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  • Is there a simple automatic backup system for Visual Studio projects?

    - by Jelly Amma
    Hello, I'm using Visual Studio 2008 Express and I would like Visual Studio (or perhaps an Add-in) to save my whole project to some sort of auto-incrementing archive or whatever would help me recover from disasters. I don't have much need for SVN or complex versioning systems. I'm just looking for something simple and lean. Any help would be much appreciated. Jenny PS : I looked into the built-in AutoRecover feature but it doesn't seem to save more than a few files.

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  • Why does my simple hello world console app use so much memory?

    - by CodingThunder
    Looking in Process Explorer it uses; Virtual Size: 550,000k , Working Set: 28000k Why does my simple hello world console app use so much memory? I take it the difference between the Working Set and Virtual Size means that difference will be paged to disk? /I am running 64 bit XP. Thanks class Program { static void Main(string[] args) { Console.WriteLine("Hello world"); Console.ReadLine(); } }

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  • How to read a simple file to fill a HyperTable with column qualifiers?

    - by Wajih
    I had been looking at the wiki/docs to see for a sample which teaches how to load column qualifiers from a file. Is there any such sample. Could you guide me on the following simple table? Specifically the field FromID which should contain, say FromID: person1 FromID: person2 create table stats (ID1,ID2,To,FromID, QUALIFIER INDEX FromID) My TSV file looks like this #SeQ ID1 ID2 To FromID 01099 3 4 me ---What would the entries be here?? Thank you for the guidance, Wajih

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  • how to build a simple gridview only for insertion (no bind)?

    - by user359706
    Hello, how do I implement a rapid gridview with two columns, with ability to add rows by clicking the add button, and remove in the same way. I look for plugins, but often contain many unnecessary option in my case, I'm looking for something fairly simple. solution jquery, ajax.. I'm using ASP.net (C #) someone has an idea ? thank you

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  • Wordpress: Does exists some simple code to retrieve the page when inserting the ID to form input?

    - by Daniel
    I would like to know if there's exists some simple code to get to the page i know its ID , I would like to create small input (no matter where in templates)from where the people can easily get to the page if they know it's page ID . I have the girls portfolio in wordpress - portfolio=pages x jobs in clubs offers=posts , I would like the girls portfolios to be easily searchable by ID(s) , if possible the same for the posts=jobs in clubs

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  • Creating my own simple CMS for asp.net-mvc?

    - by coure06
    I want to create a simple CMS for my asp.net-mvc site. Needs some help to start. Will i save my whole page to db? what if my page contain links like Url.Content("~/somepage") When the admin will edit the page he will get the plain link not the Url.Content. How i can handle this in CMS?

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  • Simple way to play a single frequency in java?

    - by alleywayjack
    I just want to play a very simple, straight forward note by giving my computer a certain frequency as an integer, and from there I can figure out how to make it play the note longer or shorter. It does not necessarily have to come out of the actual sound card - if it's generated and output by the internal speaker that's okay. I looked at the midi libraries that java has included, and they are way more than what I want to do. This just needs to be very basic.

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  • Does anyone know of a simple (free?) feature request tracking system we could use internally for sales people?

    - by Ryan
    I sometimes hear about pain points of customers using our app from sales people, but there really isn't a good way for us to currently keep track of these. I was going to write one myself but figured I would ask first. I was thinking something so simple it would literally just be a small form for adding a new feature, and then it would appear in the list, like stackexchange questions. Then users can upvote them, or even record each time a user complains about something related to the request so we can order them in priority based on real data. Then I can easily go look every few days and see what's going on. That's really it, nothing more complicated than that. Know of anything?

