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  • Stop CDATA tags from being output-escaped when writing to XML in C#

    - by Smallgods
    We're creating a system outputting some data to an XML schema. Some of the fields in this schema need their formatting preserved, as it will be parsed by the end system into potentially a Word doc layout. To do this we're using <![CDATA[Some formatted text]]> tags inside of the App.Config file, then putting that into an appropriate property field in a xsd.exe generated class from our schema. Ideally the formatting wouldn't be out problem, but unfortunately thats just how the system is going. The App.Config section looks as follows: <header> <![CDATA[Some sample formatted data]]> </header> The data assignment looks as follows: HeaderSection header = ConfigurationManager.GetSection("header") as HeaderSection; report.header = "<[CDATA[" + header.Header + "]]>"; Finally, the Xml output is handled as follows: xs = new XmlSerializer(typeof(report)); fs = new FileStream (reportLocation, FileMode.Create); xs.Serialize(fs, report); fs.Flush(); fs.Close(); This should in theory produce in the final Xml a section that has information with CDATA tags around it. However, the angled brackets are being converted into &lt; and &gt; I've looked at ways of disabling Outout Escaping, but so far can only find references to XSLT sheets. I've also tried @"<[CDATA[" with the strings, but again no luck. Any help would be appreciated!

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  • Linq to xml, retreaving generic interface-based list

    - by Rita
    I have an xml document that looks like this <Elements> <Element> <DisplayName /> <Type /> </Element> </Elements> I have an interface, interface IElement { string DisplayName {get;} } and a couple of derived classes: public class AElement: IElement public class BElement: IElement What I want to do, is to write the most efficient query to iterate through the xml and create a list of IElement, containing AElement or BElement, based on the 'Type' property in the xml. So far I have this: IEnumerable<AElement> elements = from xmlElement in XElement.Load(path).Elements("Element") where xmlElement.Element("type").Value == "AElement" select new AElement(xmlElement.Element("DisplayName").Value); return elements.Cast<IElement>().ToList(); But this is only for AElement, is there a way to add BElement in the same query, and also make it generic IEnumerable? Or would I have to run this query once for each derived type?

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  • Creating a 'flexible' XML schema

    - by Fiona Holder
    I need to create a schema for an XML file that is pretty flexible. It has to meet the following requirements: Validate some elements that we require to be present, and know the exact structure of Validate some elements that are optional, and we know the exact structure of Allow any other elements Allow them in any order Quick example: XML <person> <age></age> <lastname></lastname> <height></height> </person> My attempt at an XSD: <xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xs:element name="person"> <xs:complexType> <xs:sequence> <xs:element name="firstname" minOccurs="0" type="xs:string"/> <xs:element name="lastname" type="xs:string"/> <xs:any processContents="lax" minOccurs="0" maxOccurs="unbounded" /> </xs:sequence> </xs:complexType> </xs:element> </xs:schema> Now my XSD satisfies requirements 1 and 3. It is not a valid schema however, if both firstname and lastname were optional, so it doesn't satisfy requirement 2, and the order is fixed, which fails requirement 4. Now all I need is something to validate my XML. I'm open to suggestions on any way of doing this, either programmatically in .NET 3.5, another type of schema etc. Can anyone think of a solution to satisfy all 4 requirements?

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  • Help with an RSS Feed

    - by Pete Herbert Penito
    Hi Everyone! I've spent ages on this, all I'm trying to do is extract the "title" contents from an rss feed, everything else can be ignored. I've looked into simplepie, magpie and all that stuff, but I feel its kind of overkill for what I need to do. I realise there are google gadgets that are made that can do this, but I didn't want all the google logo stuff, and I wanted to personally make this. theres a whole bunch of unneeded tags thats coming in from the rss feed all I need is the title tag, it looks like this <title> My Title 3.0 </title> My server has PHP 5+ so I know I can use some of these simple xml functions which look promising. so far I've got <?php $blogfeed = file_get_contents("http://myblog.blogspot.com/feeds/posts/default?alt=rss"); echo $blogfeed; ?> And it gives me all the data, I was thinking of running through it with strpos and searching for <title> but is there any easier way to do this?? Thanks alot!

