Search Results

Search found 18132 results on 726 pages for 'connection timeout'.

Page 443/726 | < Previous Page | 439 440 441 442 443 444 445 446 447 448 449 450  | Next Page >

  • .NET datetime issue with SQL stored procedure

    - by DanO
    I am getting the below error when executing my application on a Windows XP machine with .NET 2.0 installed. On my computer Windows 7 .NET 2.0 - 3.5 I am not having any issues. The target SQL server version is 2005. This error started occurring when I added the datetime to the stored procedure. I have been reading alot about using .NET datetime with SQL datetime and I still have not figured this out. If someone can point me in the right direction I would appreciate it. Here is the where I believe the error is coming from. private static void InsertRecon(string computerName, int EncryptState, TimeSpan FindTime, Int64 EncryptSize, DateTime timeWritten) { SqlConnection DBC = new SqlConnection("server=server;UID=InventoryServer;Password=pass;database=Inventory;connection timeout=30"); SqlCommand CMD = new SqlCommand(); try { CMD.Connection = DBC; CMD.CommandType = CommandType.StoredProcedure; CMD.CommandText = "InsertReconData"; CMD.Parameters.Add("@CNAME", SqlDbType.NVarChar); CMD.Parameters.Add("@ENCRYPTEXIST", SqlDbType.Int); CMD.Parameters.Add("@RUNTIME", SqlDbType.Time); CMD.Parameters.Add("@ENCRYPTSIZE", SqlDbType.BigInt); CMD.Parameters.Add("@TIMEWRITTEN", SqlDbType.DateTime); CMD.Parameters["@CNAME"].Value = computerName; CMD.Parameters["@ENCRYPTEXIST"].Value = EncryptState; CMD.Parameters["@RUNTIME"].Value = FindTime; CMD.Parameters["@ENCRYPTSIZE"].Value = EncryptSize; CMD.Parameters["@TIMEWRITTEN"].Value = timeWritten; DBC.Open(); CMD.ExecuteNonQuery(); } catch (System.Data.SqlClient.SqlException e) { PostMessage(e.Message); } finally { DBC.Close(); CMD.Dispose(); DBC.Dispose(); } } Unhandled Exception: System.ArgumentOutOfRangeException: The SqlDbType enumeration value, 32, is invalid. Parameter name: SqlDbType at System.Data.SqlClient.MetaType.GetMetaTypeFromSqlDbType(SqlDbType target) at System.Data.SqlClient.SqlParameter.set_SqlDbType(SqlDbType value) at System.Data.SqlClient.SqlParameter..ctor(String parameterName, SqlDbType dbType) at System.Data.SqlClient.SqlParameterCollection.Add(String parameterName, SqlDbType sqlDbType) at ReconHelper.getFilesInfo.InsertRecon(String computerName, Int32 EncryptState, TimeSpan FindTime, Int64 EncryptSize, DateTime timeWritten) at ReconHelper.getFilesInfo.Main(String[] args)

    Read the article

  • Default MVC Web Application Database

    - by wows
    When setting up a new ASP.NET MVC Web Application, the default connection string inside Web.Config is something like this: connectionString="data source=.\SQLEXPRESS;Integrated Security=SSPI;AttachDBFilename=|DataDirectory|aspnetdb.mdf;User Instance=true" I'm just wanting to play around with logging in and registering, etc but when I run the app it obviously can't find a SQL database. What database with what tables do I need to setup to do this? I have SQL Server 2005 Standard installed on my system, is that enough? Thanks.

    Read the article

  • I have problems with adding rows to data-binded DataGridView in desktop app.

