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PHP - Use Just 1 Page Not Lots Of External?!
hi,
I am building a web app and need all my php on one page and links posting to same page etc, at the moment, I am linking to my townpubs.php page see code below: townpubs.php Code: [Select] <!doctype html> <?php include "../config.php"; $loggedIn = (isset($_COOKIE['loggedin']) && $_COOKIE['loggedin'] == 'true')?true:false; $town = $_REQUEST['RSTOWN']; $towns = mysql_query("SELECT * FROM pubs WHERE RSTOWN = '".$town."' ORDER BY RSTOWN ASC"); ?> <html> <head> <meta charset="UTF-8" /> <title>My Pub Space v1.0 β</title> <style type="text/css" media="screen">@import "jqtouch/jqtouch.min.css";</style> <style type="text/css" media="screen">@import "themes/jqt/theme.min.css";</style> <script src="jqtouch/jquery.1.3.2.min.js" type="text/javascript" charset="utf-8"></script> <script src="jqtouch/jqtouch.min.js" type="application/x-javascript" charset="utf-8"></script> <script type="text/javascript" charset="utf-8"> var jQT = new $.jQTouch({ icon: 'jqtouch.png', addGlossToIcon: false, startupScreen: 'jqt_startup.png', statusBar: 'black', preloadImages: [ 'themes/jqt/img/back_button.png', 'themes/jqt/img/back_button_clicked.png', 'themes/jqt/img/button_clicked.png', 'themes/jqt/img/grayButton.png', 'themes/jqt/img/whiteButton.png', 'themes/jqt/img/loading.gif' ] }); </script> </head> <body> <!-- TOWNS --> <div id="towns" class="current"> <div class="toolbar"> <h1>View Pubs</h1> <a class="back" href="index.php" rel="external">Home</a> </div> <h2>Pubs in <?php echo $town ?></h2> <ul class="plastic"> <?php while($row1 = mysql_fetch_array($towns)) { echo '<li class="arrow"><a href="pubinfo.php?PUBID='.$row1['PUBID'].'" rel="external">'.$row1['rsPubName'].'</a></li>'; } ?> </ul> </div> </body> </html> I would like to have this on my index.php page: index.php Code: [Select] <!doctype html> <?php include "../config.php"; $loggedIn = (isset($_COOKIE['loggedin']) && $_COOKIE['loggedin'] == 'true')?true:false; $query1 = "SELECT DISTINCT rsTown FROM pubs ORDER BY rsTown asc"; $result = mysql_query($query1); $towns = mysql_query("SELECT DISTINCT RSTOWN, COUNT(PUBID) As PubCount FROM pubs GROUP BY RSTOWN ORDER BY RSTOWN ASC"); $counties = mysql_query("SELECT DISTINCT RSCOUNTY, COUNT(PUBID) As PubCount1 FROM pubs GROUP BY RSCOUNTY ORDER BY RSCOUNTY ASC"); if ($loggedin == 'true'){ $locals = mysql_query("SELECT * FROM pubs INNER JOIN favepub_copy ON pubs.PUBID=favepub_copy.PUBID WHERE USERID='".$_SESSION['USERID']."' ORDER BY rsPubName asc"); } ?> <html> <head> <meta charset="UTF-8" /> <title>My Pub Space v1.0 β</title> <style type="text/css" media="screen">@import "jqtouch/jqtouch.min.css";</style> <style type="text/css" media="screen">@import "themes/jqt/theme.min.css";</style> <script src="jqtouch/jquery.1.3.2.min.js" type="text/javascript" charset="utf-8"></script> <script src="jqtouch/jqtouch.min.js" type="application/x-javascript" charset="utf-8"></script> <script type="text/javascript" charset="utf-8"> var jQT = new $.jQTouch({ icon: 'jqtouch.png', addGlossToIcon: false, startupScreen: 'jqt_startup.png', statusBar: 'black', preloadImages: [ 'themes/jqt/img/back_button.png', 'themes/jqt/img/back_button_clicked.png', 'themes/jqt/img/button_clicked.png', 'themes/jqt/img/grayButton.png', 'themes/jqt/img/whiteButton.png', 'themes/jqt/img/loading.gif' ] }); </script> </head> <body> <div id="about" class="selectable"> <p><img src="jqtouch.png" /></p> <p><strong>My Pub Space</strong><br />Version 1.0 beta<br /> <a href="http://www.jbiddulph.com" target="_blank">By John Biddulph</a></p> <p><em>Mobile Web Development</em></p> <p><a href="mailto:john.mbiddulph@gmail.com" target="_blank">E-mail</a></p> <p><a href="http://twitter.com/#!