Search Results

Search found 6679 results on 268 pages for 'echo'.

Page 191/268 | < Previous Page | 187 188 189 190 191 192 193 194 195 196 197 198  | Next Page >

  • Better explanation of $this-> in this example please

    - by Doug
    Referring to this question: http://stackoverflow.com/questions/2035449/why-is-oop-hard-for-me class Form { protected $inputs = array(); public function makeInput($type, $name) { echo '<input type="'.$type.'" name="'.$name.'">'; } public function addInput($type, $name) { $this->inputs[] = array("type" => $type, "name" => $name); } public function run() { foreach($this->inputs as $array) { $this->makeInput($array['type'], $array['name']; } } } $form = new form(); $this->addInput("text", "username"); $this->addInput("text", "password");** Can I get a better explanation of what the $this->input[] is doing in this part: public function addInput($type, $name) { $this->inputs[] = array("type" => $type, "name" => $name); }

    Read the article

  • How do I rewrite binary data in Java

    - by hayato
    I'm trying to figure out how to replace binary data using Java. below is a PHP example of replacing "foo" to "bar" from a swf file. <?php $fp = fopen("binary.swf","rb"); $size = filesize("binary.swf"); $search = bin2hex("foo"); $replace = bin2hex("bar"); $data = fread($fp, $size); $data16 = bin2hex($data); $data16 = str_replace($search, $replace, $data16); $data = pack('H*',$data16); header("Content-Type:application/x-shockwave-flash"); echo $data; ?> How do I do this in Java.

    Read the article

  • Can't get Javac to work on Mac OS X

    - by elguapo-85
    I am trying to compile with javac on Snow Leopard through the command line. I have Xcode installed. I am just using a simple Hello World file, it works in Eclipse but I can't get it to work using javac. javac -version returns javac 1.6.0_17 HelloWorld.java public class HelloWorld { public static void main(String[] args) { String message = "Welcome to Java!"; System.out.println(message); } } I type: javac HelloWorld.java and get the following error. HelloWorld.java:1: class, interface, or enum expected public class HelloWorld ^ 1 error and... javac -cp . HelloWorld.java returns the same. echo $CLASSPATH just returns blank. Thanks for the help.

    Read the article

  • How can I set the main theme-font dynamically, in WordPress

    - by windyjonas
    I have created a theme where I already have a custom options page where I let the user set text for footer, twitter user and some other things and that works well. Now i'd like to add the functionality of letting the user select which font that should be used for content on the site. How can i accomplish this? I can probably create a php file that outputs something like: <style type="text/css"> body{ font-family: <?php echo get_option('my-font');?>; } </style> and include that file in header.php, but that means that I have to hit php for every request for this css and I want to avoid that if posssible.

    Read the article

  • PHP Array To Class Static Public Values.

    - by what
    I want to make a class that has all the values of an array as a static object in that class. For example: $vars=array(... );//some array with actual values $Class_code='class MyClass{'; for($i=0; $i<count($vars); $i++){ $Class_code.='static public $'.strval($vars[i]).';'; } eval($Class_code.'}'); /* When I echo the line above it says that the vars for MyClass (MyClass::vars) are missing even with strval(). */ So, how can I get $vars[i] to be a string if strval didn't work?

    Read the article

  • change last key name from array in php

    - by robertdd
    i want to be able to change the last key from array i try with this function i made: function getlastimage($newkey){ $arr = $_SESSION['files']; $oldkey = array_pop(array_keys($arr)); $arr[$newkey] = $arr[$oldkey]; unset($arr[$oldkey]); $results = end($arr); //echo json_encode($results); print_r($arr); } if i call the function getlastimage('newkey') it change the key!but after if i print the $_SESSION the key is not changed? why this?

    Read the article

  • php with two class

    - by lolalola
    Hi Guys, my sample is good or not? I have a good connection to the database, or too should be in the class? Thanks <?php mysql_connect('localhost','root','admin'); mysql_select_db('test'); class UserDisplay { function getDisplayName() { $sql = 'select first_name, last_name, display_name from users where user_id = "3"'; $results = mysql_query($sql); $row = mysql_fetch_array($results); $this->user_id = $user_id; return $this->user_id; } } class UserInsert function InsertName($name) { mysql_query("INSERT INTO Persons (first_name)VALUES ('".$name."')"); } } $userD = new UserDisplay(); echo "User known as: " . $userD->getDisplayName() . "\n"; $userI = new UserInsert(); $userI->InsertName("Peter"); ?>

    Read the article

  • How to build unlimited level of menu through PHP and mysql

    - by Starx
    Well, to build my menu my menu I use a db similar structure like this To assign another submenu for existing submenu I simply assign its parent's id as its value of parent field. parent 0 means top menu now there is not problem while creating submenu inside another submenu now this is way I fetch the submenu for the top menu <ul class="topmenu"> <? $list = $obj -> childmenu($parentid); //this list contains the array of submenu under $parendid foreach($list as $menu) { extract($menu); echo '<li><a href="#">'.$name.'</a></li>'; } ?> </ul> What I want to do is. I want to check if a new menu has other child menu and I want to keep on checking until it searches every child menu that is available and I want to display its child menu inside its particular list item like this Home ........

