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  • sending javascript array to external php file using POST

    - by user1472224
    I am trying to send a javascript array to an external php page, but the only thing the php page is picking up is the fact that I am sending array, not the actual data within the array. javascript - var newArray = [1, 2, 3, 4, 5]; $.ajax({ type: 'POST', url: 'array.php', data: {'something': newArray}, success: function(){ alert("sent"); } }); External PHP Page - <?php echo($_POST['something'])); ?> I know this question has been asked before, but for some reason, this isn't working for me. I have spent the last couple days trying to figure this out as well. Can someone please point me in the right direction. current output (from php page) - Array (thats all the page outputs)

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  • How to perform an external request in Kohana 3?

    - by alex
    I've always used cURL for this sort of stuff, but this article got me thinking I could request another page easily using the Request object in Kohana 3. $url = 'http://www.example.com'; $update = Request::factory($url); $update->method = 'POST'; $update->post = array( 'key' => 'value' ); $update->execute(); echo $update->response; However I get the error Accessing static property Request::$method as non static From this I can assume it means that the method method is static, but that doesn't help me much. I also copied and pasted the example from that article and it threw the same error. Basically, I'm trying to POST to a new page on an external server, and do it the Kohana way. So, am I doing this correctly, or should I just use cURL (or file_get_contents() with context)?

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  • FastCGI C++ program and missing SCRIPT_NAME

    - by Simone Margaritelli
    Hi guys, i'm studying the fastcgi developement kit because i'm writing a new scripting language and i'd like to write a fastcgi version of the interpreter to run scripts as webpages. I'm using this example from the sdk, located at /srv/http/bin/echocpp So, in my httpd.conf file, i have the following lines : FastCgiServer /srv/http/bin/echocpp -idle-timeout 120 -processes 4 ScriptAlias / /srv/http/bin/echocpp/ ... ... AddHandler fastcgi-script hy Where 'hy' is the extensions of my scripts. Then, when i try browse, for instance http://localhost/~evilsocket/prime.hy I see all the environment variables as expected from the echo-cpp.cpp programm, except for the SCRIPT_NAME that is empty. Is there something i'm missing out of this? How am i supposed to obtain the full path of the script to run it inside my fastcgi version of the interpreter? Thanks

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  • How to maipulate the shell output in php

    - by Mirage
    I am trying to write php script which does some shell functions like reporting. So i am starting with diskusage report I want in following format drive path ------------total-size --------free-space Nothing else My script is $output = shell_exec('df -h -T'); echo "<pre>$output</pre>"; and its ouput is like below Filesystem Type Size Used Avail Use% Mounted on /dev/sda6 ext3 92G 6.6G 81G 8% / none devtmpfs 3.9G 216K 3.9G 1% /dev none tmpfs 4.0G 176K 4.0G 1% /dev/shm none tmpfs 4.0G 1.1M 4.0G 1% /var/run none tmpfs 4.0G 0 4.0G 0% /var/lock none tmpfs 4.0G 0 4.0G 0% /lib/init/rw /dev/sdb1 ext3 459G 232G 204G 54% /media/Server /dev/sdb2 fuseblk 466G 254G 212G 55% /media/BACKUPS /dev/sda5 fuseblk 738G 243G 495G 33% /media/virtual_machines How can i convert that ouput into my forn\matted output

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  • Escaping Problem in bash using isql

    - by latz
    Hi there, I am currently working on a little backup script from some firebird databases and I've come up with a weird escaping problem that I don't seem to be able to solve. Here's the thing in my script I create a variable called sqllog in which I would like to put the output of a chain of commands, here it is. sqllog=echo "SELECT * FROM RDB\$DATABASE;" | isql -u SYSDBA -pass mypasswd localhost:mydatabase | tail -n 2 | head -n 1 | wc -l if I try to execute this in shell I get the following error Statement failed, SQLCODE = -204 Dynamic SQL Error -SQL error code = -204 -Table unknown -RDB -At line 1, column 15. Table unknown RDB means it didn't take my try to escape the $. thx for any help :)

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  • How much time roughly it takes to learn ASP.net programming

    - by Mirage
    I am a PHP MYSQL person but now i have got webiste to do it ASP.net. I kow nothing about ASP.NET. INitially i was thinking of ASP.NET as a programming language but now i now its not , i really don't know what it is. Someone told me that i have to do c sharp to build the website. I i have to display simple data from database how much different is c sharp from php. Does it has all function like php to echo stuff etc

