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  • Profiling a Java Spring application

    - by niklassaers
    Hi guys, I have a Spring application that I believe has some bottlenecks, so I'd like to run it with a profiler to measure what functions take how much time. Any recommendations to how I should do that? I'm running STS, the project is a maven project, and I'm running Spring 3.0.1 Cheers Nik

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  • Retrieving data from a JSON sub array in javascript, where identifier starts with an integer

    - by Archie Ec
    I must be missing something simple here, but I'm having trouble retrieving data from a JSON array response. I can access objects with identifiers that start with letters, but not ones that start with numbers. For example, I can access data.item[0].specs.overview.details But I can't access data.item[0].specs.9a99.details If anyone can point me in the right direction, I'd really appreciate it. Thanks.

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  • Trying to setup externalizing properties in spring

    - by Gandalf StormCrow
    Hi all, I'm building my project with maven so according to maven way, config should be in src/main/conf , how can I say to my spring application context that that is where jdbc.properties is found? Here is example bean : <bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"> <property name="location" value="jdbc.properties" /> </bean> Spring assumens that this configuration is inside src/main/webapp/WEB-INF, I hope I've been clear if not I'll rephrase my question thank you

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  • parsing JSON in java ?

    - by bukuai
    How to parsing JSON in java? { "code": 100, "message": "SUCCESS", "result": { "list": [ { "cardNumber": "ALUFZZ5SZ1Q5", "expireTime": 1270742400, "surplusThreshold": 4, "uid": 771292, "useTime": 1270214375, "wareId": 1145 }, { "cardNumber": "ALUFZZ5SZ1QD", "expireTime": 1270828800, "surplusThreshold": 7, "uid": 784289, "useTime": 1270302200, "wareId": 1145 }, { "cardNumber": "ALUFZZ5SZ1RC", "expireTime": 1270828800, "surplusThreshold": 10, "uid": 773664, "useTime": 1270300871, "wareId": 1145 }, { "cardNumber": "ALUFZZ5SZ1UM", "expireTime": 1270828800, "surplusThreshold": 10, "uid": 779560, "useTime": 1270282841, "wareId": 1145 }, { "cardNumber": "ALUFZZ5SZ1VT", "expireTime": 1270656000, "surplusThreshold": 2, "uid": 754099, "useTime": 1270106775, "wareId": 1145 } ], "pageCount": 61, "pageIndex": 2, "pageSize": 5, "recordCount": 308 } }

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  • Question about JSON?

    - by Alex
    Hi all, I have a fairly simple question: In Javascript how can I return the boolean (if its found in the JSON) rather than the actual value? Example: var myJSON = { "foo" : "bar", "bar" : "foo" }; var keyToLookFor = "foo"; var found = myJSON[keyToLookFor]; if (found) { // I know I can test if the string exists in the if } // but is there a way I could just return something like: return found; // instead of testing found in the if statement and then returning true?

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  • convert json string into array or object...

    - by qulzam
    I get some json data form the web which is like: [{"pk": 1, "model": "stock.item", "fields": {"style": "f/s", "name": "shirt", "color": "red", "sync": 1, "fabric_code": "0012", "item_code": "001", "size": "34"}}, {"pk": 2, "model": "stock.item", "fields": {"style": "febric", "name": "Trouser", "color": "red", "sync": 1, "fabric_code": "fabric code", "item_code": "0123", "size": "44"}}] How can i use it in the C# winforms desktop application. I already get this data in the form of string. All types of answer are welcome.

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  • DataContractJsonSerializer generating Ghost string to JSON keys?