    Read the article

  • Populate FILE field with default text

    - by dclowd9901
    I'm trying to reutilize code that generates FILE fields for use when something is to be added to the database, and grayed out (and disabled) with data already in the database when the item in question is being edited or viewed in detail. However, I can't seem to get the text to fill the field. I'm using this: echo '<input type="file" name="small[]" value="' . $value_from_database . '" DISABLED><br>'; Am I missing anything? If not, are there any decent workarounds?

    Read the article

  • Dropdownmenu SELECTED value as per value return fromDB

    - by Faizan Qadri
    All i am trying to do is to set the selected value of drop down menu according to the particular value returned from the database like if person saved his gender as 'Male' and he wants to update his profile then the selected option shown on the Gender's dropdown llist should be shown as Male cause if this doesn't happen 'Poor guy becomes a female due to this small problem in my code' KINDLY HELP!!!!!!! MY Current Code: <select name="Gender" id="Gender"> <option selected="selected"><?php echo $row_Recordset1['Gender']; ?></option> <option value="Male">Male</option> <option value="Female">Female</option> </select> The above code work fine but causes repitition of values in dropdown like Male Male Female

    Read the article

  • different explanation

    - by Delirium tremens
    The following code echoes 5, not 10: $global_obj = null; class my_class { var $value; function my_class() { global $global_obj; $global_obj = &$this; } } $a = new my_class; $a->my_value = 5; $global_obj->my_value = 10; echo $a->my_value; "Upon first examination, it would seem that the constructor of my_class stores a reference to itself inside the $global_obj variable. Therefore, one would expect that, when we later change the value of $global_obj-my_value to 10, the corresponding value in $a would change as well. Unfortunately, the new operator does not return a reference, but a copy of the newly created object." Please, give me a different explanation.

    Read the article

  • Using javascript to access a json array from php

    - by celenius
    I'm trying to understand how my php script can pass an array to my javascript code. Using the following php, I pass an array: $c = array(3,2,7); echo json_encode($c); My javascript is as follows: $.post("getLatLong.php", { latitude: 500000}, function(data){ arrayData = data document.write(arrayData) document.write(arrayData[0]); document.write(arrayData[0]); document.write(arrayData[0]); }); </script> What is printed out on screen is [3,2,7][3, I'm trying to understand how json_encode works - I though I would be able to pass the array to a variable, and then access it like a normal javascript array, but it views my array as one large text string. How do ensure that it reads it like an array?

    Read the article

  • PHP & MySQL form question.

    - by peakUC
    How do I allow all users to have there username field empty without having them to enter a username when they submit the form using PHP and MySQL? Here is part my PHP and MySQL code. $username = mysqli_real_escape_string($mysqli, $purifier->purify(htmlentities(strip_tags($_POST['username'])))); if(isset($_POST['username'])) { // Make sure the username address is available: $u = "SELECT * FROM users WHERE username = '$username' AND user_id <> '$user_id'"; $r = mysqli_query ($mysqli, $u) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($mysqli)); if (mysqli_num_rows($r) == TRUE) { // Unavailable. echo '<p class="error">Your username is unavailable!</p>'; $username = NULL; } else if(mysqli_num_rows($r) == 0) { // Available. $username = mysqli_real_escape_string($mysqli, $purifier->purify(htmlentities(strip_tags($_POST['username'])))); } }

    Read the article

  • passing parameters to javacsript using php

    - by ayush
    i have the following line of code - <a href="javascript:;" onClick="tweeet('myid')">My Tweets!</a> Now while this is working perfectly fine the following line is not - <a href="javascript:;" onClick="tweeet(<?php echo 'myid'; ?>)">My Tweets!</a> Can anyone help me out why it is not working and suggest any changes. The variable i want to pass to the javascript function is a php variable. also i have tried the php with single quotes and double quotes but it is not working.

    Read the article

  • Display data requested by an ajax.load() call once complete, not during the call.