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  • MySQL ON DUPLICATE KEY UPDATE issue

    - by user644347
    Hi could some one look at this and tell me where I am going wrong. I have an SQL statement that when I echo using php I get this to screen INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '18' , 'GenreName' = 'Drama' ON DUPLICATE KEY UPDATE 'GenreName' = 'Drama' WHERE 'GenreID' = '18' INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '16' , 'GenreName' = 'Animation' ON DUPLICATE KEY UPDATE 'GenreName' = 'Animation' WHERE 'GenreID' = '16' And here is the statement $sql="INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '{$genresID[$i]}' , 'GenreName' = '{$genreName[$i]}' ON DUPLICATE KEY UPDATE 'GenreName' = '{$genreName[$i]}' WHERE 'GenreID' = '{$genresID[$i]}'"; This is the error I recieve: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''moviedb'.'genre' SET 'GenreID' = '18' , 'GenreName' = 'Drama' ON DUPLICATE KEY ' at line 1 Any help would be greatly appreciated, thanks in advance.

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  • How to assign a PHP variable that contains a mixed of PHP and html elements?

    - by Rick
    I have a PHP page that contains html and tags and PHP codes and I am using AJAX to manipulate some data and now I need to spit out the response but I am not sure if it is possible to put the mixed content into a string variable to pass through... Is this possible? For example I need to: $response = "<p>some text--<?php echo $something; ?>"; Thanks... To clear things up, I am not trying to send raw PHP code to clientside as that would be silly and won't be evaluated as it is a server side script. I am just looking for a fast way to put everything into a variable without manually re-doing the whole page stripping out the php open and close tags and echos...etc

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  • Php code not executing - dies out when trying to refer to member of static class - no error displaye

    - by Ali
    I'm having some problems with this piece of code. I've included a class declaration and trying to create an object of that class but my code dies out. It doesn't seem to be an include issue as all the files are being included even the files called for inclusion within the class file itself. However the object is not created - I tried to put an echo statement in the __construct function but nothing it just doesn't run infact doesn't create the object and the code won't continue from there - plus no error is reported or displayed and I have error reporting set to E_ALL and display errors set to true WHats happening here :( =============EDIT SOrry I checked again the error is prior to teh object creation thing - it dies out when it tries to refer to a constant in a static class like so: $v = Zend_Oauth::REQUEST_SCHEME_HEADER; THis is the class or part of it - it has largely static functions its the Zend Oauth class: class Zend_Oauth { const REQUEST_SCHEME_HEADER = 'header'; const REQUEST_SCHEME_POSTBODY = 'postbody'; const REQUEST_SCHEME_QUERYSTRING = 'querystring'; // continued LIke I said no error is being reported at all :(

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  • Changing a php "echoed" div attribute with php

    - by Zakaria
    Hi everybody, I'm using PHP to echo a content stored on my database. The content is a DIV carrying any type of data. The problem is that I don't know the ID and I have some problems with these DIVs if I try to display them more that once. So, the idea is to modify the DIV id each time I'd like to display them. Something like this: <?php modify_div_id($data,"id-456"); ?> Is there a solution for this problem? Thank you very much, regards.

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  • Why does my text has the justify effect when I didnt made it to have this effect (css/php)

    - by linkcool
    Why my text has the justify effect? In my whole site, I make echos and i dont specify a "text-align:justify;" but my text is still justifying. Justify is when you make the browser window smaller, the text moves so it fits in the window. I tryed making something like this: <?php echo "<h1>some stuff.</h1>"; ?> <html> <head> <style> h1 { text-align:center; } etc.... but it just makes the text go in the center and it keeps the justify effect. please help me =[ thanks

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  • Modify bash variables with sed

    - by Alexander Cska
    I am trying to modify a number of environmental variables containing predefined compiler flags. To do so, I tried using a bash loop that goes over all environmental variables listed with "env". for i in $(env | grep ipo | awk 'BEGIN {FS="="} ; { print $1 } ' ) do echo $(sed -e "s/-ipo/ / ; s/-axAVX/ /" <<< $i) done This is not working since the loop variable $i contains just the name of the environmental variable stored as a character string. I tried searching a method to convert a string into a variable but things started becoming unnecessary complicated. The basic problem is how to properly supply the environmental variable itself to sed. Any ideas how to properly modify my script are welcome. Thanks, Alex

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  • Set the output of a function equal to a variable

    - by ryno
    How would i set the output of a custom php function to a variable? The function is: function getRandomColor1() { global $cols; $num_cols = count($cols); $rand = array_rand($cols); $rand_col = $cols[$rand]; echo $rand_col; unset($cols[$rand]); } How would i set getRandomColor1 equal to $RandomColor1? I need it to be a variable so I can use it in css like: #boxone1 { height: 150px; width: 150px; background: <?=$RandomColor1?>; float: left; } If its not possible to set it as a variable, how else could i put the output of the function into css?