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  • REST web service keeps first POST parametrs

    - by Diego
    I have a web service in REST, designed with Java and deployed on Tomcat. This is the web service structure: @Path("Personas") public class Personas { @Context private UriInfo context; /** * Creates a new instance of ServiceResource */ public Personas() { } @GET @Produces("text/html") public String consultarEdad (@QueryParam("nombre") String nombre) { ConectorCliente c = new ConectorCliente("root", "cafe.sql", "test"); int edad = c.consultarEdad(nombre); if (edad == Integer.MIN_VALUE) return "-1"; return String.valueOf(edad); } @POST @Produces("text/html") public String insertarPersona(@QueryParam("nombre") String msg, @QueryParam("edad") int edad) { ConectorCliente c = new ConectorCliente("usr", "passwd", "dbname"); c.agregar(msg, edad); return "listo"; } } Where ConectorCliente class is MySQL connector and querying class. So, I had tested this with the @GET actually doing POST work, any user inputed data and information from ma Java FX app and it went direct to webservice's database. However, I changed so the CREATE operation was performed through a webservice responding to an actual POST HTTP request. However, when I run the client and add some info, parameters go OK, but in next time I input different parameters it'll input the same. I run this several times and I can't get the reason of it. This is the clients code: public class WebServicePersonasConsumer { private WebTarget webTarget; private Client client; private static final String BASE_URI = "http://localhost:8080/GetSomeRest/serviciosweb/"; public WebServicePersonasConsumer() { client = javax.ws.rs.client.ClientBuilder.newClient(); webTarget = client.target(BASE_URI).path("Personas"); } public <T> T insertarPersona(Class<T> responseType, String nombre, String edad) throws ClientErrorException { String[] queryParamNames = new String[]{"nombre", "edad"}; String[] queryParamValues = new String[]{nombre, edad}; ; javax.ws.rs.core.Form form = getQueryOrFormParams(queryParamNames, queryParamValues); javax.ws.rs.core.MultivaluedMap<String, String> map = form.asMap(); for (java.util.Map.Entry<String, java.util.List<String>> entry : map.entrySet()) { java.util.List<String> list = entry.getValue(); String[] values = list.toArray(new String[list.size()]); webTarget = webTarget.queryParam(entry.getKey(), (Object[]) values); } return webTarget.request().post(null, responseType); } public <T> T consultarEdad(Class<T> responseType, String nombre) throws ClientErrorException { String[] queryParamNames = new String[]{"nombre"}; String[] queryParamValues = new String[]{nombre}; ; javax.ws.rs.core.Form form = getQueryOrFormParams(queryParamNames, queryParamValues); javax.ws.rs.core.MultivaluedMap<String, String> map = form.asMap(); for (java.util.Map.Entry<String, java.util.List<String>> entry : map.entrySet()) { java.util.List<String> list = entry.getValue(); String[] values = list.toArray(new String[list.size()]); webTarget = webTarget.queryParam(entry.getKey(), (Object[]) values); } return webTarget.request(javax.ws.rs.core.MediaType.TEXT_HTML).get(responseType); } private Form getQueryOrFormParams(String[] paramNames, String[] paramValues) { Form form = new javax.ws.rs.core.Form(); for (int i = 0; i < paramNames.length; i++) { if (paramValues[i] != null) { form = form.param(paramNames[i], paramValues[i]); } } return form; } public void close() { client.close(); } } And this this the code when I perform the operations in a Java FX app: String nombre = nombreTextField.getText(); String edad = edadTextField.getText(); String insertToDatabase = consumidor.insertarPersona(String.class, nombre, edad); So, as parameters are taken from TextFields, is quite odd why second, third, fourth and so on POSTS post the SAME.

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  • How to close all tabs in Safari using AppleScript?

    - by Form
    I have made a very simple AppleScript to close all tabs in Safari. The problem is, it works, but not completely. Here's the code: tell application "Safari" repeat with aWindow in windows repeat with aTab in tabs of aWindow aTab close end repeat end repeat end tell I've also tried this script: tell application "Safari" repeat with i from 0 to the number of items in windows set aWindow to item i of windows repeat with j from 0 to the number of tabs in aWindow set aTab to item j of tabs of aWindow aTab close end repeat end repeat end tell ... but it does not work either. I tried that on my system (MacBook Pro jan 2008), as well as on a Mac Pro G5 under Tiger and the script fails on both, albeit with a much less descriptive error on Tiger. The problem is that only a couple of tabs are closed. Running the script a few times closes a few tab each time until none is left, but always fails with the same error after closing a few tabs. Under Leopard I get an out of bounds error. Since I am using fast enumeration (not using "repeat from 0 to number of items in windows") I don't see how I can get an out of bounds error with this... My goal is to use the Cocoa Scripting Bridge to close tabs in Safari from my Objective-C Cocoa application but the Scripting Bridge fails in the same manner. The non-deletable tabs show as NULL in the Xcode debugger, while the other tabs are valid objects from which I can get values back (such as their title). In fact I tried with the Scripting Bridge first then told myself why not try this directly in AppleScript and I was surprised to see the same results. I must have a glaring omission or something in there... (seems like a bug in Safari AppleScript support to me... :S) I've used repeat loops and Obj-C 2.0 fast enumeration to iterate through collections before with zero problems, so I really don't see what's wrong here. Anyone can help? Thanks in advance!