    - by niczoom
    My jQuery code (using ajax) request's data from a php script (pgiproxy.php) using the following function: function grabPage($pageURL) { $homepage = file_get_contents($pageURL); echo $homepage; } I then extract the html code i need from the returned data using jQuery and insert it into a div called #BFX, as follows: $("#btnNewLoadMethod1").click(function(){ $('#temp1').load('pgiproxy.php', { data : $("#formdata").serialize(), mode : "graph"} , function() { $('#temp').html( $('#temp1').find('center').html() ); $('#BFX').html( $('#temp').html() ); }); }); This works fine. I get the html data (which is a gif image) i need displayed on screen in the correct div. The problem is i can see the html data loading into the div (dependant on network speed), but what I want is to insert the extracted html code into #BFX ONLY when the ajax request has fully completed.

    Read the article

  • Re-execute target when specified as dependency to multiple rules

    - by andrew
    I have the following GNU makefile: .PHONY a b c d a: b c b: d c: d d: echo HI I would like the target 'd' to be run twice -- since it is specified as a dependency by both b & c. Unfortunately, the target 'd' will be executed only once. The output of running make will simply be 'HI', instead of 'HI HI'. How can I fix this? Thanks! To Clarify, the goal is something like this: subdirs = a b c build: x y x: target=build x: $(subdirs) y: target=prepare y: $(subdirs) $(subdirs): $(make) -f $@/makefile $(target)

    Read the article

  • Get only new RSS entries with PHP Script ?

    - by ArneRie
    What im trying to do: Fetch X numbers of RSS Feeds from my Blogs and echo only new entries. My Problem is, how to know wich items are already parsed? Solution so far: Fetch the Feed every 5 hours, store all titles inside an Database table or flat file. Next run check if the title is already in database if not print it and save it inside the database. But iam not sure if this is best practise to do this? If someone knows a fast way, it would be great. Sorry for my poor english.

    Read the article

  • Decoding and caching json every 60 minutes

    - by Gary
    Hi, How do I do this on a php webpage? I want to get and decode a json string and display the results as html on my page, however, I don't want it hotlinking back to the source. If I could write the decoded string to a txt file say weather.txt on the server and keep the html formatting and do it so that the page won't fetch the json script until 60 minutes has passed since the last time it was fetched regardless of how many times the page is opened during that 60 minute period and the weather.txt is viewed. All I can come up with is a simple script that hotlinks, everything else I have tried simply failed. $file = file_get_contents('http://sample.com/weather'); $out = (json_decode($file)); echo $out-mainText; Will appreciate any help with this.

    Read the article

  • Kohana controller shows blank page if missing URI segment/argument

    - by Petruza
    When accessing http://www.nowplayingnashville.com/event2/detail/440698386/Whatever the controller shows normally. But when omitting the last parameter (event's title, which can be any string), http://www.nowplayingnashville.com/event2/detail/440698386/ It shows blank. I added a default value and tried to hardcode the title on the first line of the controller, but it doesn't work anyway. It looks like a view issue, because I can add an echo at the end of the controller and it gets printed, so the controller is executing to the end, but the views aren't displayed. This problem is present on the live site, but I checked out the code on my local machine and it works perfectly without a title. That's strange. Could it be a Kohana or Apache configuration issue?

    Read the article

  • Echoing a pseudo column value after a COUNT

    - by rob - not a robber
    Hi Gang... Please don't beat me if this is elementary. I searched and found disjointed stuff relating to pseudo columns. Nothing spot on about what I need. Anyway... I have a table with some rows. Each record has a unique ID, an ID that relates to another entity and finally a comment that relates to that last entity. So, I want to COUNT these rows to basically find what entity has the most comments. Instead of me explaining the query, I'll print it SELECT entity_id, COUNT(*) AS amount FROM comments GROUP BY entity_id ORDER BY amount DESC The query does just what I want, but I want to echo the values from that pseudo column, 'amount' Can it be done, or should I use another method like mysql_num_rows? Thank you!!!

    Read the article

  • How do I move Zend Framework From Development to Production?

    - by dirtylogic
    I'm just wondering if anyone else has had problems moving the Zend Framework from development to production. I changed my docroot to the public folder, updated my library path, but it's still not working out for me. The IndexController is working just fine, but my ServiceController is giving me an internal server error. ServiceController <?php class ServiceController extends Zend_Controller_Action { public function amfAction() { require_once APPLICATION_PATH . '/models/MyClass.php'; $srv = new Zend_Amf_Server(); $srv->setClass('Model_MyClass', 'MyClass'); echo $srv->handle(); exit; } }

