PHP - Using Php To Make My Website Dynamic
Hello everyone. This is my first post, so be nice!
I am building a website that will have a lot of content, similar to a newspaper. I have some pretty good HTML/CSS pages written, but the problem is that I need a way to make things more dynamic. One of my templates has a Header, Left Column, Middle Column, Right Column, and Footer. Everything stays the same from page to page except for the Middle Column (which holds each article). As it stands now, if I had 12 articles, I would have to have 12 nearly duplicate HTML pages which isn't good! I started studying PHP a while ago, but put that on hold to learn HTML/CSS, so I've kinda forgotten how PHP can help me out! Can someone help me figure out how to use PHP to my benefit? Thanks, Debbie Similar TutorialsFolks, I need help (Php code ) to generate a Dynamic Text on a Base Image. What i want to do is, to make this Image as header on my Site and to make this Header Specific to a Site, i want to Add the Domain Name on the Lower Left of the Image. Got the Idea? Here is the Image link: Quote http://img27.imageshack.us/i/shoppingheader1.jpg/ PHP Variable that holds the Domain name is: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hello. I want to create some variables dynamically. I have this so far: for($i = 1; $i <= 4 ; $i++){ $a = 'member' . $i; $$a = array(); } Does that give me this 4 arrays?: $member1, $member2, $member3 and $member4? Of course I want some process inside of the loop, but that would depend on the variables. Would that work? thanks! since many do not know the term sticky = when a form is submitted, the data contained is not deleted but instead last checked data will be shown. I have some dynamically generated checkboxes from the database. I tried the below code to make them sticky but they are simply preselected in stead of becoming sticky. What am I doing wrong? <?php $sql2 = "SELECT relation FROM table_rel_info"; $result2 = @mysqli_query($dbc, $sql2); while($row2 = mysqli_fetch_array($result2, MYSQLI_ASSOC)) { echo "<label>" . $row2['relation'] . "</label>"; echo "<input class='box' type='checkbox' value=1"; ?> <?php if (isset($row2['relation'])) { echo 'checked="checked"';}?> <?php echo "'/>" . "<br />"; } ?> I have a product page which populates all my products in one page. I have also a detail page which gives details on a product which i wanted to know. My problem is when I am going to click on the product that I want the detail page shows incorrect product details. I just want one detail product page so that it will be easy to edit the page in the future. I am asking an Idea on how to make one detail page in all of my products.. thanks... I got a question regarding a news website content that i want to make ! my question is how do i call my contents without using a lot of page? i explain let's say i got 10 news how do i put this 10 news in different pages without using 10 pages ? ex: you can see some links having a number like this http://bbc.uk/news/murder_case-12 then the next page got http://bbc.uk/news/finance-13 the title and the number id change but the page news doesnt change thanks for your answer. Hey guys I had created a while ago a script for my friend where you can buy points and then redeem stuff with those points, i'm looking for ways to keep my site secu currently what i have done- - protected all mysql queries with mysql_real_escape_string, strip_tags, and addslashes - have a valid SSL certificate on my website - checked if emails are valid for account creation what else can I do? Thank you. How can i make/add facebook like on my website or blog? Can anyone post sample example code? Hello !
How I can make a script of my PHP code to work even if my website is not running
Can I make this with php ?
