PHP - How Do I Make A Dynamic Link?
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: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hello everyone. This is my first post, so be nice! I am building a website that will have a lot of content, similar to a newspaper. I have some pretty good HTML/CSS pages written, but the problem is that I need a way to make things more dynamic. One of my templates has a Header, Left Column, Middle Column, Right Column, and Footer. Everything stays the same from page to page except for the Middle Column (which holds each article). As it stands now, if I had 12 articles, I would have to have 12 nearly duplicate HTML pages which isn't good! I started studying PHP a while ago, but put that on hold to learn HTML/CSS, so I've kinda forgotten how PHP can help me out! Can someone help me figure out how to use PHP to my benefit? Thanks, Debbie Hello. I want to create some variables dynamically. I have this so far: for($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... This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=321416.0 Im rebuilding my website and have decided to make the dynamic images for my main photography gallery be layed out using div tags and css. I can get the images to load in and layout correctly but for some reason the links on the images only works on the last loaded in image. I check the code on the page and all the links are there but only the last one is active. The code worked when it was all wrapped in table tags. Is this a PHP or css issue stopping all the links expect for the last to be active. This is the css code I am using. #maincontent{width:1000px;margin:0 auto;padding:0 20px} #maingallery{height:382px;position:relative;width:1000px;text-align:center;margin-top:15px} #maingallery #text{position:absolute;top:50px} .gallery_item {float:left; margin: 20px 0 0 0; width: 200px; height: 67; padding: 0 10px 0 10px} .gallery_item .p {padding: 10px; font-size:12px} .gallery_item a:hover, a:active { display : block; border : none; } Any help would be greatly appreciated. Not sure if there would be any thing conflicting the links being active in either php or css. Thanks Matt 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/$title (where 'title' is pulling the 'title' from the mysql db) but I all I get is page not found. My Question: How do I tell description.php to pull the 'Title' and 'Description' and link to it from index.php? My Gratitude: Thanks in advance! Hey, What i want is that if a username inserts some text into a field like "Hey @martin how are ya" ? then it will display @martin as a link that takes you to site.com/martin, i searched php.net from preg_replace and str_replace...unfortunately for me things got so complicated that i don't even want to talk about it. Anyone who can do it will be my saviour! This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=307669.0 ok, how do I make this button go to the specific link that is set with it.......it goes to page without eventid=eventid: Code: [Select] <form action= \"editevent.php?eventid=".$row['eventid']."\" method=\"link\"><INPUT TYPE=\"submit\" VALUE=\"Edit\"></form>"; I have created a database that echos out the result on a webpage, what i want to do is create one of the echo results into a link - the code is as follows: "Web Address: $web_address <br>" . Is there an easy way to make the result a link as the input will always be a web address? Thanks in advance! I have a site, and there is an image with a link to another page, how can I make it so this is the only way to get to that page is by clicking on that image, and not by typing www.mywebsite.com/page.php into the address bar in a browser? Hi I would like to know how to make a link select certain stuff from the database. For example the link called cape town with the id 11 is selected then the link must search through the events table for the city_id 11 and take that information and display it in another page. I would like to know how to do that or can you let me know of place that I can look. I have tried google but I don't know how to put what I'm looking for in words I would like to know how to make my links show that it's active. For example if I click on a link that says overview then that link has to be highlighted. This is the code that I have: Code: [Select] <?php if (isset($this->_params['submenus']) && is_array($this->_params['submenus'])){ foreach ($this->_params['submenus'] as $submenu){ ?> <li class="active"><a href="<?php echo $this->get_uri("/{$submenu['controller']}/{$submenu['action']}") ?>" onclick="javascript:return setSubMenu(<?php echo $submenu['id'] ?>);"><?php echo strtoupper($submenu['name'])?></a></li> <?php } } ?