Not saving in my database

Question

Hi! I can't save on my database a picture. I receive this message when try ro run my webpage: No overload for method 'execution' takes '7' arguments Default.aspx.cs using System; using System.Data; using System.Configuration; using System.Data.SqlClient; using System.ComponentModel; using System.Collections; using System.Web; using System.Web.Security; using System.Web.UI; using System.Web.UI.WebControls; using System.Web.UI.WebControls.WebParts; using System.Web.UI.HtmlControls; public partial class Default2 : System.Web.UI.Page { public string GetConnectionString() { return System.Configuration.ConfigurationManager.ConnectionStrings [divin].ConnectionString; } private void execution(string fname string lname string dob string gender string fathername string contact string address Image photo) { SqlConnection conn = new SqlConnection(GetConnectionString()); string sql = INSERT INTO onlineapplication (fname lname dob gender fathername contact addressphoto) VALUES  +  (@fname @lname @dob @gender @fathername @contact @address @photo); try { conn.Open(); SqlCommand cmd = new SqlCommand(sql conn); SqlParameter[] pram = new SqlParameter[8]; pram[0] = new SqlParameter(@fname SqlDbType.VarChar 50); pram[1] = new SqlParameter(@lname SqlDbType.VarChar 50); pram[2] = new SqlParameter(@dob SqlDbType.VarChar 50); pram[3] = new SqlParameter(@gender SqlDbType.Char 10); pram[4] = new SqlParameter(@fathername SqlDbType.VarChar 50); pram[5] = new SqlParameter(@contact SqlDbType.Int 20); pram[6] = new SqlParameter(@address SqlDbType.VarChar 100); pram[7] = new SqlParameter(@photo SqlDbType.Image); pram[0].Value = fname; pram[1].Value = lname; pram[2].Value = dob; pram[3].Value = gender; pram[4].Value = fathername; pram[5].Value = contact; pram[6].Value = address; pram[7].Value = photo; for (int i = 0; i { cmd.Parameters.Add(pram[i]); } cmd.CommandType = CommandType.Text; cmd.ExecuteNonQuery(); } catch(System.Data.SqlClient.SqlException ex_msg) { string msg = Error occured while inserting; msg += ex_msg.Message; throw new Exception(msg); } finally { conn.Close(); } } protected void Page_Load(object sender EventArgs e) { } protected void submit_Click(object sender EventArgs e) { if(fname.Text==) { Response.Write(Please complete the form.); } else { //bug on this line. execution(fname.Textlname.Textdob.Textgender.Textfathername.Textcontact.Textaddress.Text); conform.Visible = true; fname.Text=; lname.Text=; dob.Text=; gender.Text=; gender.Text=; fathername.Text=; contact.Text=; address.Text=; } } } /////////////////////////////////////////////////////////////////////////////////////////////////// default.aspx http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd > http://www.w3.org/1999/xhtml  > Simple Application Form in ASP.Net Simple Application Form In ASP.Net First Name : Last Name : Date of Birth : Gender : Father's Name : Contact Number : Address : Select Photo : Width=152px> Database Details EmptyDataText=There are no data records to display.>  ProviderName=  SelectCommand=SELECT [id] [fname] [lname] [dob] [gender] [fathername] [contact] [address] [photo] FROM [onlineapplication]>

Sanjeeb Lenka · Answer

try this: private void execution(string fname string lname string dob string gender string fathername string contact string address byte[] photo )

