PHP - Which Softwares Are Required To Create A Php Page?
Hi,
Please let me know which softwares are required to create a php page? And which is the best one for professional website creation? Thanks. Similar TutorialsHello there,
I am learning PHP & MySQL and developing websites.I have a doubt whether could I develop a software using PHP & MySQL without uploading to a website.I want to execute the projects in my PC.Whether i can develop a .exe software.Please help me.
I'm having a issue at the moment where my required page is being read when it shouldnt be. This if statement is true to where the first header in the if statement activates header("Location: forgotpassword.php"); but yet for some reason it reads the required safe page at the very bottom. if i disable the required safe.php the header works. Does required files not work with the flow of a page or something? Cause thats what appears to be going on here. // if npd is in the link the the user access the page from a link in their email cause they forgot their password if (isset($_GET['npd']) && (!isset($_SESSION['user_id']))) { // grabs user name and password from the link then seperates it into an array $seperate = explode("-", $_GET['npd']); $link_name = $seperate[0]; $forgotpd = $seperate[1]; // checks if username matches a deleted or banned account $count3 = "SELECT forgotpd FROM users WHERE name= '".mysql_real_escape_string($link_name)."' AND forgotpd= '".mysql_real_escape_string($forgotpd)."'"; $count2 = mysql_query($count3) or die(mysql_error()); $count1 = mysql_num_rows($count2); if($count1 < 1){ $_SESSION['mycache'] = "That recovery password link has already been used. If you didnt change your password after reseting your password then you will need to reset it again."; header("Location: forgotpassword.php"); }else{ // Search the database and get the password, id, and login ip that belongs to the name in the username field. $users3 = "SELECT login_ip,goauld,id FROM users WHERE name= '".mysql_real_escape_string($link_name)."' AND forgotpd= '".mysql_real_escape_string($forgotpd)."'"; $users2 = mysql_query($users3) or die(mysql_error()); $users1 = mysql_fetch_array($users2); // checks to see if the login ip has an ip already if(empty($users1['login_ip'])){ $users1['login_ip'] = $_SERVER['REMOTE_ADDR']; } // if the ip is different from the ip that is on the database it will store it $ip_information = explode("-", $users1['login_ip']); if (in_array($_SERVER['REMOTE_ADDR'], $ip_information)) { $users1['login_ip'] = $users1['login_ip']; } else { $users1['login_ip'] = $users1['login_ip']."-".$_SERVER['REMOTE_ADDR']; } // creates session for user id and login time $_SESSION['user_id'] = $users1['id'];// stores the id of the user $_SESSION['goauld'] = $users1['goauld']; $_SESSION['login_time'] = time(); // stores the log in time of the user $forgotpd= "UPDATE users SET password = '".mysql_real_escape_string($forgotpd)."' , login_count= login_count+1 , userip='".($_SERVER['REMOTE_ADDR'])."', login_ip='".($users1['login_ip'])."' , forgotpd= '' WHERE name= '".mysql_real_escape_string($link_name)."' AND forgotpd = '".mysql_real_escape_string($forgotpd)."'"; mysql_query($forgotpd) or die(mysql_error()); $_SESSION['mycache'] = "Your password has been reset. You will need to change your password in order to log in again in the future."; header("Location: account_settings.php"); } } require("safe.php"); Hi all, I'm trying to scrape the contents of a page that is behind a login screen; namely: http://my.mail.ru/apps. Here's my code. It almost works, but doesn't appear to be properly logging in -- I just get a login screen on the url download. Any ideas? Thanks much. Here's my code <?php $ch=login(); $html=downloadUrl('http://my.mail.ru/apps', $ch); echo $html; function downloadUrl($Url, $ch){ curl_setopt($ch, CURLOPT_URL, $Url); curl_setopt($ch, CURLOPT_POST, 0); curl_setopt($ch, CURLOPT_REFERER, "http://my.mail.ru/cgi-bin/login?noclear=1&page=http%3a%2f%2fmy.mail.ru%2fapps%2f"); curl_setopt($ch, CURLOPT_USERAGENT, "MozillaXYZ/1.0"); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_TIMEOUT, 10); $output = curl_exec($ch); return $output; } function login(){ $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, 'http://my.mail.ru/cgi-bin/login?noclear=1&page=http%3a%2f%2fmy.mail.ru%2fapps%2f'); //login URL curl_setopt ($ch, CURLOPT_POST, 1); $postData=' page=http%3A%2F%2Fmy.mail.ru%2Fapps%2F &Login=username &Domain=mail.ru &Password=password'; curl_setopt ($ch, CURLOPT_POSTFIELDS, $postData); curl_setopt ($ch, CURLOPT_COOKIEJAR, 'cookie.txt'); curl_setopt ($ch, CURLOPT_FOLLOWLOCATION,1); curl_setopt ($ch, CURLOPT_MAXREDIRS, 10); curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); $store = curl_exec ($ch); return $ch; } ?