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  • INSERT DELAYED on locked tables blocks PHP processes to continue

    - by sw0x2A
    Our webservers write some tracking information into a MySQL database (using INSERT DELAYED into MyISAM table). When a huge SELECT query is executed on this table or when it is locked for another reason, the webserver processes (with INSERT DELAYED) are waiting for the database and in some cases the MaxServer limit is reached in Apaches, so they will stop serving requests. We use INSERT DELAYED because The DELAYED option for the INSERT statement is a MySQL extension to standard SQL that is very useful if you have clients that cannot or need not wait for the INSERT to complete. This is a common situation when you use MySQL for logging and you also periodically run SELECT and UPDATE statements that take a long time to complete. Quote from MySQL documentation. I am wondering why the Apache processes are waiting for the INSERT DELAYED to finish. And what can I do to just send the data and forget about it. (Since this is logging data, I do not care if we lose some entries.) Even when the table is locked the PHP script should just go on and should not wait for an answer of MySQL. (We do not want to setup Master-slave for this table but we are thinking about move this data to some NoSQL database. But for now I would like to know why INSERT DELAYED is not working as expected.)

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  • where are wrong in my php code ????

    - by user318068
    hi all, <td align="center" bgcolor="#FFFFFF"><?php echo '<label onclick="window.open('profilephp.php?member=$row['MemberID']','mywindow')">'.{$row['MemberName']}.'</label>';?><br /> <?php echo "<p align='center'><img width='100' height='100' src={$row['MemberImg']} alt='' /></p>";?></td></tr> Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\xampp\htdocs\home - Copy\membercopy.php on line 141 I really don't know where it went wrong. Please help,

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  • php Dollar amount Regular Expression

    - by Thildemar
    I am have completed javascript validation of a form using Regular Expressions and am now working on redundant verification server-side using PHP. I have copied this regular expression from my jscript code that finds dollar values, and reformed it to a PHP friendly format: /\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/ Specifically: if (preg_match("/\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/", $_POST["cost"])){} While the expression works great in javascript I get : Warning: preg_match() [function.preg-match]: Compilation failed: nothing to repeat at offset 1 when I run it in PHP. Anyone have a clue why this error is coming up?

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  • Can't serve HTML5 video through PHP on Safari/Mac (5.0)

    - by JKS
    I'm encountering a strange bug in Safari where, when I serve MP4 video through PHP (to obfuscate the file beneath the document root with a token-based authentication system), Safari for some reason fires the <video>'s onerror event, and the video never loads (I can't get any useful information out of the event object sent to onerror — everything is undefined). When I access the PHP script directly (i.e., the video is not embedded in a page), the video controls appear momentarily before flashing to a QuickTime question mark. When I access the MP4 file directly, it works as expected. What's bizarre is that the embedded video works perfectly in the latest version of Chrome for Mac. Here are the headers when accessed through PHP: Connection:Keep-Alive Content-Disposition:inline; filename="test.mp4" Content-Length:5558749 Content-Type:video/mp4 Date:Tue, 22 Jun 2010 01:24:25 GMT Keep-Alive:timeout=10, max=29 Server:Apache/2.2.15 (CentOS) mod_ssl/2.2.15 0.9.8l DAV/2 mod_auth_passthrough/2.1 FrontPage/5.0.2.2635 X-Powered-By:PHP/5.2.13 And here are the headers when test.mp4 is accessed directly: Accept-Ranges:bytes Connection:Keep-Alive Content-Length:5558749 Content-Type:video/mp4 Date:Tue, 22 Jun 2010 01:26:45 GMT Etag:"1c04757-54d1dd-489944c5a6400" Keep-Alive:timeout=10, max=30 Last-Modified:Tue, 22 Jun 2010 01:25:36 GMT Server:Apache/2.2.15 (CentOS) mod_ssl/2.2.15 0.9.8l DAV/2 mod_auth_passthrough/2.1 FrontPage/5.0.2.2635 The only differing headers are: Accept-Ranges (which I don't think is necessary), Etag, Last-Modified, Content-Disposition, and X-Powered-By. Not only can Chrome handle the PHP-served video fine, but when I use the same script to load the MP4 through a Flash player, it also works fine. I just can't figure out what Safari is choking on. EDIT: Also, when I change the content disposition to "attachment", Safari will download the MP4 file just fine.

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  • will php/apache ever support multi threading?

