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  • Configuring servlet for using spring

    - by dominolog
    Hello How to configure my web application in Eclipse (based on Servlets and deployed to Tomcat) to use Spring framework. I need only IoC container (Beans only and ApplicationContext), not Spring MVC. How to configure web.xml for this? Regards

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  • XSD Client in Spring

    - by wuntee
    I have an XSD document that I need to communicate with an endpoint (client side only) - is there this functionality built into spring? I have been using JAXB, but was wondering if spring has some sort of wrapper. Thanks.

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  • how to get json string value?

    - by Net205
    var responseFromServer = "{\"flag\":true,\"message\":\"\",\"result\":{\"ServicePermission\":true,\"UserGroupPermission\":true}}"; var serializer = new System.Web.Script.Serialization.JavaScriptSerializer(); var responseValue = serializer.DeserializeObject(responseFromServer); responseFromServer value is get a webservice, and then how to get the json string value, such as "flag","Servicepermission"??

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  • PHP getting Twitter API JSON file contents without OAuth (Almost have it)

    - by DexCurl
    Hey guys, I have this script working fine with OAuth, but I accidentally nuked my 350 API hits with a stupid while statement :( I'm trying to get data from the Twitter API without OAuth, I can't figure it out (still pretty new), heres what I have <html> <body> <center> <hr /> <br /> <table border="1"> <tr><td>ScreenName</td><td>Followed back?</td></tr> <?php //twitter oauth deets $consumerKey = 'x'; $consumerSecret = 'x'; $oAuthToken = 'x'; $oAuthSecret = 'x'; // Create Twitter API objsect require_once("twitteroauth.php"); $oauth = new TwitterOAuth($consumerKey, $consumerSecret, $oAuthToken, $oAuthSecret); //get home timeline tweets and it is stored as an array $youfollow = $oauth->get('http://api.twitter.com/1/friends/ids.json?screen_name=lccountdown'); $i = 0; //start loop to print our results cutely in a table while ($i <= 20){ $youfollowid = $youfollow[$i]; $resolve = "http://api.twitter.com/1/friendships/exists.json?user_a=".$youfollow[$i]."&user_b=jwhelton"; $followbacktest = $oauth->get($resolve); //$homedate= $hometimeline[$i]->created_at; //$homescreenname = $hometimeline[$i]->user->screen_name; echo "<tr><td>".$youfollowid."</td><td>".$followbacktest."</td></tr>"; $i++; } ?> </table> </center> </body> </html> Neither of the two Twitter functions require authentication, so how can I get the same results? Thanks guys, Dex

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  • equivalent of javascript class using JSON

    - by brz dot net
    See following class: function availItem(xs, s, m, l, xl) { this.xs = xs; this.s = s; this.m = m; this.l = l; this.xl = xl; } How can I declare the above class using JSON? I think It should be in following manner but problem is to pass argument. var availItem = { xs : xs, s : s, m : m, l : l, xl : xl }

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  • How can I visually format JSON data (programmatically)?

    - by Ian Robinson
    I'm working with big blobs of JSON. These blobs change slightly over time and a revision history is kept. I'd really like to be able to do a visual diff on them, but my problem is they're being stored without any formatting at all - everything is on one line, so that makes it a little hard to see what changed. Is there a good way to programatically format them ala http://jsonformat.com/ or http://jsonformatter.curiousconcept.com/?

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  • Any webservice that returns JSON

    - by roneden
    Does anyone know of free webservice that returns a list in the form of JSON? I have searched many webservice sites but they all return xml. Geonames.org is not required. List all that you know of please. thanks.