> This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=323045.0 i don't know what this would fall under, but it sounds like php..........is there a way to make a form make a new page from a template........like I have a picture page and I can upload pictures from a form with a picture name....and that name turns into a /name.php file? does this make sense? please can you help me in this question , imt confused about it
Create a dice page, that allow the user to enter an integer (from1 to 6) and the program must generate a random number (from1 to 6). if the random number equals the users’ entered number an image with smiley facemust be displayed if not another image with sad face must be displayed. Also,you have to display the integer user entered and the program generated integer in the page.
I'm using a popular PHP script for my web site, which uses main_1.htm for the header and footer, and inner_index.htm for the main part of the home page. Also, of course it has index.php. How can I set up a duplicates of these files to work on and test changes to the home page, before I actually deploy the changes on to my live site's home page? Thanks Hello, I am a C++ coder who is new to php so please excuse my n00bness. I have a form which I am then passing into a MySQL database via a php script. I am also using fwrite to try and create a .php site out of the variables entered into the form. Everything is working fine and I have managed to do this with a .html site but when I try to pass php into fwrite it doesn't work. example code $newhtml = fopen("backup/$v_uniquename.php", "w"); fwrite($newhtml, ' <html code here> '); works like a charm but when I use something like (just an example) $newhtml = fopen("backup/$v_uniquename.php", "w"); fwrite($newhtml, ' <?php while($row = mysql_fetch_array($result_date)){ $date = $row['1']; $team1 = $row['2']; $team2 = $row['3']; ?> '); I get errors and I am assuming it is because I am passing using ' ' within the php code within a fwrite function. Is this correct? and is there a work around way to make a php page using this method or something similar. Whats going on PHPfreaks? So this is what I've been trying to do for the whole weekend and just can't figure it out. I want to let users create new pages on my site like wiki style but dont want a wiki or drupal or joomla or any of that. I know its possible with: $text = $_POST['text']; $file = fopen($text . ".php","x"); fwrite($file,"Welcome to your new page!"); but I'm not sure how to integrate it. For further clarification here's an example: You come to the site and there is a list of previously made user pages all with their own user generated name. What I want to have is a create new page button that opens a new page allows the user to upload images then submit it to the database and upon refresh of the page their page is now in the list. The create new page resets and is available for another user and on and on. Any help will be appreciated. I need to get this done it's driving me crazy. Hi guyz, I am looking to create a dymanic comment box for my website where already logged in users can comment on webpages. It is quite a challenge getting to dynamically display only the comments a particular user posted on a web page. What I mean is I would like to be able to click on a page, see only the comments logged users posted on that page, and then post my own comment on that page. At the moment I created a comment box for my webpages but the issue is my posted comments on one particular page is displaying on all web pages with is comment system..This is the problem and it is very difficult to find helpful material hopefully you guyz can come to the rescue. Please guys any help or suggestions are more than welcome. Thanks Kdiamonds First of all i create the page to copy the buttons as pixels to paste into a design, it's not for a working webpage. It goes wrong here, <option value="0">Day</option> <?