/jmbiddulph" target="_blank">@jmbiddulph on Twitter</a></p> <p><br /><br /><a href="#" class="grayButton goback">Close</a></p> </div> <div id="home" class="current"> <?php if (!$loggedIn){ ?> <div class="toolbar"> <h1>My Pub Space</h1> <a href="#about" id="infoButton" class="button slideup">About</a> </div> <ul class="rounded"> <li class="forward"><a href="#signup">Register an account</a></li> <li class="forward"><a href="#login">Login</a></li> <li class="forward"><a href="#towns">View pubs by town</a></li> <li class="forward"><a href="#counties">View pubs by county</a></li> </ul> <div class="info"> <p>All Rights Reserved © 2011 mypubspace.com Created by: jbiddulph.com</p> </div> <?php } else { ?> <div class="toolbar"> <h1>Hi <?php echo $_SESSION['s_username'];?></h1> <a href="dologoff.php" rel="external" class="button">Logout</a> </div> <ul class="rounded"> <li class="forward"><a href="#locals">View Locals</a></li> <li class="forward"><a href="#towns">View Towns</a></li> <li class="forward"><a href="#counties">View Counties</a></li> </ul> <div class="info"> <p>All Rights Reserved © 2011 mypubspace.com Created by: jbiddulph.com</p> </div> <?php } ?> </div> <!-- LOCALS --> <div id="locals"> <div class="toolbar"> <h1>My Locals</h1> <a class="back" href="#home">Back</a> </div> <ul class="plastic"> <?php while($row1 = mysql_fetch_array($locals)) { echo '<li class="arrow"><a href="pubinfo.php?PUBID='.$row1['PUBID'].'" rel="external">'.$row1['rsPubName'].', '.$row1['rsTown'].'</a></li>'; } ?> </ul> </div> <!-- TOWNS --> <div id="towns"> <div class="toolbar"> <h1>View Towns</h1> <a class="back" href="#home">Back</a> </div> <ul class="edgetoedge"> <?php $ltr = ''; while($row1 = mysql_fetch_array($towns)) { $letter1 = substr($row1['RSTOWN'],0,1); if ($letter1 != $ltr) { print'<li class="sep">'. $letter1 .'</li>'; } $ltr = $letter1; echo '<li class="forward"><a href="townpubs.php?RSTOWN='.$row1['RSTOWN'].'" rel="external">'.$row1['RSTOWN'].'<small class="listcounter">'.$row1['PubCount'].'</small></a></li>'; } ?> </ul> </div> <!-- COUNTIES --> <div id="counties"> <div class="toolbar"> <h1>View Counties</h1> <a class="back" href="#home">Back</a> </div> <ul class="edgetoedge"> <?php $ltr = ''; while($row1 = mysql_fetch_array($counties)) { $letter1 = substr($row1['RSCOUNTY'],0,1); if ($letter1 != $ltr) { print'<li class="sep">'. $letter1 .'</li>'; } $ltr = $letter1; echo '<li class="forward"><a href="countypubs.php?RSCOUNTY='.$row1['RSCOUNTY'].'" rel="external">'.$row1['RSCOUNTY'].'<small class="listcounter">'.$row1['PubCount1'].'</small></a></li>'; } ?> </ul> </div> <form id="login" action="dologin.php" method="POST" class="form"> <div class="toolbar"> <h1>Login</h1> <a class="back" href="#">Back</a> </div> <ul class="rounded"> <li><input type="text" name="rsUser" value="" placeholder="Username" /></li> <li><input type="Password" name="rsPass" value="" placeholder="Password" /></li> </ul> <a style="margin:0 10px;color:rgba(0,0,0,.9)" href="#" class="submit whiteButton">Submit</a> </form> <form id="signup" action="dosignup.php" method="POST" class="form"> <div class="toolbar"> <h1>Sign up</h1> <a class="back" href="#">Back</a> </div> <ul class="rounded"> <li><select name="RSTOWN" class="postcodedrop"> <option value="">Choose your Town...</option> <?PHP while($row = mysql_fetch_array($result)) { echo '<option value="'.$row['rsTown'].'">'; echo $row['rsTown']; echo '</option>'; }?> </select></li> <li><input name="RSUSER" type="text" class="textbox" id="RSUSER" value="" placeholder="Username" /></li> <li><input type="Password" name="RSPASS" value="" placeholder="Password" /></li> <li>Male <input name="RSGENDER" type="radio" value="Male" /> Female <input name="rsGender" type="radio" value="Female" /></li> <li><input name="RSEMAIL" type="text" class="textbox" id="rsEmail" placeholder="Email" /></li> <li><input name="RSMOBILE" type="text" class="textbox" id="rsMobile" placeholder="Mobile No." /></li> <li><input name="RSAGE" type="text" class="textbox" id="datepicker" placeholder="DOB: dd/mm/yyyy" /></li> </ul> <a style="margin:0 10px;color:rgba(0,0,0,.9)" href="#" class="submit whiteButton">Submit</a> </form> </body> </html> I need to set empty variables at the top and link page to itself?! Similar TutorialsHello,
