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PHP - Php Return Continue
By calling function two will it break the loop?; or do i have to echo what is returned?
Code: [Select] functione one() { $getids = mysql_query("SELECT * FROM table"); while($row=mysql_fetch_assoc($getids); { $this->two(); } } function two() { return 'break;'; } Similar Tutorialsim building a form in php and is their away to have a submit or continue button on the same page so if ya want to fill in more it will go to the next page if not it will e-mail the info hello i need someone to take a look on this , General comments on the code process and how should i continue !!! [attachment deleted by admin] Help...I almost had this but I included the password field in my form now it is not working. What was happening before I had all the tested members added to the Welcome page at once. When I put in the password field it is not welcoming anyone so sessions is not operating. // Get the user session id variable into a local php variable for easier use in scripting $id = $_SESSION['id']; // Now let's initialize vars to be printed to page in the HTML section so our script does not return errors // they must be initialized in some server environments $firstname = ''; $lastname = ''; $country = ''; $email = ''; //Formulate Query //This is the best way to perform an SQL query $query = "SELECT id, firstname FROM `Members` WHERE id='$id'"; $result = mysql_query($query); //Check result //This shows the actual query sent to MySQL and the error. Useful for debugging. if(!$result){ $message = 'Invalid query:' . mysql_error() . "\n"; $message .= 'Whole query:' . $query; die($message); } //Use result //Attempting to print $result won't allow access to information in the resource //One of the mysql result functions must be used //See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc. while($row=mysql_fetch_assoc($result)){ echo "Welcome, {$row[firstname]}"; } //Free the resources associated with the result set mysql_free_result($result); At one point it when it was including everyone all at once it was able to view the private pages but the new update password takes me back to square one. It was showing a error message with SQL " after the WHERE clause where I had the id={$_SESSION['id']} so I just tested to see how the codes id='$id' would work and it does not have the echo Welcome, firstname. these codes are very tricky. hi i am stuck in a loop please can someone check what is miss. I wants to quit from loop and go to next title if that tile is found but cant figure out how should i go. my code: $feed_url = "http://www.myfeed.com/feed/"; $xml = simplexml_load_file($feed_url); foreach ($xml->channel->item as $item) { $ns_content = $item->children('http://purl.org/rss/1.0/modules/content/'); $title = $item->title; // opening file which contain titles already $content = file('file.txt'); foreach ($content as $aline) { if (strstr($title, $aline)) { $found = true; } } // i dont wants to print that $title which is already in file.txt so loop continue. if($found) { continue; } echo $title . "<br>" ; } Problem is with that continue statement if a title is matched from file.txt it will display me nothing, i only wants to skip that title and echo other titles. thanks for any help i'm need to retrieve a session after i've collected the data and went to PayPal (...urk...) i think i'm starting off correctly... but how the hell do i get the session name on the other side? Code: [Select] $uuid = uniqid(); session_name = $uuid; session_start(); // do some stuff // go to PayPal // come back from PayPal ....um... now i'm lost! how do i get the $uuid here?? once i come back from PayPal... this is where i get lost. how do i load a UUID-based session if i can't get the $uuid? and if there's a better way, by all means let me know! someone said i could pass the $uuid through PayPal, but i'm not seeing how yet... TIA gang! WR! This topic has been remastered in brilliant HD quality in JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=356870.0 I have spent the last few days going over Zend framework 2, and I am finding it extremely complex. Not saying it shouldn't be, but it seems to solve problems I never have come across (so to speak) .
For instance:
Routing is super complex. I usually use torophp - which is simple & gets the job done.
Dependency injection - example of problem I have never come across.
My motivations for learning zend, are -
1. To write modular reuseable code.
2. Learn advanced / modern PHP , as far as frameworks are concerned I just know codeigniter.
3. A framework with good long term support.
Is there some other framework that is easier to learn ? and that would satisfy my requirements ?
I have been programming in PHP for the last 2-3 years btw, so I know my way around
Thanks
Edited by nik_jain, 31 August 2014 - 09:56 AM. I am getting the following error when using composer: "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? Just started. I am pretty sure composer.json wasn't changed. journalctl doesn't show anything. Maybe Doctrine related, however, nothing seems to help.
Any ideas? [michael@devserver www]$ php -v PHP 7.3.4 (cli) (built: Apr 2 2019 13:48:50) ( NTS ) Copyright (c) 1997-2018 The PHP Group Zend Engine v3.3.4, Copyright (c) 1998-2018 Zend Technologies with DBG v9.1.9, (C) 2000,2018, by Dmitri Dmitrienko [michael@devserver www]$ composer -V Composer version 1.4.2 2017-05-17 08:17:52 [michael@devserver www]$ yum info composer Loaded plugins: fastestmirror Loading mirror speeds from cached hostfile * base: mirror.us.oneandone.net * epel: mirror.rnet.missouri.edu * extras: mirror.us.oneandone.net * remi-php73: mirror.bebout.net * remi-safe: mirror.bebout.net * updates: mirror.us.oneandone.net Installed Packages Name : composer Arch : noarch Version : 1.8.4 Release : 1.el7 Size : 1.8 M Repo : installed From repo : epel Summary : Dependency Manager for PHP URL : https://getcomposer.org/ License : MIT Description : Composer helps you declare, manage and install dependencies of PHP projects, : ensuring you have the right stack everywhere. : : Documentation: https://getcomposer.org/doc/ [michael@devserver www]$ composer update Loading composer repositories with package information Updating dependencies (including require-dev) [ErrorException] "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? update [--prefer-source] [--prefer-dist] [--dry-run] [--dev] [--no-dev] [--lock] [--no-custom-installers] [--no-autoloader] [--no-scripts] [--no-progress] [--no-suggest] [--with-dependencies] [-v|vv|vvv|--verbose] [-o|--optimize-autoloader] [-a|--classmap-authoritative] [--apcu-autoloader] [--ignore-platform-reqs] [--prefer-stable] [--prefer-lowest] [-i|--interactive] [--root-reqs] [--] [<packages>]... [michael@devserver www]$
return (int) $value VS return $value
What is the purpose of using int in a bracket? return (int) $value
Which is the best and safest method?