    - by fayer
    i mainly focus on the web, i think i will never create desktop applications. so i think it's better for me to focus on typical web languages like php. i know an advantage java has over php is multi threading though. will php ever support this feature in the future? thanks

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  • Learning PHP - start out using a framework or no?

    - by Kevin Torrent
    I've noticed a lot of jobs in my area for PHP. I've never used PHP before, and figure if I can get more opportunities if I pick it up then it might be a good idea. The problem is that PHP without any framework is ugly and 99% of the time really bad code. All the tutorials and books I've seen are really lousy - it never shows any kind of good programming practice but always the quick and dirty kind of way of doing things. I'm afraid that trying to learn PHP this way will just imprint these bad practices in my head and make me waste time later trying to unlearn them. I've used C# in the past so I'm familiar with OOP and software design patterns and similar. Should I be trying to learn PHP by using one of the better known frameworks for it? I've looked at CakePHP, Symfony and the Zend Framework so far; Zend seems to be the most flexible without being too constraining like Cake and Symfony (although Symfony seemed less constraining than CakePHP which is trying too hard to be Ruby on Rails), but many tutorials for Zend I've seen assume you already know PHP and want to learn to use the framework. What would be my best opportunity for learning PHP, but learning GOOD PHP that uses real software engineering techniques instead of spaghetti code? It seems all the PHP books and resources either assume you are just using raw PHP and therefore showcase bade practices, or that you already know PHP and therefore don't even touch on parts of the language.

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  • Hide URL of PHP page

    - by manoj singhal
    I want to hide the URL of my PHP page; that is, I don't want to write /register.php directly in the href tag, I want to write /register/ and have it open the register.php page directly. I want to do that for all the webpages.

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  • Reading XML or objects from a Web service

    - by Shawn
    This is my first time working with webservices and I am a bit lost. I successfully called the functions, but I only can get one value from the service. I read that the easiest way is to read xml or create objects and then call their values. Currently I use functions that return the desired value but I need to call them 3 times to get all the data witch is a waste of time and resources. I tried to call the service with the URL and use it as a website or getting the service to work the same way without importing into the program. The thing is that i cant find a way to pass the values into the url, because of that i get only blank pages. What is the fastest way to get my data from the services? I need city name, temperature and a flag if the city is valid. I need to pass the zip code to the service. Thank you. My current code wetther.Weather wether = new wetther.Weather(); string farenhait = wether.GetCityWeatherByZIP(zip).Temperature; string city = wether.GetCityWeatherByZIP(zip).City; bool correct = wether.GetCityWeatherByZIP(zip).Success; I tried it that way // Retrieve XML document XmlTextReader reader = new XmlTextReader("http://xml.weather.yahoo.com/forecastrss?p=94704"); // Skip non-significant whitespace reader.WhitespaceHandling = WhitespaceHandling.Significant; // Read nodes one at a time while (reader.Read()) { // Print out info on node Console.WriteLine("{0}: {1}", reader.NodeType.ToString(), reader.Name); } This one works for the yahoo page but not for mine. I need to use this webservice - http://wsf.cdyne.com/WeatherWS/Weather.asmx

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  • jQuery ajax request to php, how to return plain text only

    - by jyoseph
    I am making an ajax request to a php page and I cannot for the life of me get it to just return plain text. $.ajax({ type: 'GET', url: '/shipping/test.php', cache: false, dataType: 'text', data: myData, success: function(data){ console.log(data) } }); in test.php I am including this script to get UPS rates. I am then calling the function with $rate = ups($dest_zip,$service,$weight,$length,$width,$height); I am doing echo $rate; at the bottom of test.php. When viewed in a browser shows the rate, that's great. But when I request the page via ajax I get a bunch of XML. Pastie here: http://pastie.org/1416142 My question is, how do I get it so I can just return the plain text string from the ajax call, where the result data will be a number? Edit, here's what I see in Firebug- Response tab: HTML tab:

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  • PHP language specification ?