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  • HTML form to JSON without modern libraries

    - by Rok
    Does anyone know of a simple javascript library that serializes a form DOM object to JSON? I know jQuery has the serialize method but since I am writing an app for IE mobile, I can't use any of the new fancy libraries, must be oldschool simple javascript, as light as possible. Cheers

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  • C# class to parse JSON result

    - by user285677
    Hi, How should be a c# class to deserialize the folowing JSON string: { "data": [ { "id" : "id0012", "comments": { "data": [ { "id": "123", "from": { "name": "xpto", "id": "5ddd" }, "message": "tttt", "created_time": "2010-01-07T09:16:15+0000" }, { "id": "222", "from": { "name": "wwwww", "id": "343434" }, "message": "3333", "created_time": "2020-07-07T09:30:12+0000" } ], "paging": { "previous": "prevlink", "next": "nextLink" } } } ] } Thanks

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  • Instantiating custom PropertySourcesPlaceholderConfigurer from spring context

    - by mmona
    I want to define a custom PropertySourcesPlaceholderConfigurer in spring context xml. I want to use there multiple PropertySources, so that I can load part of the configuration from several property files and provide other part dynamically by my custom PropertySource implementation. The advantage is that it should be then easy to adjust the order of loading these property sources just by making modifications to the xml spring configuration. And here I run into a problem: how to define an arbitrary list of PropertySources and inject it into PropertySourcesPlaceholderConfigurer, so that it uses the sources defined by me? Seems to be a basic thing that should be provided by spring, but since yesterday I cannot find a way to do it. Using namespace would enable me to load several property files, but I also need to define the id of the PropertySourcesPlaceholderConfigurer (as other projects refer to it), and also I want to use my custom implementation. That is why I am defining the bean explicitly and not using the namespace. The most intuitive way would be to inject a list of PropertySources into PropertySourcesPlaceholderConfigurer like this: <bean id="applicationPropertyPlaceholderConfigurer" class="org.springframework.context.support.PropertySourcesPlaceholderConfigurer"> <property name="ignoreUnresolvablePlaceholders" value="true" /> <property name="ignoreResourceNotFound" value="true" /> <property name="order" value="0"/> <property name="propertySources"> <list> <!-- my PropertySource objects --> </list> </property> </bean> but unfortunately propertySources is of type PropertySources and does not accept a list. The PropertySources interface has one and only implementor which is MutablePropertySources, which indeed stores list of PropertySource objects, but has no constructor nor setter through which I can inject this list. It only has add*(PropertySource) methods. The only workaround I see now is to implement my own PropertySources class, extending MutablePropertySources, which would accept list of PropertySource objects on creation and manually add it via using add*(PropertySource) method. But why so much workaround would be needed to provide something that I thought was supposed to be the main reason of introducing the PropertySources (having flexible configuration manageable from spring configuration level). Please clarify what am I getting wrong :)

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  • json retrival failed with jquery .each

    - by user545520
    {"paging": {"pageNum":2,"action":"Next","type":"","availableCacheName":"getAllFunds","selectedCacheName":"","showFrom":101,"showTo":200,"totalRec":289,"pageSize":100}, "Data":[{"sourceCodeId":0,"radio_fund":"individua l","availableFunds":[],"fundId":288,"searchName":[],"fundName":"Asian Equity Fund A Class Income","srcFundGrpId":"PGI","firstElement":0,"las tElement":0,"totalElements":0,"pageList":[],"standardExtract":true}] I have json file with above format with two fileds,one paging and one is Data array. I able to retrieve values of paging,but i am not able to retrieve the values of data array with .each function of jquery. Any suggestions or inputs really appreciated.

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  • Upload an image from JSON with rails

    - by Dougui
    I have an application created with PhoneGap and Backbone. I upload a file as JSon and my server receive data like this : data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQABAAD/... I'm trying to write the file like this : File.open("#{Rails.root}/public/images/#{self.id}.jpg", "w+") do |f| f.write(data) end It's not working and I don't know what to do. When I'm trying to open the file I have this message "Not a JPEG file: starts with 0x64 0x61". Do you have a solution?

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  • jquery json parsing

    - by lolweb
    How do I parse this json with jQuery? DayEvents : [{"0":"886","event_id":"886","1":"5029","user_id":"5029","2":"Professional","user_type":"Professional", ...