Thank you
Hello First of all i am very weak about OOP and classes in php. I am learning now but i stuck with very silly problem. My codes are below. <?php class myTestClass { function __construct() { $this->OldName("This is Old Name"); } function OldName($VeryOld) { if($VeryOld == "This is Old Name") $this->NewName(); else $VeryOld = "Something Wrong"; return $VeryOld; } function NewName() { echo "This is Brand New Name"; } } $i = new myTestClass(); ?> if i send value to OldName "This is Old Name" then codes works fine. But if i send "This is Old Nameeee" then does not appear anything. what i want to do is print "Something Wrong" text if i enter different value. I have no idea what return do and how it do? How to retrieve/print "Something Wrong"? i can do that with echo but i want to transfer result of return to another function. Really appreciate for any help/idea. Hey all. My site has a config database table that has a column showlogin to set whether or not login functionality is enabled. I'm playing around with MySQLi, Classes, and Returns. My return is always 1. Where am I going wrong? test.php <?php // Start Session session_start(); // Production Settings ini_set('display_errors', 1); mysqli_report(MYSQLI_REPORT_ALL); include 'db.php'; $mysqli = new mysqli($host, $user, $pass, $db); class styles { function showlogin($show) { global $mysqli; $select = $mysqli->query("SELECT showlogin FROM config LIMIT 1"); $value = $select->fetch_object(); if ($value->showlogin = 1) { return 1; } return 0; } } $login = new styles(); echo $login->showlogin($show); ?> Hi Everyone,
My name is Raoul Grosser, live in The Netherlands.
I'm a front/backend developer working with AngularJS and Bootstrap and Symphone2.
I've been writing PHP code for over 4 years now my focus is on opensource projects.
If you have any question, please let me know!
If a user is on Page X, but not logged in, when they log in to my website, I want them to return back to Page X (in this example). To handle this, I am adding this code to the top of each webpage... Code: [Select] // Set current Script Name. $_SESSION['returnToPage'] = $_SERVER['SCRIPT_NAME']; And then as part of my Log In script I have... Code: [Select] // Redirect User. if (isset($_SESSION['returnToPage'])){ header("Location: " . BASE_URL . $_SESSION['returnToPage']); }else{ // Take user to Home Page. header("Location: " . BASE_URL . "index.php"); } What do you think about this approach? Thanks, Debbie Hello again. I use echo "Rating:".$json['Rating]."/10"; and get in return: 7.5/10 - it's ok. But I want to replace 7.5 with 75.png (image), 6.3 with 63.png, 9.4 with 94.png... hope u understand. Know someone how I can do this? Thanks. Say I have my login form which then when person tries to log in with the wrong user name and password and sent straight back to the log in page, how do I send back they got the password and/or username wrong. I read somewhere in some code header("location: index.php&$errorvariable"); or something like that. I haven't tested it, not even sure how to get it to work on the index, would I just use an isset on a get/post to check if it exists then echo the error? Hi all, I am kinda stuck on the return() statement in the php.net manual about control structures. (http://au.php.net/manual/en/function.return.php) All I see is notes on when not to use it and some scary examples with loads of foo and bar but, it's hard to find out when it would be nice to use it. At the moment i am thinking its another way of echo, but I bet that's wrong. If someone could enlighten me a bit on when return() is useful I would be very happy since i see it more often everyday. CHeers! Hi, I have setup my paypal payment method for a shopping cart on my site, all is well but in order for my server to flag an invoice as paid, the user must return to my website after paying and the query string in the url give the script the go ahead to mark it as paid which in turn alerts the appropriate department that its ok to source the order. ~The problem is, if the user for whatever reason skip the redirect back to my site, the invoice is remain 'unpaid' and has to be changed manually when the payment has been checked in the account. As you might imagine this is a bit of a problem because not only can it cause quite a delay, it also requires manually changing values which can lead to other problems. So if there is a better way i would appreciate the info, thanks a lot. Hi all, can you give me suggestion how to return value from function like this Code: [Select] function returnArray(){ for($i=1; $i<=3; $i++){ return $i; //how to return this as array?? } } when I code "echo" inside the function then call the function it give me output => 123 but when I code "return" inside the function, it will give output => 1 Can you suggest me how to output it as array? I am trying make my code display different things depending on how many rows are returned This is my code Code: [Select] <?PHP include 'db_con.php'; $month = date("n"); $day = date("j"); $sql = mysql_query("SELECT * FROM table WHERE month = $month AND day = $day ")or die(mysql_error()); while($row = mysql_fetch_array($sql) ) if (mysql_fetch_array($row = 1 ) ) ///THIS IS AN ERROR BUT IF ONE ROW RETURNED THEN THIS { echo "ONE THING"; } elseif (mysql_fetch_array($row >1 ) )////IF > 1 ROW RETURNED THAN THIS { echo"TWO THINGS"; } else { echo "SOMETHING ELSE "; } ?> I have tried several variations of the above code but cannot get it to work. Thanks for any help anyone can give me. |
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