OleDb returning System.Data.OleDb.OleDbException: 'No value given for one or more required parameters.' when...











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I keep getting System.Data.OleDb.OleDbException: 'No value given for one or more required parameters.' but i am giving it a value as you can see in my code:



private void PopWeapSel(string read)
{
string weapID = read;
string weapDam = "";
string weapPat = "";
string weapHeat = "";
string weapName = "";
OleDbConnection dbConn = new System.Data.OleDb.OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=BHShooterProjectDB.accdb");
string sql = "SELECT ItemWeaponType.WeaponID, ItemWeaponType.WeaponDamage, ItemWeaponType.WeaponPattern, ItemWeaponType.WeaponHeatValue, ItemWeaponType.WeaponCoolRate, ItemWeaponType.WeaponName FROM ItemWeaponType WHERE ItemWeaponTypeWeaponID = @PU;";
OleDbCommand cmd = new OleDbCommand(sql, dbConn);
foreach (string WeapID in weapID)
{
if (WeapID == "")
{
return;
}
cmd.Parameters.AddWithValue("@PU", WeapID); ///Problem Line
dbConn.Open();
OleDbDataReader reader = cmd.ExecuteReader();

if (reader.Read())
{
weapDam = reader[1].ToString();
weapPat = reader[2].ToString();
weapHeat = reader[3].ToString();
weapName = reader[4].ToString();
}
dbConn.Close();
this.DGV_WS.Rows.Add(weapDam, weapPat, weapHeat, weapName);
}
}


When i look at the locals in the locals in debug. WeapID has a value of "1" as it should. Have i missed something in terms of syntax?



Thanks!










share|improve this question
























  • I also suspect this code is going to have issues on the second iteration of the foreach - since you will have added two parameters (with the same name) to a single OleDbCommand. Did you mean to instantiate the OleDbCommand inside the loop?
    – mjwills
    Nov 8 at 11:30










  • The most obvious cause of this error is that WeapID is null. The behaviour you are seeing would be expected in that scenario.
    – mjwills
    Nov 8 at 11:31










  • @mjwills if i type "?WeapID" in the immediate window it returns "1". Also moved OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each. Still get the same error.
    – Blackclaw _
    Nov 8 at 16:55










  • ItemWeaponTypeWeaponID is missing a .. This is a typo. Voting to close.
    – mjwills
    Nov 8 at 20:13










  • @mjwills Ah thank you so much! Good spot! Looked all the way through my SQL about 10 times.......
    – Blackclaw _
    Nov 8 at 20:39















up vote
-1
down vote

favorite












I keep getting System.Data.OleDb.OleDbException: 'No value given for one or more required parameters.' but i am giving it a value as you can see in my code:



private void PopWeapSel(string read)
{
string weapID = read;
string weapDam = "";
string weapPat = "";
string weapHeat = "";
string weapName = "";
OleDbConnection dbConn = new System.Data.OleDb.OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=BHShooterProjectDB.accdb");
string sql = "SELECT ItemWeaponType.WeaponID, ItemWeaponType.WeaponDamage, ItemWeaponType.WeaponPattern, ItemWeaponType.WeaponHeatValue, ItemWeaponType.WeaponCoolRate, ItemWeaponType.WeaponName FROM ItemWeaponType WHERE ItemWeaponTypeWeaponID = @PU;";
OleDbCommand cmd = new OleDbCommand(sql, dbConn);
foreach (string WeapID in weapID)
{
if (WeapID == "")
{
return;
}
cmd.Parameters.AddWithValue("@PU", WeapID); ///Problem Line
dbConn.Open();
OleDbDataReader reader = cmd.ExecuteReader();

if (reader.Read())
{
weapDam = reader[1].ToString();
weapPat = reader[2].ToString();
weapHeat = reader[3].ToString();
weapName = reader[4].ToString();
}
dbConn.Close();
this.DGV_WS.Rows.Add(weapDam, weapPat, weapHeat, weapName);
}
}


When i look at the locals in the locals in debug. WeapID has a value of "1" as it should. Have i missed something in terms of syntax?