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  • How to navigate around a '[' in JSON.

    - by Kyle
    I'm new to JSON and moving around in it in jQuery. I'm fine until I hit a '[', as in: "gd$when": [{ "startTime": "2006-11-15", "endTime": "2006-11-17", "gd$reminder": [{"minutes": "10"}] }], I tried to do a eventTime = event["gd$when"]["startTime"]; to get to the 'startTime' (Yes, event is the variable for ajax stuff, it's all working fine until I hit the '[') Thanks for any help.

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  • PHP $array[0] in $string for JSON

    - by user1907696
    I am trying to use PHP to make a JSON file. Part of the code is as follow $array = array("hello", "world"); $string='{"person": [ { "name":'$array[0];', "age":'$array[1];' } ] }'; The file created. However, $array[0] and $array[1] doesn't return the values "hello" and "world" but as $array[0] and $array[1] Any idea? Thanks

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  • How to add reflection definition to read JSON files in web game

    - by user3728735
    I have a game which I deployed for desktop and Android. I can read JSON data and create my levels, but when it comes to reading JSON files from web app, I get an error that logs, "cannot read the json file". I researched a lot and I found out that I should add my JSON config class to configurations, so I added this line to gameName.gwt.xml, which is in core folder: <extend-configuration-property name="gdx.reflect.include" value="com.las.get.level.LevelConfig"/> But it did not work out. I have no idea where should I place this line or where I should change to make my web app work, so I can read JSON files.

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  • How to add reflection definition to read json files on web game

    - by user3728735
    I have a game which I deployed for desktop and android, I can read json data and create my levels, but the problem is, when it comes to reading json files from web app, I get an error that logs, cannot read the json file, I researched a lot and I found out that I should add my json config class to configurations, I added this line to gameName.gwt.xml, which is in core folder <extend-configuration-property name="gdx.reflect.include" value="com.las.get.level.LevelConfig"/> but it did not work out too, I have no idea where should I place this line, or where should I change to make my web app work, so I can read json files

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  • Problems using HibernateTemplate: java.lang.NoSuchMethodError: org.hibernate.SessionFactory.openSession()Lorg/hibernate/classic/Session;