    - by Mishko
    DataTable table1 = new DataTable(); double brutoUkupno1 = 0; double porezUkupno1 = 0; double doprinosUkupno1 = 0; double netoUkupno1 = 0; double doprinosTeretUkupno1 = 0; double topliObrokUkupno1 = 0; double regresUkupno1 = 0; Connection con = new Connection(); table1 = con.boundTable(month, Convert.ToInt32(year)); //This is method which returns DataTable table1.Rows.Add(null, null, null, null, null, null, null, null, null, null, null, null, null, null); table1.Rows.Add(null, null, null, null, null, null, null, null, null, null, null, null, null, null); dgv2.Visible = true; dgv2.DataSource = table1; for (int i = 0; i < dgv2.RowCount - 2; i++) { topliObrokUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[7].Value); regresUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[8].Value); brutoUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[9].Value); porezUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[10].Value); doprinosUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[11].Value); netoUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[12].Value); doprinosTeretUkupno1 += Convert.ToDouble(dgv2.Rows[i].Cells[13].Value); //Now I am having problems with this below, putting things above to dgv2 : } dgv2.Rows[dgv2.Rows.Count - 1].Cells[0].Value = "Ukupno"; dgv2.Rows[dgv2.Rows.Count - 1].Cells[3].Value = month.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[4].Value = year.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[7].Value = topliObrokUkupno1.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[8].Value = regresUkupno1.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[9].Value = brutoUkupno1.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[10].Value = porezUkupno1.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[11].Value = doprinosUkupno1.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[12].Value = netoUkupno1.ToString(); dgv2.Rows[dgv2.Rows.Count - 1].Cells[13].Value = doprinosTeretUkupno1.ToString(); dgv2.Rows[dgv2.RowCount - 2].Height = 3; dgv2.Rows[dgv2.RowCount - 2].DefaultCellStyle.BackColor = Color.Black;

    Read the article

  • Why the parent page get refreshed when I click the link to open thickbox-styled form?

    - by user333205
    Hi, all: I'm using Thickbox 3.1 to show signup form. The form content comes from jquery ajax post. The jquery lib is of version 1.4.2. I placed a "signup" link into a div area, which is a part of my other large pages, and the whole content of that div area is ajax+posted from my server. To make thickbox can work in my above arangement, I have modified the thickbox code a little like that: //add thickbox to href & area elements that have a class of .thickbox function tb_init(domChunk){ $(domChunk).live('click', function(){ var t = this.title || this.name || null; var a = this.href || this.alt; var g = this.rel || false; tb_show(t,a,g); this.blur(); return false; });} This modification is the only change against the original version. Beacause the "signup" link is placed in ajaxed content, so I Use live instead of binding the click event directly. When I tested on my pc, the thickbox works well. I can see the signup form quickly, without feeling the content of the parent page(here, is the other large pages) get refreshed. But after transmiting my site files into VHost, when I click the "signup" link, the signup form get presented very slowly. The large pages get refreshed evidently, because the borwser(ie6) are reloading images from server incessantly. These images are set as background images in CSS files. I think that's because the slow connection of network. But why the parent pages get refreshed? and why the browser reloads those images one more time? Havn't those images been placed in local computer's disk? Is there one way to stop that reloadding? Because the signup form can't get displayed sometimes due to slow connection of network. To verified the question, you can access http://www.juliantec.info/track-the-source.html and click the second link in left grey area, that is the "signup" link mentioned above. Thinks!

    Read the article

  • iphone sqlite app startup issues

    - by Futur
    Hi All, I have 4 database catalogues in my app, i will be switching the four catalogues. When i start the app, i read the NSUserDefaults and try to load the default catalogue in the memory, but in the first time this doesnt happen. Instead i get null values returned first time just because the DB connection is not successful for some unknown reason, the debugger is unable to go there too. But the app starts up the next time, the values are fetched successfully. Please help

    Read the article

  • How to connect laptop with telephone using WI-FI? (ethernet)

    - by rmaster
    What I did: 1)added new wireless network in wireless network settings and gave it SSID 2)gave laptop the IP like 192.168.0.1, mask: 255.255.255.0, gateway: 192.168.0.2 3)gave the telephone with wifi the same mask but IP and gateway rewersed But telephone can not find my new wireless network(it can find all other networks), computer also can not find it. where is error? how to make it visible and working? if my steps are wrong tell me correct way to establish this connection via wi-fi

    Read the article

  • QueryTables error

    - by ktm5124
    I have the following VBA code inside an Excel macro: With ActiveSheet.QueryTables.Add( _ Connection:=ConnStr, _ Destination:=ActiveSheet.Range("A1"), _ Sql:=SqlStr) .Refresh End With And yet I am getting the error: "Run-time error '1004': Application-defined or object-defined error." The debugger points to this code as the source of error. Assuming that my ConnStr and SqlStr are correct, why am I getting this error? I am using Excel 2007 and I am querying an Oracle database.