I've tried to get a dynamic table from an external page, and searching for entries in it, so i used a dynamic XLS file using php excel reader. I only exported the file, but i couldn't search for data.
Can i get some help please ?
Hi, I am very new to PHP (and web developing generally, as well). I have been passing through two different ways of verifying user input in a PHP webpage: 1- By calling a different script file, where the verification logic code is listed and then recall the referer $_SERVER['HTTP_REFERER'] page and pass the result using $_SESSION. (I understood this is basically done to avoid repeating the action in the code with every refreshment of the browser window). 2- By enclosing the verification logic code withing the same PHP page, so the page is a big mix of HTML & PHP. (I understood this is basically done, in order to keep the user input without using $_SESSION and it should save one trip of data transfer.) For me, I see both are working; still I want to learn the best coding practices. So your advice is appreciated, and please feel free to correct my, if I missed something about both methods. Hey guys! im currently learning javascript, PHP and SQL. I have a pretty solid understanding of HTML and CSS. I want to make a site similar to facebook (a good facebook). this is going to be a big project and i plan on moving to a bigger server system in a year or so to keep up with demands.
Heres how the site will function:
1. Basic registration/splash page. I understand that the finished form is sent to a php file on the server side, correct? (ill change my server name files of course)
2. after the registration page, while the user is logged in with their new account, there are 3 pages after that that explain what the website is all about and how to use it. the last page allows the user to setup their profile information, ask friends to join, and asks what type of things they like. After the last page, it sends the user to their main control panel, where social media feed can be seen, friends and online chats, news, advertisements, links, pages and groups (think facebook and linkedin)
3. the user will have the ability to look at their profile (not the control panel), and of course switch back to their control panel. social media, friends, groups and ads will also be on their individual profile page as well.
4. i want the site to have two views: a standard view and an enhanced view. the enhanced view will reposition divs and all that stuff so they can see a background image (either stock or one they uploaded) this image will eventually change to an animated image of a 3d environment.
5. the site is going to be heavily social media based. This means social media feeds, image uploading, a structured comment system, a friend system, search functions and targeted advertising.
This is obviously a lot to ask, but since their is so much to learn related to PHP and SQL, can someone point me to the right tutorials on how to get these things done? I am currently learning javascript, PHP and SQL on lynda.com. Expect me to be on this forum a lot and asking a lot of questions. Thanks for any help.
Edited by PHPlearner32, 10 January 2015 - 11:19 PM. First, I gotta say I am new to JavaScript, but not new to languages like PHP/C++.