    - by user2104160
    I am quite new in Spring world and I am going crazy trying to integrate Hibernate in Spring application using HibernateTemplate abstract support class I have the following class to persist on database table: package org.andrea.myexample.HibernateOnSpring.entity; import javax.persistence.Entity; import javax.persistence.GeneratedValue; import javax.persistence.GenerationType; import javax.persistence.Id; import javax.persistence.Table; @Entity @Table(name="person") public class Person { @Id @GeneratedValue(strategy=GenerationType.AUTO) private int pid; private String firstname; private String lastname; public int getPid() { return pid; } public void setPid(int pid) { this.pid = pid; } public String getFirstname() { return firstname; } public void setFirstname(String firstname) { this.firstname = firstname; } public String getLastname() { return lastname; } public void setLastname(String lastname) { this.lastname = lastname; } } Next to it I have create an interface named PersonDAO in wich I only define my CRUD method. So I have implement this interface by a class named PersonDAOImpl that also extend the Spring abstract class HibernateTemplate: package org.andrea.myexample.HibernateOnSpring.dao; import java.util.List; import org.andrea.myexample.HibernateOnSpring.entity.Person; import org.springframework.orm.hibernate3.support.HibernateDaoSupport; public class PersonDAOImpl extends HibernateDaoSupport implements PersonDAO{ public void addPerson(Person p) { getHibernateTemplate().saveOrUpdate(p); } public Person getById(int id) { // TODO Auto-generated method stub return null; } public List<Person> getPersonsList() { // TODO Auto-generated method stub return null; } public void delete(int id) { // TODO Auto-generated method stub } public void update(Person person) { // TODO Auto-generated method stub } } (at the moment I am trying to implement only the addPerson() method) Then I have create a main class to test the operation of insert a new object into the database table: package org.andrea.myexample.HibernateOnSpring; import org.andrea.myexample.HibernateOnSpring.dao.PersonDAO; import org.andrea.myexample.HibernateOnSpring.entity.Person; import org.springframework.context.ApplicationContext; import org.springframework.context.support.ClassPathXmlApplicationContext; public class MainApp { public static void main(String[] args) { ApplicationContext context = new ClassPathXmlApplicationContext("Beans.xml"); System.out.println("Contesto recuperato: " + context); Person persona1 = new Person(); persona1.setFirstname("Pippo"); persona1.setLastname("Blabla"); System.out.println("Creato persona1: " + persona1); PersonDAO dao = (PersonDAO) context.getBean("personDAOImpl"); System.out.println("Creato dao object: " + dao); dao.addPerson(persona1); System.out.println("persona1 salvata nel database"); } } As you can see the PersonDAOImpl class extends HibernateTemplate so I think that it have to contain the operation of setting of the sessionFactory... The problem is that when I try to run this MainApp class I obtain the following exception: Exception in thread "main" java.lang.NoSuchMethodError: org.hibernate.SessionFactory.openSession()Lorg/hibernate/classic/Session; at org.springframework.orm.hibernate3.SessionFactoryUtils.doGetSession(SessionFactoryUtils.java:323) at org.springframework.orm.hibernate3.SessionFactoryUtils.getSession(SessionFactoryUtils.java:235) at org.springframework.orm.hibernate3.HibernateTemplate.getSession(HibernateTemplate.java:457) at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:392) at org.springframework.orm.hibernate3.HibernateTemplate.executeWithNativeSession(HibernateTemplate.java:374) at org.springframework.orm.hibernate3.HibernateTemplate.saveOrUpdate(HibernateTemplate.java:737) at org.andrea.myexample.HibernateOnSpring.dao.PersonDAOImpl.addPerson(PersonDAOImpl.java:12) at org.andrea.myexample.HibernateOnSpring.MainApp.main(MainApp.java:26) Why I have this problem? how can I solve it? To be complete I also insert my pom.xml containing my dependencies list: <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>org.andrea.myexample</groupId> <artifactId>HibernateOnSpring</artifactId> <version>0.0.1-SNAPSHOT</version> <packaging>jar</packaging> <name>HibernateOnSpring</name> <url>http://maven.apache.org</url> <properties> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> </properties> <dependencies> <dependency> <groupId>junit</groupId> <artifactId>junit</artifactId> <version>3.8.1</version> <scope>test</scope> </dependency> <!-- Dipendenze di Spring Framework --> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-core</artifactId> <version>3.2.1.RELEASE</version> </dependency> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-beans</artifactId> <version>3.2.1.RELEASE</version> </dependency> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-context</artifactId> <version>3.2.1.RELEASE</version> </dependency> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-context-support</artifactId> <version>3.2.1.RELEASE</version> </dependency> <dependency> <!-- Usata da Hibernate 4 per LocalSessionFactoryBean --> <groupId>org.springframework</groupId> <artifactId>spring-orm</artifactId> <version>3.2.0.RELEASE</version> </dependency> <!-- Dipendenze per AOP --> <dependency> <groupId>cglib</groupId> <artifactId>cglib</artifactId> <version>2.2.2</version> </dependency> <!-- Dipendenze per Persistence Managment --> <dependency> <!-- Apache BasicDataSource --> <groupId>commons-dbcp</groupId> <artifactId>commons-dbcp</artifactId> <version>1.4</version> </dependency> <dependency> <!-- MySQL database driver --> <groupId>mysql</groupId> <artifactId>mysql-connector-java</artifactId> <version>5.1.23</version> </dependency> <dependency> <!-- Hibernate --> <groupId>org.hibernate</groupId> <artifactId>hibernate-core</artifactId> <version>4.1.9.Final</version> </dependency> </dependencies> </project>

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  • How to add custom SOAP-Header element to the generated WSDL in Spring-WS