Thanks!










share|improve this question
























  • I also suspect this code is going to have issues on the second iteration of the foreach - since you will have added two parameters (with the same name) to a single OleDbCommand. Did you mean to instantiate the OleDbCommand inside the loop?
    – mjwills
    Nov 8 at 11:30










  • The most obvious cause of this error is that WeapID is null. The behaviour you are seeing would be expected in that scenario.
    – mjwills
    Nov 8 at 11:31










  • @mjwills if i type "?WeapID" in the immediate window it returns "1". Also moved OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each. Still get the same error.
    – Blackclaw _
    Nov 8 at 16:55










  • ItemWeaponTypeWeaponID is missing a .. This is a typo. Voting to close.
    – mjwills
    Nov 8 at 20:13










  • @mjwills Ah thank you so much! Good spot! Looked all the way through my SQL about 10 times.......
    – Blackclaw _
    Nov 8 at 20:39













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I keep getting System.Data.OleDb.OleDbException: 'No value given for one or more required parameters.' but i am giving it a value as you can see in my code:



private void PopWeapSel(string read)
{
string weapID = read;
string weapDam = "";
string weapPat = "";
string weapHeat = "";
string weapName = "";
OleDbConnection dbConn = new System.Data.OleDb.OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=BHShooterProjectDB.accdb");
string sql = "SELECT ItemWeaponType.WeaponID, ItemWeaponType.WeaponDamage, ItemWeaponType.WeaponPattern, ItemWeaponType.WeaponHeatValue, ItemWeaponType.WeaponCoolRate, ItemWeaponType.WeaponName FROM ItemWeaponType WHERE ItemWeaponTypeWeaponID = @PU;";
OleDbCommand cmd = new OleDbCommand(sql, dbConn);
foreach (string WeapID in weapID)
{
if (WeapID == "")
{
return;
}
cmd.Parameters.AddWithValue("@PU", WeapID); ///Problem Line
dbConn.Open();
OleDbDataReader reader = cmd.ExecuteReader();

if (reader.Read())
{
weapDam = reader[1].ToString();
weapPat = reader[2].ToString();
weapHeat = reader[3].ToString();
weapName = reader[4].ToString();
}
dbConn.Close();
this.DGV_WS.Rows.Add(weapDam, weapPat, weapHeat, weapName);
}
}


When i look at the locals in the locals in debug. WeapID has a value of "1" as it should. Have i missed something in terms of syntax?



Thanks!










share|improve this question















I keep getting System.Data.OleDb.OleDbException: 'No value given for one or more required parameters.' but i am giving it a value as you can see in my code:



private void PopWeapSel(string read)
{
string weapID = read;
string weapDam = "";
string weapPat = "";
string weapHeat = "";
string weapName = "";
OleDbConnection dbConn = new System.Data.OleDb.OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=BHShooterProjectDB.accdb");
string sql = "SELECT ItemWeaponType.WeaponID, ItemWeaponType.WeaponDamage, ItemWeaponType.WeaponPattern, ItemWeaponType.WeaponHeatValue, ItemWeaponType.WeaponCoolRate, ItemWeaponType.WeaponName FROM ItemWeaponType WHERE ItemWeaponTypeWeaponID = @PU;";
OleDbCommand cmd = new OleDbCommand(sql, dbConn);
foreach (string WeapID in weapID)
{
if (WeapID == "")
{
return;
}
cmd.Parameters.AddWithValue("@PU", WeapID); ///Problem Line
dbConn.Open();
OleDbDataReader reader = cmd.ExecuteReader();

if (reader.Read())
{
weapDam = reader[1].ToString();
weapPat = reader[2].ToString();
weapHeat = reader[3].ToString();
weapName = reader[4].ToString();
}
dbConn.Close();
this.DGV_WS.Rows.Add(weapDam, weapPat, weapHeat, weapName);
}
}


When i look at the locals in the locals in debug. WeapID has a value of "1" as it should. Have i missed something in terms of syntax?



Thanks!







c# sql oledb






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share|improve this question




share|improve this question








edited Nov 8 at 12:07









Rob

9711021




9711021










asked Nov 8 at 11:25









Blackclaw _

43




43












  • I also suspect this code is going to have issues on the second iteration of the foreach - since you will have added two parameters (with the same name) to a single OleDbCommand. Did you mean to instantiate the OleDbCommand inside the loop?
    – mjwills
    Nov 8 at 11:30










  • The most obvious cause of this error is that WeapID is null. The behaviour you are seeing would be expected in that scenario.
    – mjwills
    Nov 8 at 11:31










  • @mjwills if i type "?WeapID" in the immediate window it returns "1". Also moved OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each. Still get the same error.
    – Blackclaw _
    Nov 8 at 16:55










  • ItemWeaponTypeWeaponID is missing a .. This is a typo. Voting to close.
    – mjwills
    Nov 8 at 20:13










  • @mjwills Ah thank you so much! Good spot! Looked all the way through my SQL about 10 times.......
    – Blackclaw _
    Nov 8 at 20:39


















  • I also suspect this code is going to have issues on the second iteration of the foreach - since you will have added two parameters (with the same name) to a single OleDbCommand. Did you mean to instantiate the OleDbCommand inside the loop?
    – mjwills
    Nov 8 at 11:30










  • The most obvious cause of this error is that WeapID is null. The behaviour you are seeing would be expected in that scenario.
    – mjwills
    Nov 8 at 11:31