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  • How add loading image using jquery?

    - by user244394
    I'm working on a form, <form id="myform" class="form"> </form> that gets submitted to the server using jquery ajax. How can I refresh the form on success to show the updated form information and add a spinner until the form loads? here is my html and jquery snippet <div class="container"> <div class="page-header"> <div class="span2"> <!--Sidebar content--> <img src="img/emc_logo.png" title="EMC" > </div> <div class="span6"> <h2 class="form-wizard-heading">Configuration</h2> </div> </div> <form id="myform" class="form"> </form> </div> <!-- /container --> <!-- Modal --> <div id="myModal" class="modal hide fade" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true"> <div class="modal-header"> <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button> <h3 id="myModalLabel">Configuration Changes</h3> <p><span class="label label-important">Please review your changes before submitting. Submitting the changes will result in rebooting the cluster</span></p> </div> <div class="modal-body"> <table class="table table-condensed table-striped" id="display"></table> </div> <div class="modal-footer"> <button id="cancel" class="btn" data-dismiss="modal" aria-hidden="true">Close</button> <button id="save" class="btn btn-primary">Save changes</button> </div> </div> //Jquery part $(document).ready(function () { $('input').hover(function () { $(this).popover('show') }); // On mouseout destroy popout $('input').mouseout(function () { $(this).popover('destroy') }); $('#myform').on('submit', function (ev) { ev.preventDefault(); var data = $(this).serializeObject(); json_data = JSON.stringify(data); $('#myModal').modal('show'); $.each(data, function (key, val) { var tablefeed = $('<tr><td>' + key + '</td><td id="' + key + '">' + val + '</td><tr>').appendTo('#display'); }); $(".modal-body").html(tablefeed); }); $("#cancel").click(function () { $("#display").empty(); }); $(function () { $("#save").click(function () { // validate and process form here alert("button submitted" + json_data); $.ajax({ type: "POST", url: "somefile.json.", data: json_data, contentType: 'application/json', success: function (data, textStatus, xhr) { console.log(arguments); console.log(xhr.status); alert("Your form has been submitted: " + textStatus + xhr.status); }, error: function (jqXHR, textStatus, errorThrown) { alert(jqXHR.responseText + " - " + errorThrown + " : " + jqXHR.status); } }); }); }); });

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  • Bypass OpenID. Please give us a simple login form.

    - by Florin
    Can I kindly ask that we're allowed to login without the OpenID nonsense? This system is so popular that stack-overflow is the only place that I use it. If it is the policy of stack-overflow to prevent people to login, they've succeeded. I am a passive reader. For some reasons, I really don't like the idea of having one Id for all sites. To me, this system is dead in the water. Unless used within organizations I will never use it. Of course, until the government decides to reign us all in. Will you give them a hand? Until then, can we simply have a login form as in 1995? Thank you for your consideration.