I am working on a large huge JS file that has a lot of things and I have a feeling that it can be made better. How is where I am at a loss. If it was PHP, I'd split that JS file into probably 20 other smaller JS files, aka "classes", and it may or may not be the right thing to do, but let me focus on one particular aspect for now, and maybe deal with others later.
Here is the piece of code that bugs the me most (truncated for clarity):
if (selection[index].selected) {
result += '<TABLE class="list_grey" style="float: left; margin-left: 20px;">';
result += '<TR>';
result += '<TH colspan="2">Data</TH>';
result += '</TR>';
var flag = "All";
if (selection[index].flag == 'P')
flag = "Premium";
result += '<TR>';
result += '<TD>Flag</TD>';
result += '<TD>' + flag + '</TD>';
result += '</TR>';
result += '</TABLE>';
}
else
{
result += '<TABLE class="list_grey" style="float: left; margin-left: 20px;">';
result += '<TR>';
result += '<TH colspan="2">Frame</TH>';
result += '</TR>';
result += '<TR>';
result += '<TD>Frame</TD>';
result += '<TD>' + selection[index].frame + '</TD>';
result += '</TR>';
result += '</TABLE>';
}
So what it does is this:
1. it gets "selection" var with eval() from a DIV block of HTML page, where that DIV has a looooong prepared-by-PHP JSON string with all the options prepopulated from various sources accessible to PHP.
var selection = eval("(" + document.getElementById("result").innerHTML + ")");
2. It then dynamically constructs and populates further HTML using logic and data found in "selection" var.
3. This particular JS implements a selection algorithm, where basically user clicking buttons can navigate "selection" var forwards and backwards, and as user does, the dynamic portion of HTML updates per behavior defined in the JS file.
My Problem / Question: I have a feeling that putting HTML the way it is done here in JS file is not proper separation of concerns and not good way to code. HTML should be. .. in HTML, when possible, but being new to JS I don't know if there is a better way. Is there a more recommended way to code what I have now ?
I am trying to get more than one row from a database at once, then explode each value into a seperate variable. is there a way of doing this? My site seems to be going well, but on the back end, the users database is going crazy with duplicate entrys. but its not doing it from an INSERT. it seems to be doing it with this code: Code: [Select] $bitquery = $bitcoin->move("bittleship$username", "bittleships", $cost, $minconfirmations); if ($bitquery == true){ $query = mysql_query("INSERT INTO `transactions` ( `id` , `username` , `type` , `amount`, `address`) VALUES ('', '$username', 'buyclicks', '$cost', '')") or die(mysql_error()); $query = mysql_query("UPDATE users SET `clicks` = `clicks`+$quantity, `balance`=`balance`-$cost WHERE `username` = '$username' ") or die(mysql_error()); include('update.php'); Hey guys, I'm having a strange problem with cURL that I'm drawing an absolute blank to. It works completely fine locally, but when I upload the files to my webhost and test it over there, it doesn't work at all - it loads for about 2 minutes, and then displays: Server error. The website encountered an error while retrieving http://www.craigwatcher.me/playground/units.php. It may be down for maintenance or configured incorrectly. I've checked the error logs but nothing new is being added into there. Basically I'm just iterating through a list of URLs and cURLing each of them. Let me show you the code that I'm working with: ini_set('max_execution_time',0); require_once('../system/utilities.php'); // This will give us our getLocations() function, which has been tested to work 100% both locally and online $categories = array('jjj', 'ggg', 'bbb', 'sss', 'hhh'); $locations = getLocations(); if (!$locations) exit; $ch = curl_init(); curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE); curl_setopt($ch, CURLOPT_FOLLOWLOCATION, TRUE); curl_setopt($ch, CURLOPT_TIMEOUT, 10); foreach ($locations as $country => $cities) { $countryCode = getCountryCode($country); foreach ($cities as $city) { foreach ($categories as $category) { $url=$city.'.en.craigslist.'.$countryCode.'/'.$category.'/index.rss'; echo $url.'<br/>'; curl_setopt($ch, CURLOPT_URL, $url); $rss = curl_exec($ch); if (!$rss) echo ' FAILED to load: '.$url.'