    Read the article

  • textbox disappears in paging - php

    - by fusion
    i'm calling the search.php page via ajax to search.html. the problem is, since i've implemented paging, the textbox with the search keyword from search.html 'disappears' when the user clicks the 'Next' button [because the page goes to search.php which has no textbox element] i'd like the textbox with the search keyword to be there, when the user goes through the records via paging. how'd i achieve this? search.html: <body> <form name="myform" class="wrapper"> <input type="text" name="q" onkeyup="showPage();" class="txt_search"/> <input type="button" name="button" onclick="showPage();" class="button"/> <p> <div id="txtHint"></div> <div id="page"></div> </form> </body> search.php [the relevant part]: $self = $_SERVER['PHP_SELF']; $limit = 5; //Number of results per page $numpages=ceil($totalrows/$limit); $query = $query." ORDER BY idQuotes LIMIT " . ($page-1)*$limit . ",$limit"; $result = mysql_query($query, $conn) or die('Error:' .mysql_error()); ?> <div class="caption">Search Results</div> <div class="center_div"> <table> <?php while ($row= mysql_fetch_array($result, MYSQL_ASSOC)) { $cQuote = highlightWords(htmlspecialchars($row['cQuotes']), $search_result); ?> <tr> <td style="text-align:right; font-size:15px;"><?php h($row['cArabic']); ?></td> <td style="font-size:16px;"><?php echo $cQuote; ?></td> <td style="font-size:12px;"><?php h($row['vAuthor']); ?></td> <td style="font-size:12px; font-style:italic; text-align:right;"><?php h($row['vReference']); ?></td> </tr> <?php } ?> </table> </div> <?php //Create and print the Navigation bar $nav=""; if($page > 1) { $nav .= "<a href=\"$self?page=" . ($page-1) . "&q=" .urlencode($search_result) . "\">< Prev</a>"; $first = "<a href=\"$self?page=1&q=" .urlencode($search_result) . "\"><< First</a>" ; } else { $nav .= "&nbsp;"; $first = "&nbsp;"; } for($i = 1 ; $i <= $numpages ; $i++) { if($i == $page) { $nav .= "<B>$i</B>"; }else{ $nav .= "<a href=\"$self?page=" . $i . "&q=" .urlencode($search_result) . "\">$i</a>"; } } if($page < $numpages) { $nav .= "<a href=\"$self?page=" . ($page+1) . "&q=" .urlencode($search_result) . "\">Next ></a>"; $last = "<a href=\"$self?page=$numpages&q=" .urlencode($search_result) . "\">Last >></a> "; } else { $nav .= "&nbsp;"; $last = "&nbsp;"; } echo "<br /><br />" . $first . $nav . $last; }

    Read the article

  • How to redirect to the current page with form in php?

    - by garcon1986
    Hello, I tried to redirect to the current page with form in my php application. Now i have met a problem. <form name="myform" action="?page=matching" method="GET"> <input id="match_button" type="submit" name="button" value="button" onClick="func_load3()" /> </form> action="?page=matching" means the current page, because i use the single entry in my php application. With the code upon, When i click the button, it redirects to the homepage. And I tried to use: <form name="myform" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="GET"> but it still doesn't work. So i have to ask for help from you. Do you have any ideas about that? How to fix it? Thanks in advance.

    Read the article

  • Bash Scripting: I want to open a set of .php files, and add line before html tag

    - by Bashn00b
    Hi guys, I have a set of .php files in a folder, I want to add text just before these lines: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en" > What i want is to insert just before these lines in the html file. So just want to prepend that file before each docType declaration. However the DOCTYPE declaration is never on line 1, as there is loads of php lines before. I have this current script (where FE is the folder containing all the scripts i want to edit): for file in ${fe}*; do echo "$file" done Thanks,

    Read the article

  • update a column in input file by taking value from Database in perl.

    - by Rahul Singh
    input file: 1,a,USA,, 2,b,UK,, 3,c,USA,, i want to update the 4th column in the input file from taking values from one of the table. my code looks like this: my $customer_dbh = DBI-connect("DBI:Oracle:$INST", $USER, $PASS ) or die "Couldn't connect to datbase $INST"; my $cust_smh; print "connected \n "; open FILE , "+$input_file" or die "can't open the input file"; print "echo \n"; while(my $line=) { my @line_a=split(/\,/,$line); my $customer_id=$line_a[3]; print "$customer_id\n"; $cust_smh = $customer_dbh-prepare("SELECT phone_no from book where number = $line_a[0]"); $cust_smh-execute() or die "Couldn't execute stmt, error : $DBI::errstr"; my $number = $cust_smh-fetchrow_array(); $line_a[3]=$number; }

    Read the article

< Previous Page | 187 188 189 190 191 192 193 194 195 196 197 198  | Next Page >