= for(var i=0; i<31; i++){?> <option value="<?=i?>"><?=i?></option> <?= } ?> </select> Code: [Select] <?php ?> <html> <head> <title>Community Development Project</title> </head> <body> <div id="container"> <form id="opties1" name="opties1" method="post" action="something.php"> <div class="forminput"> <input type="button" id="signin" name="signin" value="Sign In" /> <br/> <br/> <input type="button" id="register" name="register" value="Register" /> <br/> <br/> <input type="button" id="uploadimage" name="uploadimage" value="Upload Image" /> <br/> <br/> <input type="button" id="next" name="next" value="Next" /> <br/> <br/> <input type="button" id="addanother" name="addanother" value="Add Another" /> <br/> <br/> <input type="button" id="register" name="register" value="Register" /> <br/> <br/> <input type="button" id="create" name="create" value="Create" /> <br/> <br/> <input type="button" id="addfolder" name="addfolder" value="Add Folder" /> <br/> <br/> <input type="button" id="addimage" name="addimage" value="Add Image" /> <br/> <br/> <input type="button" id="addtask" name="addtask" value="Add Task" /> <br/> <br/> <input type="button" id="cancel" name="cancel" value="Cancel" /> <br/> <br/> <input type="button" id="watch" name="watch" value="Watch" /> <br/> <br/> </div> </form> <form id="opties2" name="opties2" method="post" action="something.php"> <div class="forminput"> <select id="day" name="day"> <option value="0">Day</option> <?= for(var i=0; i<31; i++){?> <option value="<?=i?>"><?=i?></option> <?= } ?> </select> <select id="country" name="Country"> <option value="0">Netherlands</option> <option value="1">Germany</option> <option value="2">Belgium</option> </select> </div> </form> </div> </body> </html> This is more of an SEO question. I have a site with couple hundred categories within each city. I was wondering what the best approach is to do them? www.mysite.com/browse/apples?city=new-york www.mysite.com/browse/new-york?category=apples Which one is the better way to do it for seo purposes? If it's the first method, that would mean I would have to create couple hundred pages for those categories yes? Hi, I'm trying to make a gallery, and it more or less works. But I don't like having to load so many images for the user to view.
What happens is I have a very basic method of loading the images into the index file using require.
<?php foreach (glob("*.jpg") as $filename) { echo "<div class='itemContainer'><img src='" . $filename . "' class='item' id='" . $filename . "' /></div>"; } ?>Pretty simple. but it just loads everything in the directory. What I'd like to do is make it so 30 images load, and then it creates a new "div container" and loads the next 30 in. As well as a page counter, and a next and previous button. I'm not too sure how to control the flow of information from PHP to the index file when using require. Or if this is even the right way of doing it. My thoughts are sorta like this $imageCounter = 0; $newPageStandard = 30; foreach($imageDir as $image){ // echo image; $imageCounter++ if($imageCounter == $newPageStandard){ $newPageStandard += $newPageStandard; // increase standard for next page if there are enough images //create new image container(div) //somehow re-direct the echo into this new image container } // continue echoing images }Something Along those lines. Is this at all on the right path, or should I be grabbing all the images, and stuffing them into an multidimensional array and breaking each array into segments of 30? Or perhaps another completely different method? (yeah I dont have a clue what Im doing atm) Edited by 7blake, 10 November 2014 - 06:37 AM. Hello guy , how do I write PHP page that have table to list all the data I entered to my system.