    Read the article

  • cannot access sql server after publishing site in iis7

    - by vinu
    I am created a website using visual studio 2010. On the time of the development of website I am able to access the database.. but after publishing the site using IIS7..i was unable to access the database..the exception occured during that time was "the connection is in the closed state".in IIS7 When I changed the application pool identity to localsystem, it worked. Data base is installed in the same machine. and server is SQL SERVER 2008 R2

    Read the article

  • I cnn't use Database in SQL Azure!

    - by Nahid
    Hi, I am trying to use a Database in SQL AZURE. I have installed SQL Server 2008. I can Login SQL Azure and can use master Database. But I can't use other Database and I can't see any things in my object explorer. Its Show Error: "USE statement is not supported to switch between databases. Use a new connection to connect to a different Database." How can I use other database?

    Read the article

  • Add subview to another view

    - by Martol1ni
    With two ViewControllers, MyView 1 and MyView 2, is there possible to add a subview to MyView2 from MyView1.m? I have tried: MyView2 * screen = [[MyView2 alloc]initWithNibName:nil bundle:nil]; [screen.view addSubView:mySubView]; But my new instance of MyView 2 has no connection to the 'visible' ViewController on MyView2, right? To clarify, the ViewController that is showing, is MyView1. I want MyView1 to be able to add a subview to the MyView2 view. Thanks

    Read the article

  • Improve long mysql query

    - by John Adawan
    I have a php mysql query like this $query = "SELECT * FROM articles FORCE INDEX (articleindex) WHERE category='$thiscat' and did>'$thisdid' and mid!='$thismid' and status='1' and group='$thisgroup' and pid>'$thispid' LIMIT 10"; As optimization, I've indexed all the parameters in articleindex and I use force index to force mysql to use the index, supposedly for faster processing. But it seems that this query is still quite slow and it's causing a jam and maxing out the max mysql connection limit. Let's discuss how we can improve on such long query.

    Read the article

  • Improve long mysql query

    - by John Adawan
    I have a php mysql query like this $query = "SELECT * FROM articles FORCE INDEX (articleindex) WHERE category='$thiscat' and did>'$thisdid' and mid!='$thismid' and status='1' and group='$thisgroup' and pid>'$thispid' LIMIT 10"; As optimization, I've indexed all the parameters in articleindex and I use force index to force mysql to use the index, supposedly for faster processing. But it seems that this query is still quite slow and it's causing a jam and maxing out the max mysql connection limit. Let's discuss how we can improve on such long query.

    Read the article

  • PHP syntax error “unexpected $end”

    - by Jacksta
    I have 3 files 1) show_createtable.html 2) do_showfielddef.php 3) do_showtble.php 1) First file is for creating a new table for a data base, it is a fom with 2 inputs, Table Name and Number of Fields. THIS WORKS FINE! <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <h1>Step 1: Name and Number</h1> <form method="post" action="do_showfielddef.php" /> <p><strong>Table Name:</strong><br /> <input type="text" name="table_name" size="30" /></p> <p><strong>Number of fields:</strong><br /> <input type="text" name="num_fields" size="30" /></p> <p><input type="submit" name="submit" value="go to step2" /></p> </form> </body> </html> 2) this script validates fields and createa another form to enter all the table rows. This for also WORKS FINE! <?php //validate important input if ((!$_POST[table_name]) || (!$_POST[num_fields])) { header( "location: show_createtable.html"); exit; } //begin creating form for display $form_block = " <form action=\"do_createtable.php\" method=\"post\"> <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\"> <table cellspacing=\"5\" cellpadding=\"5\"> <tr> <th>Field Name</th><th>Field Type</th><th>Table Length</th> </tr>"; //count from 0 until you reach the number fo fields for ($i = 0; $i <$_POST[num_fields]; $i++) { $form_block .=" <tr> <td align=center><input type=\"texr\" name=\"field name[]\" size=\"30\"></td> <td align=center> <select name=\"field_type[]\"> <option value=\"char\">char</option> <option value=\"date\">date</option> <option value=\"float\">float</option> <option value=\"int\">int</option> <option value=\"text\">text</option> <option value=\"varchar\">varchar</option> </select> </td> <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"> </td> </tr>"; } //finish up the form $form_block .= " <tr> <td align=center colspan=3><input type =\"submit\" value=\"create table\"> </td> </tr> </table> </form>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create a database table: Step 2</title> </head> <body> <h1>defnie fields for <? echo "$_POST[table_name]"; ?> </h1> <? echo "$form_block"; ?> </body> </html> Problem is here 3) this form creates the tables and enteres them into the database. I am getting an error on line 37 "Parse error: syntax error, unexpected $end in /home/admin/domains/domaina.com.au/public_html/do_createtable.php on line 37" <? $db_name = "testDB"; $connection = @mysql_connect("localhost", "admin_user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "CREATE TABLE $_POST[table_name]("; for ($i = 0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[field_length][$i] !="") { $sql .=" (".$_POST[field_length][$i]."),"; } else { $sql .=","; } $sql = substr($sql, 0, -1); $sql .= ")"; $result = mysql_query($sql, $connection) or die(mysql_error()); if ($result) { $msg = "<p>" .$_POST[table_name]." has been created!</p>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create A Database Table: Step 3</title> </head> <body> <h1>Adding table to <? echo "$db_name"; ?>...</h1> <? echo "$msg"; ?> </body> </html>

