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  • How to programmatically check availibilty of internet connection in Android?

    - by Fahad
    Hi! I want to check programmatically whether there is an internet connection in android phone/emulator. So that once I am sure that an internet connection is present then I'll make a call to the internet. So its like "Hey emulator! If you have an internet connection, then please open this page, else doSomeThingElse();" hope you get the idea. I would highly appreciate a quick response I need it quite early. regards Fahad Ali Shaikh

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  • Is it fine to put different category of stuff in a single domain name?

    - by Fahad Uddin
    I own a website which is regarding startups and finance. I am looking forward to work on Wordpress programming in which I would be selling wordpress themes. I thought of buying a domain name for Wordpress website but it takes quite lot of time to setup a website and then do its SEO. Is it fine(in terms of SEO and professionalism) to put the Wordpress category inside my old domain like, Domain: www.startupsandfinance.com Wordpress domain, www.wp.startupsandfinance.com

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  • Why do companies opensource their code?

    - by Fahad Uddin
    I have seen many big companies like Facebook, Twitter and LinkedIn opensource their code. I was curious to understand why would any company share their code to the world. One reason I understood that it makes the people better understand their API. Still, I am a little confused as any other company/person can use their code to find a vulnerability inside and get their site down. Why do such big companies take this risk then?

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  • FTP file access problem

    - by Fahad Uddin
    I recently got a malware on my website. I have made the backup of the website on my computer and trying to wipe off my FTP. I am trying to delete the root folder but getting this error message on all of the malicious files, Response: 550 Could not delete index.php: Permission denied I am the sole administrator of the ftp so permission should not have been an issue. My host provider seems not to suffer from this problem as his websites are running well without any malware. I have also tried to change the root to 777 to see if the file permission change could help me delete the files but still I am getting the same error. Please help out. Thanks

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  • How can I configure cowsay?

    - by Fahad Ahammed
    I have installed cowsay and fortune.....please i want to set My own talks or texts in cowsay ...but i can't configure it !!! when i open terminal there is nothing from cowsay !. I want to show cowsay when i start terminal .But this works !!! hash@ssl50:~$ cowsay -f tux "Carry on" < carry on > ---------- \ \ .--. |o_o | |:_/ | // \ \ (| | ) /'\_ _/`\ \___)=(___/

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  • Display Driver stopped responding and recovered successfully- Blue Screen - HP Elite Book

    - by Fahad Saleem
    I have HP ELiteBook , this problem has become a routine headache for me (after every 20, 25 minutes_ - While using, suddenly everything hangs and black out goes around and then a message pop at the task bar below that display driver stopped ...and successfully recovered but the hangs and black screen continues and then the Blue Death Screen comes" and then i have no other option but to restart Is this a driver problem ? Hardware problem ? what should I do Please be specific , no stories and theories

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  • Report Builder 2.0 - Creating DataSet - User Not Authorized

    - by Fahad
    Hello, we are currently using SQL Server Reporting Services and we would like to use Report Builder so our customers can create reports themselves. I have created a User on the server. I have added this user to the SQLServerMSSQLUser and SQLServerReportServerUser groups. I have given this User db_datareader access to the required database and to the Reporting Services database. I've also tried giving the User db_owner access to the Reporting Services db's. And on the Report Manager, this User is a System_User, but has all access (every checkbox is checked). When I connect using Report Builder, I can select the Report Model to create a DataSource, but when I try to create a DataSet, I get the following error: An error occurred while connection to datasource 'DataSource1'. The details are: 'User Not Authorized'. Does anyone know what server permissions I forgot to set? I'm assuming it's a Windows permissions issue because I do not see any database login errors in the event logs.

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  • Why doesn't my Administrator account have access privileges?

    - by Fahad
    We are running windows server 2003.I log in as Administrator but I don't have any access rights. I tried to create a new Administrator account but when logging into that one I still face the same issue. Is there a problem with the Group Administrators? If so is there a way to restore it's default settings? Note: This problem started after I uninstalled the Active directory role from the server. Thanks in advance.

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  • How to echo if field is not found?

    - by Fahad
    Hi I'm trying to figure out how to echo back if the value entered does not match when a database lookup is done. I'm using ajax to run the request and php to do the lookup ajax.js: function showResult(str) { if (str=="") { document.getElementById("description").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("description").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","getuser.php?voucher="+str,true); xmlhttp.send(null); } and getuser.php: <?php $q=$_GET["voucher"]; $con = mysql_connect('localhost', 'root', ''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $sql="SELECT * FROM redemption WHERE voucher = '".$q."'"; $result = mysql_query($sql); echo "<table> <tr> <th>Name</th> <th>Product</th> <th>Address</th> <th>Status</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['product'] . "</td>"; echo "<td>" . $row['address'] ." ".$row['city'] ." ".$row['province'] ." ".$row['postal'] . "</td>"; echo "<td>" . $row['status'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> What I would like to do is that once the person enters an invalid or a voucher number that is not found I would like to return an error that "Voucher number is not found". There is also a column in the db that stores the status such as "redeemed" or "not redeemed". How could I check for both whether the voucher number exists and if it has already been redeemed? I assume it'd have to be a syntax such as $sql="SELECT * FROM redemption WHERE voucher = '".$q."'" AND status = 'not redeemed' and then use an else or case statement perhaps? Thanks in advance

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  • How to modify style for an asp:textbox in codebehind ?