This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=349726.0 This topic has been Ctrl+X/Ctrl+V'd to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=347400.0 Hi all I need to combine these two scripts: Firstly, the following decides which out of the following list is selected based on its value in the mySQL table: <select name="pack_choice"> <option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option> <option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option> <option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option> </select> Although this works OK, I need it also to show dynamic values like this: select name="category"> <?php $listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC"); while($categoryReturned=mysql_fetch_array($listCategories)) { echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>"; } ?> </select> I'm not sure if this is possible? Many thanks for your help. Pete Okay so my news script is set to view only 10 pieces of news. But I want it so that it starts a new page once I have more than 10 pieces of news. Code: [Select] <?php require("functions.php"); include("dbconnect.php"); session_start(); head1(); body1(); new_temp(); sotw(); navbar(); $start = 0; $display = 10; $query = "SELECT * FROM news ORDER BY id DESC LIMIT $start, $display"; $result = mysql_query( $query ); if ($result) { while( $row = @mysql_fetch_array( $result, MYSQL_ASSOC ) ) { news_box( $row['news'], $row['title'], $row['user'], $row['date'], $row['id'] ); } mysql_free_result($result); } else { news_box( 'Could not retrieve news entries!', 'Error', 'Error', 'Error'); } footer(); mysql_close($link); ?> I tried a few things but they failed....miserably. I'm trying to make a League of Legends (a video game) community website, both as a personal project and for practice. Now the game has a lot of champions, each of whom have 5 unique abilities. Now, I thought about manually inputting all the details about each champion into a MySQL database, but that would long and tedious, and I don't really have the time for it now. Also, the game patches very oftern (like, once every 2 weeks) which changes many of the stats, etc. of the champion, and it is not possible for me to keep manually updating these every time there is a patch. Fortunately, there is a League of Legends Wiki which has all the data I need in their specific champion pages, which they keep updated per patch. So I was wondering if there was any way to get the data from the divs in the wiki, and have it display on my site. What I want to do in my website is that whenever someone types a champion's name (in a post or whatever), I want it to display a hover-over dialog with some of the champions details. And a lot of other features such as that. In plain English I need a way to : > Tell PHP to go to the wiki's source code on a specific page > Find a specific div container > Get X data from there > Pass X data into a function to display the hover-over I think this way, I would not have to maintain a database as I can leech off the wiki's data. I have not coded anything like this before, so I would like a few pointers as to how to achieve this. Any help will be appreciated! hello My database is in a same server with seperate domain name , then I want to insert from website1 mysql data on website2 mysql data. can anyone help me? I tried searching on google but couldn't find any relevant information, please redirect me to relevant source or help me with the code. I want to pass a domain name in text field which will be scanned and then the script will display entire site map. Not external links or links on a page. Sorry it is not easy for me to explain. Eg. if i pass abc.com Script will display abc.com/12/adn.php abc.com/asd/asd/ etc Whatever their url format is. All the links on that domain. Is it possible to download files from a website to my online websites root directory? reason I'm asking is because I have been downloading large video files to my computer which take about 20 minutes!! Then I use FTP to upload them to my site but it takes about 2-3 hours per video!! I was looking for a faster way! All help would be great! $looponce = 1; foreach ($this->info as $k => $v) { if ( (($k == 'subject') && ($v['required'])) && (!$this->settings['customSubject'])) { for ($i = 0; $i <= 0; $i++) { $output[] = $this->display_errors('error_system_subject'); } } if ( (($k == 'name') && (!$v['required'])) || ((!array_key_exists("name", $this->info)) && ($looponce == 1)) ) { $output[] = $this->display_errors('error_system_name'); $looponce++; } } That is my loop that i check things in. What i'm more interested in is the name if statement. If currently checks for a name key in an array(below) and makes sure that it is set to required. If it's not set to required, print out a system error and die()'s. I'm looking for ways to remove the need for program errors and just change them to what is needed to run. What i know how to do is, get into the inner array, what i don't know is how to edit the data and put it back exactly as it was given to me. The inner array data can be put back in any order, but the outer array must be in exact order as it was given. above code $this->info = $formdata; $formdata = array( 'name' => array('name'=>"Full Name", 'required'=>false, 'type'=>'text'), # This needs to be required=>true, but i can't trust the user, which is why i have the error. 'telephone' => array('name'=>"Telephone", 'required'=>false, 'type'=>'phone'), ); Any help is greatly appreciated, also am i doing the foreach loop in the code above in an efficient manner or is there another way? Hi there, As the question says i tried several things but i can't work it out and my knowledge about php isn't that well. Hello! If you search Google, you'll notice the URL: http://www.google.com/webhp?hl=en#hl=en&source=hp&q=php+freaks&aq=f&aqi=g10&aql=&oq=&gs_rfai=CtdaTdtdRTNnPD5HuzASZtMWiCgAAAKoEBU_Q0ZT5&pbx=1&fp=19d754eee0b4f223 You can copy that URL anywhere you like... and the user will still see the same results. So basically the URL dynamically does an action, and accesses the database. How on do I make a URL like this? Meaning when you change the URL parameter values, it request a slightly different database query? Does that make sense? Similarly with NexTag.com: http://www.nextag.com/serv/main/buyer/ProductCompare.jsp?search=camera&page=0&node=500001&psort=%2FDigital-Cameras--zzcameraz500001zB6z5---html&zipcode=&cptitle=657166355&cptitle=656751324&cptitle=620051906&cptitle=705150048 That was a Dynamically made URL (I selected from check boxes which products to compare) and it makes that URL so anyone can see those products. Any ideas on how to do this? More examples on the URL: http://www.cars.com/go/compare/modelCompare.jsp?myids=9721,11439 (i select the cars, it generates that URL. Notice the IDs 9721,11439) |