<li class="list-group-item">
<a href="/user/Michael" class="thumb-sm pull-left m-r-sm">
<img src="http://www.gravatar.com/avatar/8b7a9ba3cbf958009080f6da12a55029?&d=mm&r=g?&d=mm&r=g&s=215" class="img-circle">
</a>
<a href="user/Michael" class="clear">
<strong class="block">
<?php include '/includes/connection.php';?>
<?php echo $products['Title'] ; ?>
</strong>
<?php include '/includes/connection.php';?>
<small><?php echo $products['Followers'] ; ?> Followers </small>
</a>
</li>
<li class="list-group-item">
<a href="/user/Steven" class="thumb-sm pull-left m-r-sm">
<img src="http://www.gravatar.com/avatar/a5fb2decd550cdf33cbb8ce7566ba772?&d=mm&r=g?&d=mm&r=g&s=215" class="img-circle">
</a>
<a href="/user/Steven" class="clear">
<strong class="block">
<?php include '/includes/connection.php';?>
<?php echo $products['Title'] ; ?>
</strong>
<small><?php echo $products['Followers'] ; ?> Followers</small>
</a>
</li>
do i need to manually insert for every content?
I have over 100 contents how can i automatically insert in every line like content 1 display row 1 followers and content box 2 display row 2 followers and so on
The trick here is to make a loop within the code. This means you have to generate all content from out of the database or it's not gonna work. Let me show you an example:
<ul>
<?php
$sql = "SELECT ProjectId, ProjectTitel, ProjectExpertise
FROM project";
$stmt1 = mysqli_prepare($con, $sql);
mysqli_stmt_execute($stmt1);
mysqli_stmt_bind_result($stmt1,$ProjectId,$ProjectTitel,$ProjectExpertise);
while (mysqli_stmt_fetch($stmt1)){
?>
<li class="wow fadeInLeft" data-wow-offset="30" data-wow-duration="1.5s" data-wow-delay="0.15s">
<a href="inc/elements/project.php?id=<?php echo $ProjectId; ?>" class="meer">
<img src="img/portfolio/<?php echo $ProjectId; ?>/thumbnail/1.jpg" alt="<?php echo $ProjectTitel; ?> project">
<div class="project-info">
<div class="project-details">
<h5 class="witte-text blauwe-streep-onder">
<?php echo $ProjectTitel; ?>
</h5>
<div class="details witte-text">
<?php echo $ProjectExpertise; ?>
</div>
</div>
</div>
</a>
</li>
<?php
}
?>
</ul>
I start an UL outside of the PHP code and then I start my PHP query, as you see, i select the items i need from the database. Here i create a while loop and run through my setup of the list element. As you see, i use my items within the project id link (the link is an ajax call to another page), the thumbnail needed to show the picture, the title and the expertise i used.
In your case, the while loop you need requires more than just the the title and followers. I think you need as well an userid / username so you can take in account which user it is. Your loop would now proces all on the same user, since its staticly defined in your code. Also the avatar picture is staticly defined. Let me try to resolve some for you.
<?php
include '/includes/connection.php';
$sql = "SELECT products.title, products.followers
FROM products"; // as example query
$stmt1 = mysqli_prepare($con, $sql);
mysqli_stmt_execute($stmt1);
mysqli_stmt_bind_result($stmt1,$title, $followers);
while (mysqli_stmt_fetch($stmt1)){
?>
<li class="list-group-item">
<a href="/user/Michael" class="thumb-sm pull-left m-r-sm"> <!-- make the username also to be pulled out of a database. -->
<img src="http://www.gravatar.com/avatar/8b7a9ba3cbf958009080f6da12a55029?&d=mm&r=g?&d=mm&r=g&s=215" class="img-circle">
</a> <!-- you should save the avatar link into a database as well -->
<a href="user/Michael" class="clear"> <!-- make the username also to be pulled out of a database. -->
<strong class="block">
<?php echo $title; ?>
</strong>
<small>
<?php echo $followers; ?>
Followers
</small>
</a>
</li>
<?php
}
?>
This as example. As you see, i added notitions behind some code, to explain this should also be dynamic, since it will help you Remember. being a programmer is to find shortcuts on how you display your information. Code that repeats itself constantly with a changing variable should always be looped, to make sure you dont type unneccesairy code.
Since I dont know how your database is made, i can only guess, Michaels name is probably linked to an ID in the same table your products are, which you can use then in to pull out of the database, by searching in your user table. I hope I make sense here. For instance, your products.userid should be as well in your user table as user.userid. Most likely the userid will have a name linked to it in the user table.
$sql = "SELECT products.title, products.followers, products.userid, user.userid, user.username
FROM products, user
WHERE user.userid = product.userid";
So now you have in each row as well the name of the person of who's title it is. And that you can echo out again in the code i put up. (make sure you bind the result in the same order as you pulled them up)
Writing code is all about making it easier to display your information. Loops is and stays the keyword here, as NadirDev explains.
I hope I helped you getting on the right track.
Related
Actually I am beginner programmer in HTML, CSS and PHP. I have simple website for add and register in courses. The user should be add course into website and the courses should be posted on the site.so users can browse and register.
Actually my problem is how to call the course name from database and how to format it with HTML code as I want.