    - by Anil Namde
    DataContractJsonSerializer this is nice class added in the .net framework which can be used to serialize/desirealize object into JSON. Now following is the example i am trying [Serializable] class User { public string name; public string userId; } Now following is the output generated Output : Notice structure where only "name" is expected instead of k__BackingField Now this is the problem after digging so much i am not sure from where < and _BackingField is coming ? { "<name>k__BackingField":"test user", "<userId>k__BackingField":100001}

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  • using json object data in jquery select

    - by Ratna
    {"names": [ {"patientName": "Ratna"}, {"patientName": "raju" }, {"patientName": "krishna"}, {"patientName": "kishore"}, {"patientName": "Kishore1"}, {"patientName": "mahesh"} ]} this is the JSON object i'm getting from Ajax call so now i want to add all patientName values to select box through jquery can any one tell me how to accomplish this ?? here i'm using $.ajax() function for ajax call thanks in advance

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  • Need to Post json data using jQuery.ajax

    - by Anil Bhat
    I have a json of the following format which I need to send in the Ajax request through POST method: { “gbus”: [ { "code": "*" } ], “regions”: [ { "code": "*" } ], “offices”: [ { "code": "*" } ], “contracttypes”: [ { "code": "*" } ], “jobnumbers”: [ { "code": "*" } ], “disciplines”: [ { "code": "*" } ] } Its not working for me, giving 500 error always when I try to submit it. Please suggest if you have any idea.

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  • How to get data source for Spring Security Web in Java code

    - by user1443689
    I'm creating a ZK Web application which uses Spring Security for authentication and I'm trying to implement a create user function, where the administrator supplies the details and the user account is created. I've got to the part where I want to put this data into the database, but now I'm thinking I shouldn't hard code the connection to the database, there must be a way to get the connection details from the Spring Security configuration. Is there? If so how?

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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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  • Spring Framework 3.0.5 MVC Issue

    - by user578923
    I know that this may be absolutely dumb but for the life of me I cannot figure out why I'm getting these errors in my Spring Project, it is basically from the MVC tutorial with a few modifications. This is the error I get when running my tomcat server. `Caused by: java.lang.NoClassDefFoundError: org/springframework/web/servlet/mvc/SimpleFormController at java.lang.ClassLoader.defineClass1(Native Method) at java.lang.ClassLoader.defineClass(ClassLoader.java:634) at java.security.SecureClassLoader.defineClass(SecureClassLoader.java:142) at java.net.URLClassLoader.defineClass(URLClassLoader.java:277) at java.net.URLClassLoader.access$000(URLClassLoader.java:73) at java.net.URLClassLoader$1.run(URLClassLoader.java:212) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:205) at java.lang.ClassLoader.loadClass(ClassLoader.java:321) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:294) at java.lang.ClassLoader.loadClass(ClassLoader.java:266) at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1581) at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1511) at org.springframework.util.ClassUtils.forName(ClassUtils.java:257) at org.springframework.beans.factory.support.AbstractBeanDefinition.resolveBeanClass(AbstractBeanDefinition.java:408) at org.springframework.beans.factory.support.AbstractBeanFactory.doResolveBeanClass(AbstractBeanFactory.java:1271) at org.springframework.beans.factory.support.AbstractBeanFactory.resolveBeanClass(AbstractBeanFactory.java:1242) ... 54 more Caused by: java.lang.ClassNotFoundException: org.springframework.web.servlet.mvc.SimpleFormController at java.net.URLClassLoader$1.run(URLClassLoader.java:217) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:205) at java.lang.ClassLoader.loadClass(ClassLoader.java:321) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:294) at java.lang.ClassLoader.loadClass(ClassLoader.java:266) ... 71 more` I just cannot figure out the issue with my classpath...I would appreciate any help. Here are all the jars in my classpath. I know that the class is inside the web-servlet jar but it's not seeing it. Is there a conflict? aopalliance.jar aspectjweaver.jar commons-codec.jar commons-dbcp.jar commons-logging.jar commons-pool.jar jstl.jar org.springframework.aop-3.0.5.RELEASE.jar org.springframework.asm-3.0.5.RELEASE.jar org.springframework.aspects-3.0.5.RELEASE.jar org.springframework.beans-3.0.5.RELEASE.jar org.springframework.context.support-3.0.5.RELEASE.jar org.springframework.context-3.0.5.RELEASE.jar org.springframework.core-3.0.5.RELEASE.jar org.springframework.expression-3.0.5.RELEASE.jar org.springframework.instrument.tomcat-3.0.5.RELEASE.jar org.springframework.instrument-3.0.5.RELEASE.jar org.springframework.jdbc-3.0.5.RELEASE.jar org.springframework.jms-3.0.5.RELEASE.jar org.springframework.orm-3.0.5.RELEASE.jar org.springframework.oxm-3.0.5.RELEASE.jar org.springframework.test-3.0.5.RELEASE.jar org.springframework.transaction-3.0.5.RELEASE.jar org.springframework.web.portlet-3.0.5.RELEASE.jar org.springframework.web.servlet-3.0.5.RELEASE.jar org.springframework.web.struts-3.0.5.RELEASE.jar org.springframework.web-3.0.5.RELEASE.jar postgresql-9.0-801.jdbc3.jar servlet-api.jar spring-security-config-3.0.5.RELEASE.jar spring-security-core-3.0.5.RELEASE.jar spring-security-web-3.0.5.RELEASE.jar standard.jar

