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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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  • How can I run a batch file silently?

    - by Mike Pateras
    I have a batch file with some commands that I need to run with my installer, but I'd rather a console not appear (in Windows). I'm executing the batch file from a WiX installer, via a custom action. I tried adding an @ECHO OFF to the top of the file, but that didn't seem to do anything. Is there a way that I can run this batch file silently?

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  • find: missing argument to -exec in bash script

    - by Mephi_stofel
    The following works fine when I type it exactly in the command line: find /<some_path>/{epson,epson_laser,epson_inkjet} -iname "*.ppd" -exec grep "\*ModelName\:" {} \; | sed 's/.*\"\(.*\)\"/\1/' However, when I try to call the following from a bash script I get find: missing argument to -exec'. I have also tried the following (in many variants): eval find "$1" -iname "*.ppd" -exec 'bash -c grep "\*ModelName\:" "$1" | sed "s/.*\"\(.*\)\"/\1/" \; as was mentioned in find-exec-echo-missing-argument-to-exec.

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  • Creating a "virtual" path (/bla) in Mac OS X and Linux that calls custom code

    - by Michael Stum
    Just something I'd like to play with, I would like to create a "virtual" file/directory in the File System of Linux or Mac OS X (Not sure if I can share the same code - does POSIX help?), for example /foo and then perform custom code when something is read or written to it. Similar how /dev/null allows for stuff like echo "Hello!" > /dev/null I don't care if it's in /dev, /proc or anywhere else, as said it's mainly something to play with...

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  • I am having trouble using jquery to submit a form. It was working before

    - by noah
    When a user clicks a link it uses jquery ajax to submit a form to go to paypal. Not working for some reason. Really appreciate any help... LINK TO CLICK I put this in an href for onClick: javascript:go_paypal(); CODE TO EXECUTE ON CLICK function go_paypal() { data = 'req_paypal=1'; $.blockUI({ message: '<h1> Going to Paypal...</h1>',css:{background:'#000'} }); $.ajax({ type: "POST", url: "index.php", data: data, success: function(data) { $("#paypal_form").html(data); $("#payPalForm").submit(); } , error: function() {$.unblockUI(); alert('Unable to communicate to server.'); } }); return false; } CODE TO GO ON SUBMIT if(isset($_POST['req_paypal']) && $_POST['req_paypal'] == 1 ) { $sql = 'INSERT INTO `transactions` (id,type,ip,time,ammount,status) VALUES (NULL,1,\''.$_SERVER['REMOTE_ADDR'].'\',\''.time().'\',\''.$global['paypal_prod_amount'].'\',0) '; echo $sql; mysql_query($sql); $id = mysql_insert_id(); $html = ' <form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post" id="payPalForm"> <input type="hidden" name="item_number" value="One Year of Imgur Pro"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="no_note" value="1"> <input type="hidden" name="business" value="'.$global['paypal_email'].'"> <input type="hidden" name="custom" value="'.base64_encode($id).'"> <input type="hidden" name="currency_code" value="USD"> <input type="hidden" name="return" value="'.$global['paypal_return'].'"> <input name="item_name" type="hidden" id="item_name" value="One Year of Imgur Pro" > <input name="amount" type="hidden" id="amount" value="'.$global['paypal_prod_amount'].'" > </form> '; echo $html;exit; }

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  • How to optimize foreach loop in PHP

    - by vanneto
    First off, I know the title is generic and not fitting. I just couldn't think of a title that could describe my problem. I have a table Recipients in MySQL structured like this: id | email | status 1 foo@bar S 2 bar@baz S 3 abc@def R 4 sta@cko B I need to convert the data into the following XML, depending on the status field. For example: <Recipients> <RecipientsSent> <!-- Have the 'S' status --> <recipient>foo@bar</recipient> <recipient>bar@baz</recipient> </RecipientsSent> <RecipientsRegexError> <recipient>abc@def</recipient> </RecipientsRegexError> <RecipientsBlocked> <recipient>sta@cko</recipient> </RecipientsBlocked> </Recipients> I have this PHP code to implement this ($recipients contains an associative array of the db table): <Recipients> <RecipientsSent> <?php foreach ($recipients as $recipient): if ($recipient['status'] == 'S'): echo "<recipient>" . $recipient['email'] . "</recipient>"; endif; endforeach; ?> </RecipientsSent> <RecipientsRegexError> <?php foreach ($recipients as $recipient): if ($recipient['status'] == 'R'): echo "<recipient>" . $recipient['email'] . "</recipient>"; endif; endforeach; ?> </RecipientsRegexError> <?php /** same loop for the B status */ ?> </Recipients> So, this means that if I have 1000 entries in the table and 4 different status' that can be checked, it means that there will be 4 loops, each one executing 1000 times. How can this be done in a more efficient manner? I thought about fetching four different sets from the database, meaning 4 different queries, would that be more efficient? I'm thinking it could be done with one loop but but I can't come up with a solution. Any way this could be done with only one loop?

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  • git repository sync between computers, when moving around?