    Read the article

  • Script to create a table and fields in SQL wont work

    - by Jacksta
    Warning this is lenghty! attack if you knowledagble. well at least more then a newb beginner like me. This script uses three files as detailed below. It is suppoed to create the database and fields from the form input. It gets to the end and shows my_contacts has been created!. But when i go into phpMyadmin the table has not been created. I have a file named show_createtable.html which is used to create a table in MySQL <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <h1>Step 1: Name and Number</h1> <form method="post" action="do_showfielddef.php" /> <p><strong>Table Name:</strong><br /> <input type="text" name="table_name" size="30" /></p> <p><strong>Number of fields:</strong><br /> <input type="text" name="num_fields" size="30" /></p> <p><input type="submit" name="submit" value="go to step2" /></p> </form> </body> </html> This Form Posts to do_showfielddef.php <?php //validate important input if ((!$_POST[table_name]) || (!$_POST[num_fields])) { header( "location: show_createtable.html"); exit; } //begin creating form for display $form_block = " <form action=\"do_createtable.php\" method=\"post\"> <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\"> <table cellspacing=\"5\" cellpadding=\"5\"> <tr> <th>Field Name</th><th>Field Type</th><th>Table Length</th><th>Primary Key?</th><th>Auto-Increment?</th> </tr>"; //count from 0 until you reach the number fo fields for ($i = 0; $i <$_POST[num_fields]; $i++) { $form_block .=" <tr> <td align=center><input type=\"texr\" name=\"field name[]\" size=\"30\"></td> <td align=center> <select name=\"field_type[]\"> <option value=\"char\">char</option> <option value=\"date\">date</option> <option value=\"float\">float</option> <option value=\"int\">int</option> <option value=\"text\">text</option> <option value=\"varchar\">varchar</option> </select> </td> <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"></td> <td aligh=center><input type=\"checkbox\" name=\"primary[]\" value=\"Y\"></td> <td aligh=center><input type=\"checkbox\" name=\"auto_increment[]\" value=\"Y\"></td> </tr>"; } //finish up the form $form_block .= " <tr> <td align=center colspan=3><input type =\"submit\" value=\"create table\"> </td> </tr> </table> </form>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create a database table: Step 2</title> </head> <body> <h1>defnie fields for <? echo "$_POST[table_name]"; ?> </h1> <? echo "$form_block"; ?> </body> </html> Which in turn creates the table and fields with this file do_showfielddef.php //connect to database $connection = @mysql_connect("localhost", "user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); //start creating the SQL statement $sql = "CREATE TABLE $_POST[table_name]("; //continue the SQL statement for each new field for ($i = 0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[auto_increment][$i] =="Y") { $additional = "NOT NULL auto_increment"; } else { $additional = ""; } if ($_POST[primary][$i] =="Y") { $additional .= ", primary key (".$_POST[field_name][$i].")"; } else { $additional = ""; } if ($_POST[field_length][$i] !="") { $sql .= " (".$_POST[field_length][$i].") $additional ,"; } else { $sql .=" $additional ,"; } } //clean up the end of the string $sql = substr($sql, 0, -1); $sql .= ")"; //execute the query $result = mysql_query($sql, $connection) or die(mysql_error()); //get a giid message for display upon success if ($result) { $msg = "<p>" .$_POST[table_name]." has been created!</p>"; } ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create A Database Table: Step 3</title> </head> <body> <h1>Adding table to <? echo "$db_name"; ?>...</h1> <? echo "$msg"; ?> </body> </html>

    Read the article

  • Grails query not using GORM

    - by Tihom
    What is the best way to query for something without using GORM in grails? I have query that doesn't seem to fit in the GORM model, the query has a subquery and a computed field. I posted on stackoverflow already with no response so I decided to take a different approach. I want to query for something not using GORM within a grails application. Is there an easy way to get the connection and go through the result set?