    - by Petr Macek
    Hi, we are migrating from WebLogic web-services to Spring-WS (1.5.X). There is currently one issue we are facing: We need to pass a context object (on WLS it is passed as SOAP-Header element) to other services that are still running on WLS from the Spring-WS powered service. The header element is still formulated on client side and the newly created WS (Spring-WS) should just pass it to other services. I can imagine how the custom element would be passed: override the doWithMessage(WebServiceMessage message) method... Is there a way to generate the wsdl with the help of DefaultWsdl11Definition to contain that custom header element? See the example: <wsdl:operation name="GetSomeInformation"> <soap:operation soapAction="http://www.dummyservice.com/InformationService/GetSomeInformation" /> <wsdl:input> <soap:body use="literal" /> <soap:header message="ctx:ServiceContextMessage" part="serviceContext" use="literal" /> </wsdl:input> <wsdl:output> <soap:body use="literal" /> </wsdl:output> <wsdl:fault name="Error"> <soap:fault name="Error" use="literal" /> </wsdl:fault> </wsdl:operation> Thanks for help

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  • How to parse Json object in ASP classic passed from jQuery

    - by Michael Itzoe
    Using a jQuery dialog, on clicking OK I call $.post( "save.asp", { id: 1, value: "abcxyz" } ); to pass the values to my ASP classic file that will update the database. I don't need a return value (unless it fails). I'm a relative noob to jQuery, so I'm assuming I'm using JSON to pass the values to the ASP file. I just don't know what to do with them in ASP (using VBScript). I've seen things like ASP Extreme, but I'm not clear on how to use them. I've tried referencing values via the Request collection, but no luck. All I want to do is take the values passed, parse them out, then save them to the database. Sorry if this is a duplicate, but this just isn't clicking for me.

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  • jsonSerializer.DeserializeObject and arrays

    - by Chin
    I have a column in the a database with values like the below. [{"noteText":"Today was sunny.","noteDate":"2010-03-30 10:06:22"},{"noteDate":"2010-04-06 13:21:36","noteText":"Today was windy."}] I think they are from an array of objects serialized via flash to Json. What I need to do is pull out the noteText and noteDate values only and record them back to the database as a normal string. I was hoping to just deserialize back to objects and build up a string from there, however, due to my unfamiliarity with c# and .Net I've hit a brick wall trying to deserialize the string. var obj = jsonSerializer.DeserializeObject(ns); Am I going in the right direction or should I be looking at doing some string manipulation? Any pointers much appreciated.

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  • Validating result of JsonConvert.DeserializeObject (think "try parse") using JSON.Net

    - by Riri
    I have incoming messages that I need to try and parse in my own objects structure. SOme of these are well formed JSON obejcts and some are just nonsense. I use JsonConvert.DeserializeObject<MyObject>(incmoingString); to do this. This however sometimes gives me a exception when the incoming is total garbage. Other times I get a non-complete object structure when the incoming string is kind of OK - and finally it sometimes work. I've wrapped the conversion in a try/catch and than manually validate that I've gotten the properties I need to the deserialized result. Is there a better way to do this?

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  • Create a JSON Array using Java

    - by Ankur
    Hi I want to create a JSON array. I have tried using: JSONArray jArray = new JSONArray(); while(itr.hasNext()){ int objId = itr.next(); jArray.put(objId, odao.getObjectName(objId)); } results = jArray.toString(); Note: odao.getObjectName(objId) retrieves a name based on the "object Id" which is called objId However I get a very funny looking array like [null,null,null,"SomeValue",null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,"AnotherValue",null,null,null,null,null,null,null,null,null,null,"SomethingElse","AnotherOne","LastOne"] With only "LastOne" being displayed when I retrieve it using jQuery. The Array SHould look like {["3":"SomeValue"],["40":"AnotherValue"],["23":"SomethingElse"],["9":"AnotherOne"],["1":"LastOne"]} The numbers aren't showing up at all for some reason in the array that I am getting

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