Divin Mfwamba · Answer

It's sound good. But in runing time I received these messages: The best overloaded method match for 'Default2.execution(string stringstringstringstringstring System.Web.UI.WebControls.Image)' has some invalid arguments Argument '8': cannot convert from 'byte [] ' to 'System.Web.UI.WebControls.Image' using System; using System.Data; using System.Configuration; using System.Data.SqlClient; using System.ComponentModel; using System.Collections; using System.Web; using System.Web.Security; using System.Web.UI; using System.Web.UI.WebControls; using System.Web.UI.WebControls.WebParts; using System.Web.UI.HtmlControls; using System.IO; public partial class Default2 : System.Web.UI.Page { public string GetConnectionString() { //we will set up the configuration which will call our //web.config file to provide the database details because //in configuration file we have created the //in the process we draged and droped. It creates automatically. //We normally put the database details in web.config file or //machine.config file because it is very sensitive information //usually there IP address of remote database passwords and //user names are stored. return System.Configuration.ConfigurationManager.ConnectionStrings [onlineapplicationformConnectionString1].ConnectionString; //in above line onlineapplicationformConnectionString1 is //our configuration name which is inside the web.config file. } //private void execution(string fname string lname string dob string gender string fathername string contact string address) private void execution(string fname string lname string dob string gender string fathername string contact string address Image photo) { //In above line we declaring different variables same as backend SqlConnection conn = new SqlConnection(GetConnectionString()); //In above line we are calling connection //string function which is defined already on top string sql = INSERT INTO onlineapplication (fname lname dob gender fathername contact address photo) VALUES  //+  (@fname @lname @dob @gender @fathername @contact @address); + (@fname @lname @dob @gender @fathername @contact @address @photo); //In above lines we are just storing the sql commands which //will insert value in onlineapplication named table //using variable named sql. try { conn.Open(); SqlCommand cmd = new SqlCommand(sql conn); //In above lines we are opening the connection to work and //also storing connection name and sql command in cmd variable //which has 'SqlCommand' type. SqlParameter[] pram = new SqlParameter[8]; //In above lines we are defining 7 sql parameters will be use //In below lines we will not disscuss about id column pram[0] = new SqlParameter(@fname SqlDbType.VarChar 50); pram[1] = new SqlParameter(@lname SqlDbType.VarChar 50); pram[2] = new SqlParameter(@dob SqlDbType.VarChar 50); pram[3] = new SqlParameter(@gender SqlDbType.Char 10); pram[4] = new SqlParameter(@fathername SqlDbType.VarChar 50); pram[5] = new SqlParameter(@contact SqlDbType.Int 20); pram[6] = new SqlParameter(@address SqlDbType.VarChar 100); pram[7] = new SqlParameter(@photo SqlDbType.Image 100); //Now we set-uped all fiels in database in above lines //Now we will set-up form fields pram[0].Value = fname; pram[1].Value = lname; pram[2].Value = dob; pram[3].Value = gender; pram[4].Value = fathername; pram[5].Value = contact; pram[6].Value = address; pram[7].Value = photo; //Now create loop to insert for (int i = 0; i { cmd.Parameters.Add(pram[i]); } cmd.CommandType = CommandType.Text; cmd.ExecuteNonQuery(); } catch(System.Data.SqlClient.SqlException ex_msg) { //Here will be catch elements string msg = Error occured while inserting; msg += ex_msg.Message; throw new Exception(msg); } finally { //Here will be fially elements conn.Close(); } } protected void Page_Load(object sender EventArgs e) { } protected void submit_Click(object sender EventArgs e) { //Here is the command inside the click event of button if(fname.Text==) { Response.Write(Please complete the form.); } else { int length = FileUpload1.PostedFile.ContentLength; //create a byte array to store the binary image data byte[] imgbyte = new byte[length]; //store the currently selected file memory HttpPostedFile img = FileUpload1.PostedFile; //set the binary data img.InputStream.Read(imgbyte 0 length); //execution(fname.Textlname.Textdob.Textgender.Textfathername.Textcontact.Textaddress.Text); execution(fname.Text lname.Text dob.Text gender.Text fathername.Text contact.Text address.Text imgbyte ); conform.Visible = true; fname.Text=; lname.Text=; dob.Text=; gender.Text=; gender.Text=; fathername.Text=; contact.Text=; address.Text=; } } }

Sanjeeb Lenka · Answer

hi you can't directly store image from file upload to database. for that you have to convert it to byte format then store it in database. try to do like this: int length = FileUpload1.PostedFile.ContentLength; //create a byte array to store the binary image data byte[] imgbyte = new byte[length]; //store the currently selected file in memeory HttpPostedFile img = FileUpload1.PostedFile; //set the binary data img.InputStream.Read(imgbyte 0 length); execution(fname.Text lname.Text dob.Text gender.Text fathername.Text contact.Text address.Text imgbyte);