<br/>'; } } } curl_close($ch); I've already tested everything else (eg. the getLocations() and getCountryCode(), and they prove to work 100% fine both locally and online. What could it POSSIBLY be!? I'm pulling my hair out over here, my mind is boggling and I'm completely lost at what could possibly be going wrong. Hi i followed a guid to create a login area to my site however i believe its a little outdated and therefore not working, i have done a little bit but being quite new i cannot find all the problems at the moment i have: Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/a9855336/public_html/checklogin.php on line 12 here is my script line 12 is marked <?php $host="*************"; // Host name $username="a9855**6_root"; // Mysql username $password="************"; // Mysql password $db_name="****3***36_mail; // Database name // Connect to server and select databse. mysql_connect($host, $username, $password) mysql_select_db($db_name); // username and password sent from form $myusername=$_POST['myusername']; // line 12 $mypassword=$_POST['mypassword']; // To protect MySQL injection (more detail about MySQL injection) $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); // encrypt password $encrypted_mypassword=md5($mypassword); $sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$encrypted_mypassword'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); // If result matched $myusername and $mypassword, table row must be 1 row if($count==1){ // Register $myusername, $mypassword and redirect to file "login_success.php" session_register("myusername"); session_register("mypassword"); header("location:login_success.php"); } else { echo "Wrong Username or Password"; } ?> if you help please tell me the error so i can learn to debug scripts myself in the future thanks blink359 We have a database that shows for example: 2500-5000 5000-10000 10000-15000 15000-20000 I need to show in a screen what people are have 3000 followers, and as such as grouped into that 2500-5000. The best way I can see how to do this, is to run through the database table for these values. Then for each value, put the first and second set of numbers into two respective variables, say $low, $high. Then find those people who are on a particular platform with followers between those two values. So how do I turn $row->followsrange (which might be for example: "2500-5000"), into:
$low = 2500 Many thanks. ps I have a feeling the hyphen might play a bit part, as it cannot be done by the amount of characters.......... Hi there, I'm creating a shopping cart and to work out postage cost I need to add all the items weights together. The weight of each item is in the database. I currently have this code: Code: [Select] $basket = $_SESSION['basket']; if ($basket) { $items = explode(',',$basket); $contents = array(); foreach ($items as $item) { $contents[$item] = (isset($contents[$item])) ? $contents[$item] + 1 : 1; } foreach ($contents as $id=>$qty) { $sql = "SELECT * FROM store WHERE id LIKE '$id' AND live LIKE '0'"; $result = mysql_query($sql); while ($rows = mysql_fetch_array($result)) { extract($row); so to work out each item's weight I can do $rows['weight'] * $qty. but how could i add all the item's weight together? i hope that makes some sense. thanks, jack I am having a problem I have encounted like no other. I am running a MySQL query from PHP and for some bizarre reason, its just not working... Ok, that sounds really general. lol To start off, here is my database: http://i56.tinypic.com/2hh0up1.png I am writing my own user interface with HTML/PHP/MySQL for every day catalog management: http://i54.tinypic.com/axxsb7.png To catalog a product is done from here (this page works): http://i51.tinypic.com/2e0qs7s.png ETC ... <FORM action="result.html" enctype="multipart/form-data" method="post"> ... ETC ETC ... Name: <INPUT name="name" type="text"> <BR> Brand: <INPUT name="brand" type="text"> <BR> Country of origin: <INPUT name="country" type="text"> <BR> Material: <INPUT name="material" type="text"> <BR> Primary colour: <INPUT name="primarycolour" type="text"> ... ETC result.html: ETC ... $query = "INSERT INTO products(name, brand, country, material, primarycolour) VALUES('".$_POST['name']."', '".$_POST['brand']."', '".$_POST['country']."', '".$_POST['material']."', '".