    - by Johan
    Hi Let's say that I have a desktop pc and a laptop, and sometimes I work on the desktop and sometimes I work on the laptop. What is the easiest way to move a git repository back and forth? I want the git repositories to be identical, so that I can continue where I left of at the other computer. I would like to make sure that I have the same branches and tags on both of the computers. Thanks Johan Note: I know how to do this with SubVersion, but I'm curious on how this would work with git. If it is easier, I can use a third pc as classical server that the two pc:s can sync against. Note: Both computers are running Linux. Update: So let's try XANI:s idea with a bare git repo on a server, and the push command syntax from KingCrunch. In this example there is two clients and one server. So let's create the server part first. ssh user@server mkdir -p ~/git_test/workspace cd ~/git_test/workspace git --bare init So then from one of the other computers I try to get a copy of the repo with clone: git clone user@server:~/git_test/workspace/ Initialized empty Git repository in /home/user/git_test/repo1/workspace/.git/ warning: You appear to have cloned an empty repository. Then go into that repo and add a file: cd workspace/ echo "test1" > testfile1.txt git add testfile1.txt git commit testfile1.txt -m "Added file testfile1.txt" git push origin master Now the server is updated with testfile1.txt. Anyway, let's see if we can get this file from the other computer. mkdir -p ~/git_test/repo2 cd ~/git_test/repo2 git clone user@server:~/git_test/workspace/ cd workspace/ git pull And now we can see the testfile. At this point we can edit it with some more content and update the server again. echo "test2" >> testfile1.txt git add testfile1.txt git commit -m "Test2" git push origin master Then we go back to the first client and do a git pull to see the updated file. And now I can move back and forth between the two computers, and add a third if I like to.

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  • Why Netbeans stop completing my php code?

    - by T1000
    It seems that Netbeans don't know about functions like mysql_query and complete only globals like $_SESSION or $_POST and stuff like echo and print (don't know about print_r)... Screenshot: http://img163.imageshack.us/img163/4290/clipboard03vr.png

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  • mysql_fetch_array() expects parameter 1 to be resource problem

    - by user225269
    I don't get it, I see no mistakes in this code but there is this error, please help: mysql_fetch_array() expects parameter 1 to be resource problem <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']); ?> <?php while ($row = mysql_fetch_array($result)) { ?> <table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3"> <tr> <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td> </tr> <tr> <td width="30" height="35"><font size="2">*Year:</td> <td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td> <?php } ?> I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

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  • php function output ina table row... how?

    - by user352868
    I want to run this function but show the result in a table row strlen($cat_row['sms'] This is what i came up with but it is not working: echo "<tr><td bgcolor=#626E7A>".**strlen($cat_row['cars']**."</td></tr>"; How do you show result of a php function in a table row? i am trying to format the output some how. please help

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  • Get previous and next row from current id

    - by Hukr
    How can I do to get the next row in a table? `image_id` int(11) NOT NULL auto_increment `image_title` varchar(255) NOT NULL `image_text` mediumtext NOT NULL `image_date` datetime NOT NULL `image_filename` varchar(255) NOT NULL If the current image is 3 for example and the next one is 7 etc. this won’t work: $query = mysql_query("SELECT * FROM images WHERE image_id = ".intval($_GET['id'])); echo $_GET['id']+1; How should I do? thanks

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  • Zend Framework: View variable in layout script is always null

    - by understack
    I set a view variable in someAction function like this: $this->view->type = "some type"; When I access this variable inside layout script like this: <?php echo $this->type ?> it prints nothing. What's wrong? My application.ini settings related to layout resources.layout.layoutPath = APPLICATION_PATH "/layouts/scripts/" resources.layout.layout = "layout" ; changed 'default' to 'layout'

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  • What is wrong with this PHP function.

    - by [email protected]
    I am new to PHP and regular expression. I was going thorugh some online examples and came with this example: <?php echo preg_replace_callback('~-([a-z])~', function ($match) { return strtoupper($match[1]); }, 'hello-world'); // outputs helloWorld ?> in php.net but to my surprise it does not work and keep getting error: PHP Parse error: parse error, unexpected T_FUNCTION Why get error ?

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  • What is the difference?

    - by Eragonio
    I try out CakePHP. I follow this Tutorial and can't find out why this code doesn't work. The code from the tutorial works fine. echo $html -> link('Löschen', array('action' => 'delete', 'id' => $post['Post']['id']), null, 'Sind Sie sicher?' );

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  • Zend Framework in a subfolder, but images taken from site root

    - by Pentium10
    I have a dev project setup in a subfolder on my testing machine and it must stay there. However all the Zend frameworks views are linked to server root. CSS are linked like: <link type="text/css" href="<?php echo $this->baseUrl('/css/frontend.css') ?>" rel="Stylesheet" /> Which must be stayed this way, but it should link to localhost/a/b/c/prj1/css/frontend.css How can I setup a global subdirectory for this?

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  • How to transfer json data to html with php?

    - by cj333
    How to transfer json data to html with php? $url="http://api.nytimes.com/svc/search/v1/article?format=json&query=usa&rank=newest&api-key=mykey" when I type the url in browser, it return {"offset" : "0" , "results" : [{"body" : "A guide to cultural and recreational goings-on in and around the Hudson Valley. ...}]} how to put the json body data into html? I mean like this echo '<div class="body"></div>';

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  • MySQL Queries using Doctrine & CodeIgniter

    - by 01010011
    Hi, How do I write plane SQL queries using Doctrine connection object and display the results? For example, how do I perform: SELECT * FROM table_name WHERE column_name LIKE '%anything_similar_to_this%'; using Doctrine something like this (this example does not work) $search_key = 'search_for_this'; $conn = Doctrine_Manager::connection(); $conn->execute('SELECT * FROM table_name WHERE column_name LIKE ?)', $search_key); echo $conn;

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