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  • @Transactional in Spring+Hibernate

    - by Arun Kumar
    I an using Spring 3.1 + Hibernate 4.x in my web application. In my DAO, i am saving User type object as following sessionFactory.getCurrentSession().save(user); But getting following exception: org.hibernate.HibernateException: save is not valid without active transaction I googled and found similar question on SO, with following solution: Session session=getSessionFactory().getCurrentSession(); Transaction trans=session.beginTransaction(); session.save(entity); trans.commit(); That solves the problem. But in that solution, there is lot of mess of beginning and committing the transactions manually. Can't i use sessionFactory.getCurrentSession().save(user); directly without begin/commit of transactions manually? I try to use @Transactional on my service/dao methods too, but the problem persists. EDIT : Here is my Hibernate Config File: <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p" xmlns:aop="http://www.springframework.org/schema/aop" xmlns:tx="http://www.springframework.org/schema/tx" xsi:schemaLocation=" http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.1.xsd"> <!-- enable the configuration of transactional behavior based on annotations --> <tx:annotation-driven transaction-manager="txManager"/> <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource" p:driverClassName="${db.driverClassName}" p:url="${db.url}" p:username="${db.username}" p:password="${db.password}" /> <bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean"> <property name="dataSource" ref="dataSource" /> <property name="packagesToScan" value="com.myapp.entities" /> <property name="hibernateProperties"> <props> <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop> <prop key="hibernate.show_sql">true</prop> </props> </property> </bean> <!--Transaction Manager Added --> <bean id="txManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager"> <property name="sessionFactory"> <ref bean="sessionFactory" /> </property> </bean> </beans> Please help.

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  • Python | How to send a JSON response with name assign to it

    - by MMRUser
    How can I return an response (lets say an array) to the client with a name assign to it form a python script. echo '{"jsonValidateReturn":'.json_encode($arrayToJs).'}'; in this scenario it returns an array with the name(jsonValidateReturn) assign to it also this can be accessed by jsonValidateReturn[1],so I want to do the same using a python script. I tried it once but it didn't go well array_to_js = [vld_id, vld_error, False] array_to_js[2] = False jsonValidateReturn = simplejson.dumps(array_to_js) return HttpResponse(jsonValidateReturn, mimetype='application/json') Thanks.

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  • Status messages on the Spring MVC-based site (annotation controller)

    - by linpulee
    What is the best way to organize status messages on the Spring MVC-based site? I mean messages which, for example, returns when user has sent What is the best way to organize status messages ("Your data has been successfully saved/added/deleted") on the Spring MVC-based site using annotation controller? So, the issue is in the way of sending the message from POST-method in contoller.

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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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  • 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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  • MVC JSON ViewData Question

    - by user325142
    Can ViewData be set & returned in an ActionResult that returns Json() ? Im asking because It's not working for me but when I put the same ViewData code into a Action method that returns a View() it works? 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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