    - by Rolf
    Hi, as I know there is an official document for Java (JLS), I'd like to know if it's also the case of PHP language. I found the "Language Reference" section on the PHP manual, but it doesn't look as detailed as the JLS. The thing is I have a good practical knowledge of PHP but I'm miserably clueless about what REALLY happens under the hood. If there isn't any official document, could you recommend me some good books to read ? Thanks in advance ! Rolf

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  • PHP & WP: Render Certain Markup Based on True False Condition

    - by rob
    So, I'm working on a site where on the top of certain pages I'd like to display a static graphic and on some pages I would like to display an scrolling banner. So far I set up the condition as follows: <?php $regBanner = true; $regBannerURL = get_bloginfo('stylesheet_directory'); //grabbing WP site URL ?> and in my markup: <div id="banner"> <?php if ($regBanner) { echo "<img src='" . $regBannerURL . "/style/images/main_site/home_page/mock_banner.jpg' />"; } else { echo 'Slider!'; } ?> </div><!-- end banner --> In my else statement, where I'm echoing 'Slider!' I would like to output the markup for my slider: <div id="slider"> <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/1.jpg" alt="" /> <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/2.jpg" alt="" /> <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/3.jpg" alt="" /> ............. </div> My question is how can I throw the div and all those images into my else echo statement? I'm having trouble escaping the quotes and my slider markup isn't rendering.

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  • PHP echo query result in Class??

    - by Jerry
    Hi all I have a question about PHP Class. I am trying to get the result from Mysql via PHP. I would like to know if the best practice is to display the result inside the Class or store the result and handle it in html. For example, display result inside the Class class Schedule { public $currentWeek; function teamQuery($currentWeek){ $this->currentWeek=$currentWeek; } function getSchedule(){ $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } $scheduleQuery=mysql_query("SELECT guest, home, time, winner, pickEnable FROM $this->currentWeek ORDER BY time", $connection); if (!$scheduleQuery){ die("database has errors: ".mysql_error()); } while($row=mysql_fetch_array($scheduleQuery, MYSQL_NUMS)){ //display the result..ex: echo $row['winner']; } mysql_close($scheduleQuery); //no returns } } Or return the query result as a variable and handle in php class Schedule { public $currentWeek; function teamQuery($currentWeek){ $this->currentWeek=$currentWeek; } function getSchedule(){ $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } $scheduleQuery=mysql_query("SELECT guest, home, time, winner, pickEnable FROM $this->currentWeek ORDER BY time", $connection); if (!$scheduleQuery){ die("database has errors: ".mysql_error()); // create an array } $ret = array(); while($row=mysql_fetch_array($scheduleQuery, MYSQL_NUMS)){ $ret[]=$row; } mysql_close($scheduleQuery); return $ret; // and handle the return value in php } } Two things here: I found that returned variable in php is a little bit complex to play with since it is two dimension array. I am not sure what the best practice is and would like to ask you experts opinions. Every time I create a new method, I have to recreate the $connection variable: see below $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } It seems like redundant to me. Can I only do it once instead of calling it anytime I need a query? I am new to php class. hope you guys can help me. thanks.

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  • PHP app.yaml, new to GAE, not sure of how to setup

    - by chetweewax
    I have a single page website built on bootstrap 3, that I am trying to move to Google Apps Engine. I Scaffold my sites using php, and all the content is showing but not the styles and javascript. My site is basically set up as follows _/js/bootstrap.js _/js/custom.js _/fonts/glypicon ...etc _/css/bootstrap.css _/css/custom.css _/php/ .. all my php files go here ... index.php can someone help me setup my app.yaml for this? I am new to GAE, and am a little confused by this.

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  • PHP socket UDP communication

    - by Ghedeon
    Server works fine, but the problem is the client doesn't receive anything. server.php <?php $buf_size = 1024; $socket = stream_socket_server("udp://127.0.0.1:3127", $errno, $errstr, STREAM_SERVER_BIND); do { $str = stream_socket_recvfrom($socket, $buf_size, 0, $peer); $str = "abc"; stream_socket_sendto($socket, $str, strlen($str), 0, $peer); } while (true); ?> client.php <?php $fp = stream_socket_client("udp://127.0.0.1:3127", $errno, $errstr); if (!$fp) { echo "$errno - $errstr<br />\n"; } else { fwrite($fp, "1 2 3"); echo fread($fp, 15); fclose($fp); } ?>

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  • Structure map and generics (in XML config)