Divin Mfwamba · Answer

Hi Before I coulnot reach the database just because the path was wrong. I recorrect it. Now it's fine. But when I run my webpage it's give these message (I am sure that I putted incorrect values for photo). Then where I colored in red it's where I am trying to correct my code. But unfortunatly nothing: The event 'System.Web.UI.Control.Load' can only appear on the left hand side of += or -0 using System; using System.Data; using System.Configuration; using System.Data.SqlClient; using System.ComponentModel; using System.Collections; using System.Web; using System.Web.Security; using System.Web.UI; using System.Web.UI.WebControls; using System.Web.UI.WebControls.WebParts; using System.Web.UI.HtmlControls; using System.IO; public partial class Default2 : System.Web.UI.Page { public string GetConnectionString() { //we will set up the configuration which will call our //web.config file to provide the database details because //in configuration file we have created the //in the process we draged and droped. It creates automatically. //We normally put the database details in web.config file or //machine.config file because it is very sensitive information //usually there IP address of remote database passwords and //user names are stored. return System.Configuration.ConfigurationManager.ConnectionStrings [onlineapplicationformConnectionString1].ConnectionString; //in above line onlineapplicationformConnectionString1 is //our configuration name which is inside the web.config file. } //private void execution(string fname string lname string dob string gender string fathername string contact string address) private void execution(string fname string lname string dob string gender string fathername string contact string address Image photo) { //In above line we declaring different variables same as backend SqlConnection conn = new SqlConnection(GetConnectionString()); //In above line we are calling connection //string function which is defined already on top string sql = INSERT INTO onlineapplication (fname lname dob gender fathername contact address photo) VALUES  //+  (@fname @lname @dob @gender @fathername @contact @address); + (@fname @lname @dob @gender @fathername @contact @address @photo); //In above lines we are just storing the sql commands which //will insert value in onlineapplication named table //using variable named sql. try { conn.Open(); SqlCommand cmd = new SqlCommand(sql conn); //In above lines we are opening the connection to work and //also storing connection name and sql command in cmd variable //which has 'SqlCommand' type. SqlParameter[] pram = new SqlParameter[8]; //In above lines we are defining 7 sql parameters will be use //In below lines we will not disscuss about id column pram[0] = new SqlParameter(@fname SqlDbType.VarChar 50); pram[1] = new SqlParameter(@lname SqlDbType.VarChar 50); pram[2] = new SqlParameter(@dob SqlDbType.VarChar 50); pram[3] = new SqlParameter(@gender SqlDbType.Char 10); pram[4] = new SqlParameter(@fathername SqlDbType.VarChar 50); pram[5] = new SqlParameter(@contact SqlDbType.Int 20); pram[6] = new SqlParameter(@address SqlDbType.VarChar 100); pram[7] = new SqlParameter(@photo SqlDbType.Image 100); //Now we set-uped all fiels in database in above lines //Now we will set-up form fields pram[0].Value = fname; pram[1].Value = lname; pram[2].Value = dob; pram[3].Value = gender; pram[4].Value = fathername; pram[5].Value = contact; pram[6].Value = address; pram[7].Value = photo; //Now create loop to insert for (int i = 0; i { cmd.Parameters.Add(pram[i]); } cmd.CommandType = CommandType.Text; cmd.ExecuteNonQuery(); } catch(System.Data.SqlClient.SqlException ex_msg) { //Here will be catch elements string msg = Error occured while inserting; msg += ex_msg.Message; throw new Exception(msg); } finally { //Here will be fially elements conn.Close(); } } protected void Page_Load(object sender EventArgs e) { } protected void submit_Click(object sender EventArgs e) { //Here is the command inside the click event of button if(fname.Text==) { Response.Write(Please complete the form.); } else { //execution(fname.Textlname.Textdob.Textgender.Textfathername.Textcontact.Textaddress.Text); execution(fname.Text lname.Text dob.Text gender.Text fathername.Text contact.Text address.Text FileUpload1.Load ); conform.Visible = true; fname.Text=; lname.Text=; dob.Text=; gender.Text=; gender.Text=; fathername.Text=; contact.Text=; address.Text=; FileUpload1.Load = ; } } }

Sanjeeb Lenka · Answer

for photo just save the photo path into database and store the photo in a application folder. check this link http://www.aspdotnet-suresh.com/2011/03/how-to-save-images-into-folder-and.html

Divin Mfwamba · Answer

I see. I just take out photo values in all my code. It's working well without that. Then my question is? How can I recorrect my codes now. Because I need to save photo in my DB from these codes.

Sanjeeb Lenka · Answer

hi your method is having 8 argument in it but you are passing 7 argument thats why it showing error. private void execution(string fname string lname string dob string gender string fathername string contact string address Image photo ) in this calling method only 7 photo is missing. execution(fname.Textlname.Textdob.Textgender.Textfathername.Text contact.Textaddress.Text);