$_POST['primarycolour']."')"; ... ETC To alter a product, you enter in a product ID: http://i56.tinypic.com/2lcsqch.png Code: [Select] <FORM action="dataentry.html" method="post"> <DIV class="drop">Alter product ID: <INPUT name="id" type="text">    <INPUT type="submit" value="Submit"></DIV> The data entry page pulls all the values from the MySQL database and populates them into the INPUT fields, so the user does not have to write them all again: http://i56.tinypic.com/2zh0hgn.png ... ETC echo '<FORM action="result.html" enctype="multipart/form-data" method="post">'; ETC ... ... ETC echo 'Name:'; $query = "select name from products where id=".$_POST['id'].""; $result = mysql_query($query); $row = mysql_fetch_array($result); echo ' <INPUT name="name" type="text" value="'.$row['name'].'">'; echo $query; ETC ... I will now change the value of "Name" from 'a' to 'c' and submit the changes: http://i55.tinypic.com/dzi9hc.png http://i54.tinypic.com/ab5lyg.png Now, you would think the result has been inserted into my database yeah? It failed... but when I enter the same command directly into MySQL. Success! http://i51.tinypic.com/29z5y5x.png Why the F is this happening?! Hi, in Mysql have a row with FLOAT 10,2 In form put the number 20.20 and after reload in form show: 20.200000762939 Hi there im not quite sure how to do something and was wondering if anyone can help me, I am going to create a large database with lots of fields and these are going to be displayed on a page i want to limit it to 20 per page and be able to sort them aswell by ID (default) and then a-z or z-a, i dont know how to do this and i also need to know how to create the next previous and last and first links im guessing theres a count++ involved ? im not quite sure but anyway if someone can help it will be great. Thanks, Blink359 Hello! I am executing an external command from a PHP script, using the exec function. Since this program can take more than 1-2 minutes to run, I thought I should use a "Loading page...please wait". What I need is to be able to get the process id from the external program that is being run and, when this finishes, I'll start outputting the results. Is there a way to do this? Quesion: Show each movie in the database on its own page, and give the user links in a "page 1, Page 2, Page 3" - type navigation system. Hint: Use LIMIT to control which movie is on which page. I have provided 3 files: 1st: configure DB, 2nd: insert data, 3rd: my code for the question. I would appreciate the help. I am a noob by the way. First set up everything for DB: <?php //connect to MySQL $db = mysql_connect('localhost', 'root', '000') or die ('Unable to connect. Check your connection parameters.'); //create the main database if it doesn't already exist $query = 'CREATE DATABASE IF NOT EXISTS moviesite'; mysql_query($query, $db) or die(mysql_error($db)); //make sure our recently created database is the active one mysql_select_db('moviesite', $db) or die(mysql_error($db)); //create the movie table $query = 'CREATE TABLE movie ( movie_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, movie_name VARCHAR(255) NOT NULL, movie_type TINYINT NOT NULL DEFAULT 0, movie_year SMALLINT UNSIGNED NOT NULL DEFAULT 0, movie_leadactor INTEGER UNSIGNED NOT NULL DEFAULT 0, movie_director INTEGER UNSIGNED NOT NULL DEFAULT 0, PRIMARY KEY (movie_id), KEY movie_type (movie_type, movie_year) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); //create the movietype table $query = 'CREATE TABLE movietype ( movietype_id TINYINT UNSIGNED NOT NULL AUTO_INCREMENT, movietype_label VARCHAR(100) NOT NULL, PRIMARY KEY (movietype_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die(mysql_error($db)); //create the people table $query = 'CREATE TABLE people ( people_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, people_fullname VARCHAR(255) NOT NULL, people_isactor TINYINT(1) UNSIGNED NOT NULL DEFAULT 0, people_isdirector TINYINT(1) UNSIGNED NOT NULL DEFAULT 0, PRIMARY KEY (people_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die(mysql_error($db)); echo 'Movie database successfully created!'; ?