    - by Fahad
    Hi all, I am trying to set the style of an asp:TextBox in codebehind, the textbox is style is set initially to style="display:none" when I set the dispaly to block in codebehind the textbox appears for a moment and then it's gone. I don't know what this problem is, when it's done in javascript it works fine Here is the code: asp.net code: <asp:TextBox ID="txtError" style="display:none" runat="server" ReadOnly="True" Width="95%"></asp:TextBox> codebehind: txtError.Style["display"] = "block"; Am I doing anything wrong? Thanks in advance.

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  • Fixed div background

    - by Fahad
    I want to create a layout where I want to display an image to the left and content on the right. The image should stay constant when the content scrolls. The css I'm using: <style type="text/css"> html, body { margin:0; padding:0; } #page-container { margin:auto; width:900px; background-color:Black; background-image:url('images/desired_layout.png'); background-attachment: fixed; background-repeat:no-repeat; } #main-image { float:left; width:250px; height:687px; background-image:url('images/desired_layout.png'); background-attachment:fixed; background-repeat:no-repeat; } #content { margin-left:250px; background-color:Olive; height:800px; width:650px; } </style> The HTML: <div id="page-container"> <div id="main-image"></div> <div id="content"></div> </div> Alot of time on this site and I have understood that background-attachment:fixed positions the image in the entire viewport and not the element it is applied to. My question is how do I go about creating that kind of layout? I do not want to give that image as a background image, as if the window is resized, it might get hidden. I want scrollbars to appear if the window size is less than 900px( my page width) so that the image can be viewed at all times. That happens with this code, however I would like the image to start at my element instead. How do I go about doing this?? Thanks in Advance :)

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  • jquery show hidden div

    - by Fahad
    Firstly, I'm sort of embarrassed asking about this, so many people have already asked this question but even after having gone through so many posts, I'm unable to achieve what I want. Basically, a div, initially hidden, has to be displayed on a button click. I tried hiding the div using display:none and hide() and then displaying it using show(), toggle(), and css("display","block"). Using all sorts of combinations of the above, I was still unable to get the result. Code: <html xmlns="http://www.w3.org/1999/xhtml"> <head runat="server"> <title></title> <link href="css/smoothness/jquery-ui-1.9.2.custom.min.css" rel="stylesheet" type="text/css" /> <script src="jQuery/jquery-1.8.3.min.js" type="text/javascript"></script> <script src="jQuery/jquery-ui-1.9.2.custom.min.js" type="text/javascript"></script> <script type="text/javascript"> $(document).ready(function () { $('#one').hide(); $('#Button1').click(function () { $('#one').toggle(500); }); }); </script> </head> <body> <form id="form1" runat="server"> <div id="one" style="height: 20px;width:200px; background-color: Red; "> </div> <asp:Button ID="Button1" runat="server" Text="Show" /> </form> </body> </html> On button click, the div is shown for a brief second before it disappears again. The same thing happens if I use show() instead of toggle() in the above code. Again the same thing if I set style="display:none" to the div instead of using hide() and then use show() or toggle(). I also tried using $('#one').css("display","block"); but again, the same result. Can anyone please tell me where I'm going wrong. Just started learning jQuery and it is really frustrating when something apparently so simple will not work. Thanks in advance. :)

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  • Is there any way to add a MouseListener to a Graphic object ?

    - by Fahad
    Hi, Is there any way to add a MouseListener to a Graphic object. I have this simple GUI that draw an oval. What I want is handling the event when the user clicks on the oval import java.awt.*; import java.awt.event.MouseEvent; import java.awt.event.MouseListener; import javax.swing.*; public class Gui2 extends JFrame { JFrame frame = new JFrame(); MyDrawPanel drawpanel = new MyDrawPanel(); public static void main(String[] args) { Gui2 gui = new Gui2(); gui.go(); } public void go() { frame.getContentPane().add(drawpanel); // frame.addMouseListener(this); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.setSize(300, 300); frame.setVisible(true); } } class MyDrawPanel extends JComponent implements MouseListener { public void paintComponent(Graphics g) { int red = (int) (Math.random() * 255); int green = (int) (Math.random() * 255); int blue = (int) (Math.random() * 255); Color startrandomColor = new Color(red, green, blue); red = (int) (Math.random() * 255); green = (int) (Math.random() * 255); blue = (int) (Math.random() * 255); Color endrandomColor = new Color(red, green, blue); Graphics2D g2d = (Graphics2D) g; this.addMouseListener(this); GradientPaint gradient = new GradientPaint(70, 70, startrandomColor, 150, 150, endrandomColor); g2d.setPaint(gradient); g2d.fillOval(70, 70, 100, 100); } @Override public void mouseClicked(MouseEvent e) { if ((e.getButton() == 1) && (e.getX() >= 70 && e.getX() <= 170 && e.getY() >= 70 && e .getY() <= 170)) { this.repaint(); // JOptionPane.showMessageDialog(null,e.getX()+ "\n" + e.getY()); } } @Override public void mouseEntered(MouseEvent e) { // TODO Auto-generated method stub } @Override public void mouseExited(MouseEvent e) { // TODO Auto-generated method stub } @Override public void mousePressed(MouseEvent e) { // TODO Auto-generated method stub } @Override public void mouseReleased(MouseEvent e) { // TODO Auto-generated method stub } } This Works Except it fires when the click is within a virtual box around the oval. Could anyone help me to have it fire when the click is EXACTLY on the oval. Thanks in advance.

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