Hi guys.. 1st post here. I'm pretty new to php.. just a few weeks in. I've gotten pretty decent at making mysql connections and extracting data, but now I'm wanting to take one of my urls that I echo to my page and create a new page. Hoping someone can help... Here's the pictu MY MySQL Setup: I have 1 table called table1. It has 2 columns which a 'Title' & 'Description' My Index.php Page: I connect to mysql. I pull the 'Title' from MySQL and loop it to produce 100 rows of data that fill my index page, as expected. My Problem: I now want to be able to click one of those rows and dynamically create a new page (description.php) that includes the 'Title' & 'Description'. My Progress: I've managed to click a row and create a hyperlink to mysite.com/$title (where 'title' is pulling the 'title' from the mysql db) but I all I get is page not found. My Question: How do I tell description.php to pull the 'Title' and 'Description' and link to it from index.php? My Gratitude: Thanks in advance! This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=318572.0 The result pages is supposed to have pagination like google help me please
hello - dear phpfreaks,
i portet over a wordpress site to localhost. (an opensuse linux-box) all went nice and was very smooth to do so. after porting over the files and the db; i added the db-name and user-name etc. but nothing more. then i looked at the site http://localhost/mysite i saw the site - but only once - that is very very interesting. i read that i have to do more. Probably these changes - mentioned below are mandantory in order to avoid a blank page cf: https://managewp.com...#comment-148613 The two fields you need to edit are “siteurl” (highlighted above) and “home” (which you may need to navigate to the second page to find). Just click the “Edit” buttons next to each field, and replace the URL contained in “option_value” with “http://localhost/yourfoldername/”. That’s it! If you now navigate to “http://localhost/yourfoldername/”, your site should load up in all its glory. Please note that if you use custom permalinks, you will need to change them to default (in the WordPress > Settings > Permalinks screen) in order for internal links on your site to work. You can of course change the permalinks back to their custom form at any time. question: is this true? Do i need to make these changes to avoid a blank page!? Hello everyone, I am working on a form that is similar to a shopping cart system and I am thinking of creating a button that submits the checked value and saves them to a $_SESSION variable. And also a link that links to a cart.html that takes the values of a $_SESSION variable. I am have trouble figuring what tag/attribute should I use in order to achieve that.
Right now my code attached below submits the checked values to cart.html directly. However I want my submit button to save the checked box to a $_SESSION variable and STAY on the same page. And then I will implement a <a> to link to the cart.php.
I researched a little bit about this subject and I know it's somewhat related to ajax/jquery. I just wanted to know more about it from you guys. I appreciate your attention for reading the post and Thanks!
Below is the form that I currently have:
<form name= "finalForm" method="POST" action="cart.php"> <input type="Submit" name="finalSelected"/> <?php foreach($FinalName as $key => $item) {?> <tr> <td><input type="checkbox" name="fSelected[]" value="<?php echo htmlspecialchars($FinalID[$key])?>" /> <?php echo "$FinalID[$key] & $item";?> </td> </tr> <?php } ;?>Below is the code for cart.php <?php require ('connect_db.php'); if(isset($_POST['finalSelected'])) { if(!empty($_POST['fSelected'])) { $chosen = $_POST['fSelected']; foreach ($chosen as $item) echo "aID selected: $item </br>"; $delimit = implode(", ", $chosen); print_r($delimit); } } if(isset($delimit)) { $cartSQL = "SELECT * from article where aID in ($delimit)"; $cartQuery = mysqli_query($dbc, $cartSQL) or die (mysqli_error($dbc)); while($row = mysqli_fetch_array($cartQuery, MYSQLI_BOTH)) { $aTitle[] = $row[ 'name' ]; } } ?> <table> <?php if(isset($delimit)) { $c=0; foreach($aTitle as $item) {?> <tr> <td> <?php echo $aTitle[$c]; $c++;?> </td> </tr> <?php }}?> </table> Hi Friends, I am trying to do an API with oracle database. The JSON request from 3rd party will look like below. { "contact_id": "1", "serial_no": "100", "name": "baby", "inv_date": "2018-06-27", "due_date": "2018-06-27", "currency": "KD", "subtotal": "143", "tax_total": "13", "shipment_data": [ { "serial_no": "33", "master_no": "55", "house_no": "77", "cost_revenue_items": [ { "charge_ref": "rr", "currency": "INR", "quantity": "2", "selling_rate": "45", "exchange_rate": "7", "taxes": [ { "serial_no": "1", "ref": "INR", "voiding_remarks": "oo" },{ "serial_no": "2", "ref": "KWD", "voiding_remarks": "asd" } ] } ] } ] }
how to handle request for JSON in which master is updated (PUT) and child record is (Either INSERTED, UPDATED or DELETED) ? |