> ******************************************************************** *********************************************************************** second file to load info into DB: <?php // connect to MySQL $db = mysql_connect('localhost', 'root', '000') or die ('Unable to connect. Check your connection parameters.'); //make sure you're using the correct database mysql_select_db('moviesite', $db) or die(mysql_error($db)); // insert data into the movie table $query = 'INSERT INTO movie (movie_id, movie_name, movie_type, movie_year, movie_leadactor, movie_director) VALUES (1, "Bruce Almighty", 5, 2003, 1, 2), (2, "Office Space", 5, 1999, 5, 6), (3, "Grand Canyon", 2, 1991, 4, 3)'; mysql_query($query, $db) or die(mysql_error($db)); // insert data into the movietype table $query = 'INSERT INTO movietype (movietype_id, movietype_label) VALUES (1,"Sci Fi"), (2, "Drama"), (3, "Adventure"), (4, "War"), (5, "Comedy"), (6, "Horror"), (7, "Action"), (8, "Kids")'; mysql_query($query, $db) or die(mysql_error($db)); // insert data into the people table $query = 'INSERT INTO people (people_id, people_fullname, people_isactor, people_isdirector) VALUES (1, "Jim Carrey", 1, 0), (2, "Tom Shadyac", 0, 1), (3, "Lawrence Kasdan", 0, 1), (4, "Kevin Kline", 1, 0), (5, "Ron Livingston", 1, 0), (6, "Mike Judge", 0, 1)'; mysql_query($query, $db) or die(mysql_error($db)); echo 'Data inserted successfully!'; ?> ************************************************************** **************************************************************** MY CODE FOR THE QUESTION: <?php $db = mysql_connect('localhost', 'root', '000') or die ('Unable to connect. Check your connection parameters.'); mysql_select_db('moviesite', $db) or die(mysql_error($db)); //get our starting point for the query from the URL if (isset($_GET['offset'])) { $offset = $_GET['offset']; } else { $offset = 0; } //get the movie $query = 'SELECT movie_name, movie_year FROM movie ORDER BY movie_name LIMIT ' . $offset . ' , 1'; $result = mysql_query($query, $db) or die(mysql_error($db)); $row = mysql_fetch_assoc($result); ?> <html> <head> <title><?php echo $row['movie_name']; ?></title> </head> <body> <table border = "1"> <tr> <th>Movie Name</th> <th>Year</th> </tr><tr> <td><?php echo $row['movie_name']; ?></td> <td><?php echo $row['movie_year']; ?></td> </tr> </table> <p> <a href="page.php?offset=0">Page 1</a>, <a href="page.php?offset=1">Page 2</a>, <a href="page.php?offset=2">Page 3</a> </p> </body> </html> I would like a part of my script to link to an external sites script It doesn't seem to be doing it though Othersite.com - index.php?name=hello&status=1 I would like my post.php to run that above. Can it be done? I was just wondering if it was possible to have an external php file that can be included in the head of a web page, like a .js file. If this isn't possible maybe have a .js file containing php code that can be executed regarding the JavaScript around it... Hi, I have a peice of code which publishes an image with a link from my database. However I cant get it to use external links. My code is: echo "<a href=\"" .$link . "\"> <img src=\"" .$image ."\" /> </a><BR />"; I have tried all the options I can think of but I cant get it work. Can anyone advise please? I'm building a program in php that will be able to view YouTube videos from the URL. So if I type: http://www.youtube.com/watch?v=(Video ID) Then my program works and it will return the Video ID! But then some idiot clicks on a related video and this URL is generated: http://www.youtube.com/watch?v=(Video ID)&feature=related And my program strips the first bit by replacing "http://www.youtube.com/watch?v=" with an empty string "" but then I'm still left with the "&feature=related" I thought about just replacing that with an empty string as well but sometimes there can be a URL like this: &feature=g-vrec&context=G28d9eecRVAAAAAAAABA Which has a different unique code each time. So I thought it'd be much simpler if I could use $_GET[] with an external URL, so the user types in: http://www.youtube.com/watch?v=(Video ID)&feature=related It just gets the "v" value rather than my buggy replace thing. Thanks. Hi all,
So im trying to improve on my PHP as my knowledge isn't that great.
lets say for argument sake my ip is:
192.168.0.1
im trying to set a external variable to set this as the url.
so on each page i use, for each link instead of typing out the ip address i can simply type:
<a href="{url}/index.php">Home</a>
and then if it is possible, to do something like this:
<link rel="stylesheet" href="{url}/main.css" type="text/css">
all in one file so in my php pages i can simple include 1 php file and it will have all of the relevant stylesheets ect linked to it.
Edited by srwright, 30 May 2014 - 01:18 PM. |
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