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  • Why does git hash-object return a different hash than openssl sha1?

    - by user657606
    Context: I downloaded a file (Audirvana 0.7.1.zip) from code.google to my Macbook Pro (Mac OS X 10.6.6). (current url: http://code.google.com/p/audirvana/downloads/detail?name=Audirvana%200.7.1.zip&can=2&q= ) I wanted to verify the checksum, which for that particular file is posted as 862456662a11e2f386ff0b24fdabcb4f6c1c446a (SHA-1). git hash-object gave me a different hash, but openssl sha1 returned the expected 862456662a11e2f386ff0b24fdabcb4f6c1c446a. The following experiment seems to rule out any possible download corruption or newline differences and to indicate that there are actually two different algorithms at play: $ echo A > foo.txt $ cat foo.txt A $ git hash-object foo.txt f70f10e4db19068f79bc43844b49f3eece45c4e8 $ openssl sha1 foo.txt SHA1(foo.txt)= 7d157d7c000ae27db146575c08ce30df893d3a64 What's going on?

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  • Getting count() of class static array

    - by xylar
    Is it possible to get the count of a class defined static array? For example: class Model_Example { const VALUE_1 = 1; const VALUE_2 = 2; const VALUE_3 = 3; public static $value_array = array( self::VALUE_1 => 'boing', self::VALUE_2 => 'boingboing', self::VALUE_3 => 'boingboingboing', ); public function countit() { // count number $total = count(self::$value_array ); echo ': '; die($total); } } At the moment calling the countit() method returns :

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  • Codeigniter Template library, add_js() method

    - by Karl
    using This Template Library when i try and use the add_js() function it errors out. I add the regions $_scripts to the template header file but when i load the page it says undefined variable _scripts. any thoughts? changed <?= $_scripts ?> to <?= (isset($_scripts)) ? $_scripts : “”; ?> and obviously i lose the error but the js file still isnt loading. I did an echo at each step though the add_js() method and the output is correct. I also did an output of $this-js and they both had the corret output. So it gets through the get_js method fine and sets the class variable fine but i dont get anything added to the actual page.

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  • How to write different implicit rules for different file names for GNU Make

    - by anupamsr
    Hi! I have a directory in which I keep adding different C++ source files, and generic Makefile to compile them. This is the content of the Makefile: .PHONY: all clean CXXFLAGS = -pipe -Wall -Wextra -Weffc++ -pedantic -ggdb SRCS = $(wildcard *.cxx) OBJS = $(patsubst %.cxx,%.out,$(SRCS)) all: $(OBJS) clean: rm -fv $(OBJS) %.out: %.cxx $(CXX) $(CXXFLAGS) $^ -o $@ NOTE: As is obvious from above, I am using *.out for executable file extensions (and not for object file). Also, there are some files which are compiled together: g++ file_main.cxx file.cxx -o file_main.out To compile such files, until now I have been adding explicit rules in the Makefile: file_main.out: file_main.cxx file.cxx file.out: file_main.out @echo "Skipping $@" But now my Makefile has a lot of explicit rules, and I would like to replace them with a simpler implicit rule. Any idea how to do it?

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  • parsing a xml to get some values

    - by Joan Silverstone
    Hello, i have this xml: http://www.managerleague.com/export_data.pl?data=transfers&output=xml&hide_header=0 These are player sales from a browser game. I want to save some fields from these sales. I am fetching that xml with curl and storing on my server. Then do the following: $xml_str = file_get_contents('salespage.xml'); $xml = new SimpleXMLElement($xml_str); $items = $xml->xpath('*/transfer'); print_r($items); foreach($items as $item) { echo $item['buyerTeamname'], ': ', $item['sellerTeamname'], "\n"; } The array is empty and i cant seem to get anything from it. What am i doing wrong?