    - by James D
    Hi I'm using the latest StructureMap (2.5.4.264), and I need to define some instances in the xml configuration for StructureMap using generics. However I get the following 103 error: Unhandled Exception: StructureMap.Exceptions.StructureMapConfigurationException: StructureMap configuration failures: Error: 103 Source: Requested PluginType MyTest.ITest`1[[MyTest.Test,MyTest]] configured in Xml cannot be found Could not create a Type for 'MyTest.ITest`1[[MyTest.Test,MyTest]]' System.ApplicationException: Could not create a Type for 'MyTest.ITest`1[[MyTest.Test,MyTest]]' ---> System.TypeLoadException: Could not loa d type 'MyTest.ITest`1' from assembly 'StructureMap, Version=2.5.4.264, Culture=neutral, PublicKeyToken=e60ad81abae3c223'. at System.RuntimeTypeHandle._GetTypeByName(String name, Boolean throwOnError, Boolean ignoreCase, Boolean reflectionOnly, StackCrawlMark& stackMark, Boolean loadTypeFromPartialName) at System.RuntimeTypeHandle.GetTypeByName(String name, Boolean throwOnError, Boolean ignoreCase, Boolean reflectionOnly, StackCrawlMark& stackMark) at System.RuntimeType.PrivateGetType(String typeName, Boolean throwOnError, Boolean ignoreCase, Boolean reflectionOnly, StackCrawlMark& s tackMark) at System.Type.GetType(String typeName, Boolean throwOnError) at StructureMap.Graph.TypePath.FindType() --- End of inner exception stack trace --- at StructureMap.Graph.TypePath.FindType() at StructureMap.Configuration.GraphBuilder.ConfigureFamily(TypePath pluginTypePath, Action`1 action) A simply replication of the code is as follows: public interface ITest<T> { } public class Test { } public class Concrete : ITest<Test> { } Which I then wish to define in the XML configuration something as follows: <DefaultInstance PluginType="MyTest.ITest`1[[MyTest.Test,MyTest]],MyTest" PluggedType="MyTest.Concrete,MyTest" Scope="Singleton" /> I've been racking my brain, however I can't see what I'm doing wrong - I've used Type.GetType to verify the type actually is valid which it is. Anyone have any ideas? Thanks !

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  • Never getting a JSON response when running server-side PHP proxy script but I do with others

    - by Dohk
    I'm on PHP 5.3.4 and Apache 2.2 btw So I'm using (or trying to use) Simple PHP Proxy (Simple PHP Proxy) I enter a URL at his example page at SPP Example Page and it works fine, I see the JSON response and all the headers. However, when I copy the exact URL, only changing the URL to now have localhost, I get both empty headers and no JSON. Assuming that the script on his site is the same I downloaded, could this be due to a multitude of things or a setting in Apache and/or the PHP ini? So for example: benalman.com/code/projects/php-simple-proxy/ba-simple-proxy.php?url=http://github.com/&full_headers=1&full_status=1 That will get me a ton of info back Now changing to localhost http://localhost/ba-simple-proxy.php?url=http://github.com/&full_headers=1&full_status=1 {"headers":[],"status":{"url":"https:\/\/github.com\/","content_type":"text\/html","http_code":301,"header_size":194,"request_size":182,"filetime":-1,"ssl_verify_result":0,"redirect_count":1,"total_time":0.094,"namelookup_time":0,"connect_time":0.047,"pretransfer_time":0,"size_upload":0,"size_download":185,"speed_download":1968,"speed_upload":0,"download_content_length":185,"upload_content_length":0,"starttransfer_time":0,"redirect_time":0.047,"certinfo":[]},"contents":null} I even went basic and just used some curl and of course, empty objects being returned other than false for my content and the url I set in my JSON. Any help is deeply appreciated or any ideas.

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  • PHP if statement - select two different get variables?

    - by arsoneffect
    Below is my example script: <li><a <?php if ($_GET['page']=='photos' && $_GET['view']!=="projects"||!=="forsale") { echo ("href=\"#\" class=\"active\""); } else { echo ("href=\"/?page=photos\""); } ?>>Photos</a></li> <li><a <?php if ($_GET['view']=='projects') { echo ("href=\"#\" class=\"active\""); } else { echo ("href=\"/?page=photos&view=projects\""); } ?>>Projects</a></li> <li><a <?php if ($_GET['view']=='forsale') { echo ("href=\"#\" class=\"active\""); } else { echo ("href=\"/?page=photos&view=forsale\""); } ?>>For Sale</a></li> I want the PHP to echo the "href="#" class="active" only when it is not on the two pages: ?page=photos&view=forsale or ?page=photos&view=projects

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