  • @mjwills if i type "?WeapID" in the immediate window it returns "1". Also moved OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each. Still get the same error.
    – Blackclaw _
    Nov 8 at 16:55










  • ItemWeaponTypeWeaponID is missing a .. This is a typo. Voting to close.
    – mjwills
    Nov 8 at 20:13










  • @mjwills Ah thank you so much! Good spot! Looked all the way through my SQL about 10 times.......
    – Blackclaw _
    Nov 8 at 20:39
















I also suspect this code is going to have issues on the second iteration of the foreach - since you will have added two parameters (with the same name) to a single OleDbCommand. Did you mean to instantiate the OleDbCommand inside the loop?
– mjwills
Nov 8 at 11:30




I also suspect this code is going to have issues on the second iteration of the foreach - since you will have added two parameters (with the same name) to a single OleDbCommand. Did you mean to instantiate the OleDbCommand inside the loop?
– mjwills
Nov 8 at 11:30












The most obvious cause of this error is that WeapID is null. The behaviour you are seeing would be expected in that scenario.
– mjwills
Nov 8 at 11:31




The most obvious cause of this error is that WeapID is null. The behaviour you are seeing would be expected in that scenario.
– mjwills
Nov 8 at 11:31












@mjwills if i type "?WeapID" in the immediate window it returns "1". Also moved OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each. Still get the same error.
– Blackclaw _
Nov 8 at 16:55




@mjwills if i type "?WeapID" in the immediate window it returns "1". Also moved OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each. Still get the same error.
– Blackclaw _
Nov 8 at 16:55












ItemWeaponTypeWeaponID is missing a .. This is a typo. Voting to close.
– mjwills
Nov 8 at 20:13




ItemWeaponTypeWeaponID is missing a .. This is a typo. Voting to close.
– mjwills
Nov 8 at 20:13












@mjwills Ah thank you so much! Good spot! Looked all the way through my SQL about 10 times.......
– Blackclaw _
Nov 8 at 20:39




@mjwills Ah thank you so much! Good spot! Looked all the way through my SQL about 10 times.......
– Blackclaw _
Nov 8 at 20:39












1 Answer
1






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up vote
0
down vote













Move your OleDbCommand cmd = new OleDbCommand(sql, dbConn); line inside the foreach statement.



You are using the same command over and over, and adding more parameters than the query has.






share|improve this answer





















  • I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
    – Blackclaw _
    Nov 8 at 16:57










  • Have you debugged all the way through? When does the exceptions is thrown?
    – Gonzalo Lorieto
    Nov 8 at 17:06











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
0
down vote













Move your OleDbCommand cmd = new OleDbCommand(sql, dbConn); line inside the foreach statement.



You are using the same command over and over, and adding more parameters than the query has.






share|improve this answer





















  • I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
    – Blackclaw _
    Nov 8 at 16:57










  • Have you debugged all the way through? When does the exceptions is thrown?
    – Gonzalo Lorieto
    Nov 8 at 17:06















up vote
0
down vote













Move your OleDbCommand cmd = new OleDbCommand(sql, dbConn); line inside the foreach statement.



You are using the same command over and over, and adding more parameters than the query has.






share|improve this answer





















  • I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
    – Blackclaw _
    Nov 8 at 16:57










  • Have you debugged all the way through? When does the exceptions is thrown?
    – Gonzalo Lorieto
    Nov 8 at 17:06













up vote
0
down vote










up vote
0
down vote









Move your OleDbCommand cmd = new OleDbCommand(sql, dbConn); line inside the foreach statement.



You are using the same command over and over, and adding more parameters than the query has.






share|improve this answer












Move your OleDbCommand cmd = new OleDbCommand(sql, dbConn); line inside the foreach statement.



You are using the same command over and over, and adding more parameters than the query has.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 8 at 12:37









Gonzalo Lorieto

521212




521212












  • I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
    – Blackclaw _
    Nov 8 at 16:57










  • Have you debugged all the way through? When does the exceptions is thrown?
    – Gonzalo Lorieto
    Nov 8 at 17:06


















  • I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
    – Blackclaw _
    Nov 8 at 16:57










  • Have you debugged all the way through? When does the exceptions is thrown?
    – Gonzalo Lorieto
    Nov 8 at 17:06
















I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
– Blackclaw _
Nov 8 at 16:57




I tried moving OleDbCommand cmd = new OleDbCommand(sql, dbConn); inside of the for each and the same error still occurs.
– Blackclaw _
Nov 8 at 16:57












Have you debugged all the way through? When does the exceptions is thrown?
– Gonzalo Lorieto
Nov 8 at 17:06




Have you debugged all the way through? When does the exceptions is thrown?
– Gonzalo Lorieto
Nov 8 at 17:06


















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