    Read the article

  • Unable to get project info and source control from a particular machine

    - by Gerard
    From one particular machine, accounts are unable to get tfs project info and source control via the visual studio tfs client. The connection with the tfs server is made but the project content remains empty. Web access is possible from this machine. Note: from all other machines everything works normally, so it must be a local machine issue. What might be misconfigured on this machine? Otherwise there seem to be no problems on this XP machine.

    Read the article

  • What design pattern should be used to create an emulator?

    - by Facon
    I have programmed an emulator, but I have some doubts about how to organizate it properly, because, I see that it has some problems about classes connection (CPU <- Machine Board). For example: I/O ports, interruptions, communication between two or more CPU, etc. I need for the emulator to has the best performance and good understanding of the code. PD: Sorry for my bad English.

    Read the article

  • When using Oracle load balancing and ADO.NET, how do you know which host the command executed on?

    - by Leeks and Leaks
    It's possible to use Microsoft's OracleClient assembly to connect to an Oracle database, and using Oracle's connection string format, set it up to use load balancing, provided your Oracle environment supports it. The question I have is how do you know after the fact, which db host the command actually executed against? Is there a way to retrieve the machine name from the load balanced set of machines?

    Read the article

  • PHP apache_request_headers() diagrees with reality (as confirmed by Firebug): why?

    - by Peter Howe
    I have written a web app in PHP which makes use of Ajax requests (made using YUI.util.Connect.asyncRequest). Most of the time, this works fine. The request is sent with an X-Requested-With value of XMLHttpRequest. My PHP controller code uses apache_request_headers() to check whether an incoming request is Ajax or not and all works well. But not always. Intermittently, I'm getting a situation where the Ajax request is sent (and Firebug confirms for me that the headers on the request include an X-Requested-With of XMLHttpRequest) but apache_request_headers() is not returning that header in its list. The output from when I var_dump the apache_request_headers() is as follows (note the lack of X- 'Host' => string 'peterh.labs.example.com' (length=26) 'User-Agent' => string 'Mozilla/5.0 (X11; U; Linux i686; en-GB; rv:1.9.0.3) Gecko/2008101315 Ubuntu/8.10 (intrepid) Firefox/3.0.3' (length=105) 'Accept' => string 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8' (length=63) 'Accept-Language' => string 'en-gb,en;q=0.5' (length=14) 'Accept-Encoding' => string 'gzip,deflate' (length=12) 'Accept-Charset' => string 'ISO-8859-1,utf-8;q=0.7,*;q=0.7' (length=30) 'Keep-Alive' => string '300' (length=3) 'Connection' => string 'keep-alive' (length=10) 'Referer' => string 'http://peterh.labs.example.com/qmail/' (length=40) 'Cookie' => string 'WORKFLOW_SESSION=55f9aff2051746851de453c1f776ad10745354f6' (length=57) 'Pragma' => string 'no-cache' (length=8) 'Cache-Control' => string 'no-cache' (length=8) But Firebug tells me: Request Headers: Host peterh.labs.example.com User-Agent Mozilla/5.0 (X11; U; Linux i686; en-GB; rv:1.9.0.3) Gecko/2008101315 Ubuntu/8.10 (intrepid) Firefox/3.0.3 Accept text/html,application/xhtml+xml,application/xml;q=0.9,**;q=0.8 Accept-Language en-gb,en;q=0.5 Accept-Encoding gzip,deflate Accept-Charset ISO-8859-1,utf-8;q=0.7,*;q=0.7 Keep-Alive 300 Connection keep-alive X-Requested-With XMLHttpRequest Referer http://peterh.labs.example.com/qmail/ Cookie WORKFLOW_SESSION=55f9aff2051746851de453c1f776ad10745354f6 This mismatch is (apparently) intermittent when executing the same code. But I don't believe in "intermittent" when it comes to software! Help!

    Read the article

< Previous Page | 439 440 441 442 443 444 445 446 447 448 449 450  | Next Page >