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  • Why ((Integer) weightModel.getObject()).intValue(); throws exception

    - by yakup
    I am learning Wicket by "Enjoying Web Development with Wicket" book. And in an example: int weight = ((Integer) weightModel.getObject()).intValue(); is used. When I click Submit button it throws exception. But after changed the code to: int weight=Integer.parseInt( (String) weightModel.getObject()); It works fine. What is the reason for throwing the exception? The full code: GetRequest.java package myapp.postage; import java.util.HashMap; import java.util.Map; import org.apache.wicket.markup.html.WebPage; import org.apache.wicket.markup.html.form.Form; import org.apache.wicket.markup.html.form.TextField; import org.apache.wicket.model.Model; @SuppressWarnings("unchecked") public class GetRequest extends WebPage { private Model weightModel=new Model(); private Model patronCodeModel=new Model(); private Map patronCodeToDiscount; public GetRequest(){ patronCodeToDiscount=new HashMap(); patronCodeToDiscount.put("p1", new Integer(90)); patronCodeToDiscount.put("p2", new Integer(95)); Form form=new Form("form"){ @Override protected void onSubmit(){ int weight = ((Integer) weightModel.getObject()).intValue(); Integer discount=(Integer)patronCodeToDiscount.get(patronCodeModel.getObject()); int postagePerKg=10; int postage=weight*postagePerKg; if(discount!=null){ postage=postage*discount.intValue()/100; } ShowPostage showPostage=new ShowPostage(postage); setResponsePage(showPostage); } }; TextField weight=new TextField("weight",weightModel); form.add(weight); TextField patronCode=new TextField("patronCode",patronCodeModel); form.add(patronCode); add(form); } } The html file GetRequest.html: <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <form wicket:id="form"> <table> <tr> <td>Weight</td> <td><input type="text" wicket:id="weight"/></td> </tr> <tr> <td>Patron code:</td> <td><input type="text" wicket:id="patronCode"/></td> </tr> <tr> <td></td> <td><input type="submit"/></td> </tr> </table> </form> </html>

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  • Submiting a Form without Refreshing the page with jQuery and Ajax Not updating MySQL database.

    - by HEEEEEEELP
    I'm a newbie to JQuery and have a problem, when I click submit button on the form everything says registration was successful but my MYSQL database was not updated everything worked fine until I tried to add the JQuery to the picture. Can someone help me fix this problem so my database is updated? Thanks Here is the JQuery code. $(function() { $(".save-button").click(function() { var address = $("#address").val(); var address_two = $("#address_two").val(); var city_town = $("#city_town").val(); var state_province = $("#state_province").val(); var zipcode = $("#zipcode").val(); var country = $("#country").val(); var email = $("#email").val(); var dataString = 'address='+ address + '&address_two=' + address_two + '&city_town=' + city_town + '&state_province=' + state_province + '&zipcode=' + zipcode + '&country=' + country + '$email=' + email; if(address=='' || address_two=='' || city_town=='' || state_province=='' || zipcode=='' || country=='' || email=='') { $('.success').fadeOut(200).hide(); $('.error').fadeOut(200).show(); } else { $.ajax({ type: "POST", url: "http://localhost/New%20Project/home/index.php", data: dataString, success: function(){ $('.success').fadeIn(200).show(); $('.error').fadeOut(200).hide(); } }); } return false; }); }); Here is the PHP code. if (isset($_POST['contact_info_submitted'])) { // Handle the form. // Query member data from the database and ready it for display $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.*, contact_info.* FROM users INNER JOIN contact_info ON contact_info.user_id = users.user_id WHERE users.user_id=3"); $user_id = mysqli_real_escape_string($mysqli, htmlentities('3')); $address = mysqli_real_escape_string($mysqli, htmlentities($_POST['address'])); $address_two = mysqli_real_escape_string($mysqli, htmlentities($_POST['address_two'])); $city_town = mysqli_real_escape_string($mysqli, htmlentities($_POST['city_town'])); $state_province = mysqli_real_escape_string($mysqli, htmlentities($_POST['state_province'])); $zipcode = mysqli_real_escape_string($mysqli, htmlentities($_POST['zipcode'])); $country = mysqli_real_escape_string($mysqli, htmlentities($_POST['country'])); $email = mysqli_real_escape_string($mysqli, strip_tags($_POST['email'])); //If the table is not found add it to the database if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO contact_info (user_id, address, address_two, city_town, state_province, zipcode, country, email) VALUES ('$user_id', '$address', '$address_two', '$city_town', '$state_province', '$zipcode', '$country', '$email')"); } //If the table is in the database update each field when needed if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE contact_info SET address = '$address', address_two = '$address_two', city_town = '$city_town', state_province = '$state_province', zipcode = '$zipcode', country = '$country', email = '$email' WHERE user_id = '$user_id'"); } if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); return; } } Here is the XHTML code. <form method="post" action="index.php"> <fieldset> <ul> <li><label for="address">Address 1: </label><input type="text" name="address" id="address" size="25" class="input-size" value="<?php if (isset($_POST['address'])) { echo $_POST['address']; } else if(!empty($address)) { echo $address; } ?>" /></li> <li><label for="address_two">Address 2: </label><input type="text" name="address_two" id="address_two" size="25" class="input-size" value="<?php if (isset($_POST['address_two'])) { echo $_POST['address_two']; } else if(!empty($address_two)) { echo $address_two; } ?>" /></li> <li><label for="city_town">City/Town: </label><input type="text" name="city_town" id="city_town" size="25" class="input-size" value="<?php if (isset($_POST['city_town'])) { echo $_POST['city_town']; } else if(!empty($city_town)) { echo $city_town; } ?>" /></li> <li><label for="state_province">State/Province: </label> <?php echo '<select name="state_province" id="state_province">' . "\n"; foreach($state_options as $option) { if ($option == $state_province) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="zipcode">Zip/Post Code: </label><input type="text" name="zipcode" id="zipcode" size="5" class="input-size" value="<?php if (isset($_POST['zipcode'])) { echo $_POST['zipcode']; } else if(!empty($zipcode)) { echo $zipcode; } ?>" /></li> <li><label for="country">Country: </label> <?php echo '<select name="country" id="country">' . "\n"; foreach($countries as $option) { if ($option == $country) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else if($option == "-------------") { echo '<option value="' . $option . '" disabled="disabled">' . $option . '</option>'; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="email">Email Address: </label><input type="text" name="email" id="email" size="25" class="input-size" value="<?php if (isset($_POST['email'])) { echo $_POST['email']; } else if(!empty($email)) { echo $email; } ?>" /><br /><span>We don't spam or share your email with third parties. We respect your privacy.</span></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="contact_info_submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • How can I cache a Subversion password on a server, without storing it in unencrypted form?