> Unfortunately with this code all the links are selected as active. I have a table where it displays some data. How do I make this Code: [Select] echo "<td>" . $row['title'] . "</td>"; into a link? Hi People. I am doing quite well with my www.airfieldcards.com website which is information for pilots in the UK and ireland should they wish to visit another airfield. It is in the form of a flight guide for free. Each card is held in a database and the cards are shown on the site by loading the following script /php/results.php?id=132 In the above example the card for c-moor in Ireland would be loaded. I want to be able to add a link within the cards that will give the user "the code" to enter that card within their own website. Sort of a "click here to get the code to add to your website" Then opening a separate box in a new window that had something like this in it. Copy and Paste the following code into your website Code: [Select] <iframe width="720" height="1500" frameborder="0" scrolling="no" marginheight="0" marginwidth="0" src="http://airfieldcards.com/php/results.php[b][i]"what goes here?"[/i][/b] What would I put into the bit "what goes here?" to dynamically load that specific card? and give the user the correct code to add into their own site? Thanks in advance, you guys on here have always been brilliant. Lets say that I want to make a certain word (or words) on a page a link. How would I make it a link without having to code in a link. Some site have adds on certain words; one day and not the next day. I am looking for similar feature, but less intrusive. No pop-up, just a link. Seems to be a basic question, but I couldn't find an answer nor figure it out on my own. Basically I have a script that takes out specific data out of the database, the script works on its own, now I just need a way to make the user execute it with a link or a button. Example: Category: [Smileys] - [Category2] - [Category3] - etc. As soon as the user clicks on [Smileys] all data in the database which contains the word smileys in the category field gets selected and outputted as a list. Again the script works, I just need to be able to execute it with a button. If I understood it correctly I have to run the script by adding a if (isset($_POST['Smileys'])) { in front of the script. But how do I build the connection with the text link? I basically have a PHP Search Form, and when a user fills in a form it outputs the results. Each result displays a image of a property, how could i make them images have their own unique link which will take them directly to the page of the property being shown? Im using PHP and mySQL tables Any help is appreciated, Thank You. Heres the PHP that outputs the results: Code: [Select] <?php require_once 'mstr_ref.php'; function san($input){ if(get_magic_quotes_gpc()){ $input=stripcslashes($input); } $output = mysql_real_escape_string($input); return $output; } if(isset($_POST['submit'])){ $pVars = array('area'=>$_POST['areas'], 'propType'=>$_POST['prop_type'], 'saleType'=>$_POST['ptype'], 'minB'=>$_POST['min_bedrooms'], 'maxB'=>$_POST['max_bedrooms'], 'minP'=>$_POST['min_price'], 'maxP'=>$_POST['max_price']); foreach ($pVars as $k=>$v){ $v = san($v); } $sql = new makeQuery(); $sql->manAdd('location_id', $pVars['area']); if($pVars['propType'] != 'Any'){ $sql->manAdd('catagory_id', $pVars['propType']); } if ($pVars['maxB'] > 0){ $sql->manAdd('bedrooms', $pVars['maxB'], '<='); } if($pVars['minB'] > 0){ $sql->manAdd('bedrooms',$pVars['minB'],'>='); } if($pVars['saleType'] != 'Any'){ if($pVars['saleType'] == "forsale"){ $sql->manAdd('market_type', 'sale'); if($pVars['minP'] != 0){ $pVars['minP'] = $pVars['minP'] * 1000; } if($pVars['maxP'] != 0){ $pVars['maxP'] = $pVars['maxP'] * 1000; } } if($pVars['saleType'] == 'forrent'){ $sql->manAdd('market_type', 'rent'); } } $qry = $sql->sqlStart.$sql->stmt.'Group By property.id'; $results = mysql_query($qry) or die (mysql_error()."<br />|-|-|-|-|-|-|-|-|-|-|-|-<br />$qry"); if(mysql_num_rows($results) < 1){ die ("Sorry, No Results Match Your Search."); } while($row = mysql_fetch_assoc($results)){ echo '<div class="container" style="float:left;">'; echo '<div class="imageholder" style="float:left;">'; echo "<img class='image1' src='{$row['image_path']}' alt='{$row['summary']}'> <br />"; echo '</div>'; echo '<div class="textholder" style="font-family:helvetica; font-size:14px; float:left; padding-top:10px;">'; echo "{$row['summary']}"; echo "<span style=\"color:#63be21;\"><br><br><b>{$row['bedrooms']} bedroom(s) {$row['bathrooms']} bathroom(s) {$row['receptions']} reception room(s)</b></span>"; if($row['parking'] != null){ echo "<span style=\"color:#63be21;\"><b> {$row['parking']} parking space(s)</b></span>"; } echo '</div>'; echo '<div style="clear:both"></div>'; } } else{ echo "There was a problem, please click<a href='index.php'> Here </a>to return to the main page and try again"; } ?> |