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  • Accessing a variable passed by jquery using php

    - by celenius
    How does php access a variable passed by JQuery? I have the following code: $.post("getLatLong.php", { latitude: 500000}, function(data){ alert("Data Loaded: " + data); }); but I don't understand how to access this value using php. I've tried $userLat=$_GET["latitude"]; and $userLat=$_GET[latitude]; and then I try to print out the value (to check that it worked) using: echo $userLat; but it does not return anything. I can't figure out how to access what is being passed to it.

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  • HTML input text box vs CakePHP Automagic Form Elements

    - by kwokwai
    Hi all, I was manually creating a simple form with one input text box field like this: <form action="/user/add" method="post"> <input type="text" name="data[user_id]" value="1"> But when I call $this->model->save($this->data) in the Controller, nothing was saved to the Table. Only when I used this and the data in the field was written to the database successfully: $form->create(null, array('url' => '/user/add')); echo $form->input('user_id', array('label' => 'User ID', 'value' => '1'));

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  • Problem with foreach loop and $_GET

    - by phpExe
    I have a very simple foreach loop foreach($tv as $id => $channel) { $ID = $_GET['ID']; if($ID == $id){$class = "currentt";} echo '<a href="http://www.mysite.com/tst.php?ID='.$id.'" class="'.$class.'">'.$channel.'</a><br>'; } With url query, with every click the current class repeated. How can avoid this? Thanks alot.

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  • How to order by results from 2 seperate tables in PHP and MySQL.

    - by Vafello
    I am trying to output results of 2 sql queries to one JSON file. The problem is that I would like to order them ascending by distance which is the result of equation that takes homelat and homelon from the users table and lat, lng from locations table.(basically it takes lattitude and longitude of one point and another and computes the distance between these points). Is it possible to take some parameters from both select queries, compute it and output the result in ascending order? $wynik = mysql_query("SELECT homelat, homelon FROM users WHERE guid='2'") or die(mysql_error()); ; $query = "SELECT * FROM locations WHERE timestamp"; $result = map_query($query); $points = array(); while ($aaa = mysql_fetch_assoc($wynik)) { while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { array_push($points, array('name'=>$row['name'], 'lat'=>$row['lat'], 'lng'=>$row['lng'], 'description'=>$row['description'], 'eventType'=>$row['eventType'], 'date'=>$row['date'], 'isotime'=>date('c', ($row['timestamp'])), 'homelat'=>$aaa['homelat'], 'homelon'=>$aaa['homelon'])); } echo json_encode(array("Locations"=>$points));

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  • PHP Path issue running backticks/exec()

    - by Lee
    Hey all I'm trying to run a java jar file from the command line and within the the execution it gives a path. Withing this path their are spaces and this is causing the issue. ie $f = `java -jar /OCR/ocr.jar /Folder/Sub Folder/filetoocr.pdf /ocr/output.txt`; echo "<pre>$output</pre>"; If you can see the space in between the Sub Folder name causes the issue. By command line it would be (which works) java -jar /OCR/ocr.jar /Folder/Sub\ Folder/filetoocr.pdf /ocr/output.txt any suggestions how I can resolve this ?? Hope you can advise

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  • function file_exists not working in php

    - by Rakesh R Nair
    I have been trying to find if a file_exist in the directory. If not i want to use a different image. But as i am using the file_exists function it always returns false. The code i used is while($r=mysql_fetch_row($res)) { if(!file_exists('http://localhost/dropbox/lib/admin/'.$r[5])) { $file='http://localhost/dropbox/lib/admin/images/noimage.gif'; } else $file='http://localhost/dropbox/lib/admin/'.$r[5];} But the function always return false even if the file exits. I checked that by using <img src="<?php echo 'http://localhost/dropbox/lib/admin/'.$r[5]; ?>" /> This displayed the image correctly. Please Someone Help Me

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  • mysqli prepare statment error?

    - by user310850
    Hi all, $mysqli = new mysqli("localhost", "root", "", "test"); $mysqli->query('PREPARE mid FROM "SELECT name FROM test_user WHERE id = ?"'); //$mysqli->query('PREPARE mid FROM "SELECT name FROM test_user" '); $res = $mysqli->query( 'EXECUTE mid 1;') or die(mysqli_error($mysqli)); while($resu = $res->fetch_object()) { echo '<br>' .$resu->name; } Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 my php version is PHP Version 5.3.0 and mysql mysqlnd 5.0.5-dev - 081106 - $Revision: 1.3.2.27 $

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