    - by Zilk
    My Subversion server only provides access via HTTPS; support for svn+ssh has been dropped because we wanted to avoid creating system users on that machine just for SVN access. Now I'm trying to provide a way for users to cache their passwords for a while, without leaving them stored on the filesystem in unencrypted form. This is no problem for Gnome or KDE users, because they can use gnome-keyring and kwallet, respectively. IIRC, TortoiseSVN has a similar caching mechanism, too. But what about users on a non-GUI system? Some context: in this case, we have a development/testing server where one project has been checked out into the Apache htdocs directory. Development for this project is almost complete, and only minor text/layout changes are performed directly on this server. Nevertheless, the changes should be checked into the repository. There's no kwallet and no gnome-keyring on this system, and the ssh-agent can't help because the repository is accessed via https instead of svn+ssh. As far as I know, that leaves them the choice of entering the password every time they talk to the SVN server, or storing it in an insecure way. Is there any way to get something like what gnome-keyring and kwallet provide in a non-GUI environment?

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  • How to convert aspell dictionary to simple list of words?

    - by rafalmag
    I want to get list of all words from aspell dictionary. I downloaded aspell and aspell polish dictionary, then unziped it using: preunzip pl.cwl I got pl.wl: ... hippie hippies hippiesowski/bXxYc hippika/MNn hippis/NOqsT hippisiara/MnN hippiska/mMN hippisowski/bXxYc ... but they appear with sufix like /bXxYc or /MNn. These suffixes are defined in pl_affix.dat, which looks like ... SFX n Y 5 SFX n a 0 [^ij]a SFX n ja yj [^aeijoóuy]ja SFX n a 0 [aeijoóuy]ja SFX n ia ij [^drt]ia SFX n ia yj [drt]ia ... It is connected to the declination and conjugation. How can I add to the first list all forms (with all corresponding suffixes as defined in .dat file ) ? BTW: I need this list to spell-checker jazzy.

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