926130
>>Actually, this is the fix:

Odd - I don't see why that would make any difference between the code and the direct approach (unless you didn't use the identical SQL statement that is when testing it directly)
0
 
LVL 75

Assisted Solution

by:käµfm³d 👽
käµfm³d   👽 earned 250 total points
ID: 39926838
Why are you using a question mark in your parameter name? I've always seen an @ symbol used.

i.e.

http://dev.mysql.com/doc/connector-net/en/connector-net-tutorials-parameters.html
0
 
LVL 2

Author Comment

by:itnifl
ID: 39928865
When I test the query directly I write as follows:
SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE 'OHTHATSERVERNAME%';

When I in the code use:
SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName%
it won't accept ?nodeName as a parameter.

I don't remember or know why I use a question mark as a parameter mark, this code was written some time ago. I am guessing the choice of a parameter mark is optional?

Obviously the wild card has to be treated as a part of the parameter value. I can't say that this is very obvious or intuitive.
0
 
LVL 45

Assisted Solution

by:AndyAinscow
AndyAinscow earned 750 total points
ID: 39929004
>>When I test the query directly I write as follows:
SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE 'OHTHATSERVERNAME%';
When I in the code use:
>>SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName%


That isn't what you posted in the question - there was no trailing % in the code based SQL query.

So, basically you got different results because you used different queries.
0
 
LVL 2

Author Comment

by:itnifl
ID: 39931191
This does not work in the code:
SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName%

And this does not work in the code:
SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName

However, this works:
SELECT NODE_ID FROM server_info_table WHERE NODE_NAME LIKE ?nodeName
and then adding the % with the parameter value:
tempTable.Add("?nodeName", nodeName + "%");

In the question I am not seeking why there are differences, I am asking how to get the result I want(all results from the query). The answer is to supply % with the parameter value, for some reason.
0
 
LVL 2

Author Closing Comment

by:itnifl
ID: 39941781
Tanks for participating! :)
0

Featured Post

Get quick recovery of individual SharePoint items

Free tool – Veeam Explorer for Microsoft SharePoint, enables fast, easy restores of SharePoint sites, documents, libraries and lists — all with no agents to manage and no additional licenses to buy.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

In this article, I’ll look at how you can use a backup to start a secondary instance for MongoDB.
What we learned in Webroot's webinar on multi-vector protection.
Video by: Steve
Using examples as well as descriptions, step through each of the common simple join types, explaining differences in syntax, differences in expected outputs and showing how the queries run along with the actual outputs based upon a simple set of dem…
Despite its rising prevalence in the business world, "the cloud" is still misunderstood. Some companies still believe common misconceptions about lack of security in cloud solutions and many misuses of cloud storage options still occur every day. …
Suggested Courses

872 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question