This is the page of courses which is content the list of available courses in website ( please note it is only HTML code, I do that to see how the page will be )
Screenshot of page:
So as you see, the first page include many this HTML code to add course list into website with the following code:
<div class="card card-1">
<a href="http://127.0.0.1/project2/course details/course1.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% "></a> <div class="container">
<h4 class="textstyle"><b>Operating System</b> </h4>
<p class="textstyle">Free Course</p>
</div>
</div>
what i want do with PHP?
I want to write a PHP code to replace the P and h4 with the course name, cost of courses from my database for each available course.
Please note: the pic for each course it will be from my pc, no need to call the pic from database.
I tried to write PHP code like below:
<div>
<div class="card card-1">
<a href="http://127.0.0.1/project2/course details/course1.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% "></a> <div class="container">
<?php
include_once("db.php");
$result = mysqli_query(OpenCon(), "SELECT Course_Name,cost FROM `course`");
//while($res = mysql_fetch_array($result)) { // mysql_fetch_array is deprecated, we need to use mysqli_fetch_array
while($res = mysqli_fetch_array($result)) {
echo "<p>".$res['Course_Name']."</p>";
echo "<p>".$res['cost']."</p>";
}
?>
</div>
</div>
</div>
This is my result:
It's okay but I want the style to be like the first screenshot. each course should have picture.
After that when the user click on course name. I want move to another page which is content the course details ( for the same course that user clicked ) also it's called for my database
like this:
I hope any one help my to solve this problem only, I should solve this problem within 2 days only. and sorry if my explanation is bad.
Thanks in advance for everyone.
Put the code in a PHP loop.....
So, this
<div class="card card-1">
<a href="http://127.0.0.1/project2/course details/course1.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% ">
</a>
<div class="container">
<h4 class="textstyle"><b>Operating System</b> </h4>
<p class="textstyle">Free Course</p>
</div>
</div>
Becomes (after cleaning up the code a bit - I think you didn't mean to use two <p> in there, but I left them so you can see it. Note that using different lines for the segments makes it a lot easier to see what you have.)
include_once("db.php");
$result = mysqli_query(OpenCon(), "SELECT Course_Name,cost FROM `course`");
$count = 0;
while($res = mysqli_fetch_array($result)) {
$count ++;
// NOTE: Here is the LOOP! - not outside the query, but INSIDE it
// First you 'jump out' of PHP, going back to HTML
?> <!-- now you are in HTML (when you need PHP again, you 'jump in' and 'jump out' as needed - see the code below....) -->
<div class="card card-<?php echo $count;?>">
<a href="http://127.0.0.1/project2/course details/course<?php echo $count;?>.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% ">
</a>
<div class="container">
<h4 class="textstyle">
<b><p><?php echo $res['Course_Name'];?></p></b>
</h4>
<p class="textstyle">
<p><?php echo $res['cost'];?></p>
</p>
</div>
</div>
<?php // we are in PHP again....
}
That should do what you asked for - though I would go a step (well, more than one...) further and make as much of this dynamic as you can.
For this I will presume that:
your database table has a column called 'id' (if it doesn't, you should have) and it relates to the course number (you could make a course number column if they don't match up, but I'm keeping it simple)
you have all your pictures labeled 'coursepicX' where the X is the course number.
We'll use 'coursepic' as a default in case there isn't a picture yet...
Now, the code is more dynamic!
include_once("db.php");
$result = mysqli_query(OpenCon(), "SELECT id,Course_Name,cost FROM `course`");
while($res = mysqli_fetch_array($result)) {
// NOTE: Here is the LOOP! - not outside the query, but INSIDE it
// First you 'jump out' of PHP, going back to HTML
?> <!-- now you are in HTML (when you need PHP again, you 'jump in' and 'jump out' as needed - see the code below....) -->
<div class="card card-<?php echo $res['id']?>">
<a href="http://127.0.0.1/project2/course details/course<?php echo $res['id']?>.php">
<?php
$pic = "http://127.0.0.1/project2/icons/coursepic.jpg";
if(file_exists("http://127.0.0.1/project2/icons/course" . $res['id'] . ".jpg") {
$pic = "http://127.0.0.1/project2/icons/course" . $res['id'] . ".jpg";
}
<img src="<?php echo $pic; ?>" alt="Avatar" style="width:101% ">
</a>
<div class="container">
<h4 class="textstyle">
<b><p><?php echo $res['Course_Name'];?></p></b>
</h4>
<p class="textstyle">
<p><?php echo $res['cost'];?></p>
</p>
</div>
</div>
<?php // we are in PHP again....
}
Note that this is the basic 'shopping cart' sort of program - you will likely use it many (many) times in your career.
Happy Coding!
I'm making a website where users can post stories etc and they get put into the MySQL database and the front page pulls the top-rated stories from the database and fills in the HTML template I made to display.
So I pull the top rated post and put it inside the article tags that I've made inside my HTML. The question is when it displays it displays outside of the box. Like so
The actual display is supposed to look like this
How do I format the PHP code, Or is there some better way to go about doing this?
<article class="post featured">
<header class="major">
<span class="date">April 25, 2017</span>
<h2>
And this is a<br />massive headline
</h2>
<?php echo '<p>' . htmlspecialchars($row['post'], ENT_QUOTES, 'UTF-8') . '</p>'; ?>
</header>
<img src="images/pic01.jpg" alt="" />
<ul class="actions special">
<li>Full Story</li>
</ul>
<span class="tags badge-pill badge-primary">Otero</span>
<span class="tags badge-pill badge-info">SCR</span>
<span class="tags badge-pill badge-success">Class of '22'</span>
<span class="tags badge-pill badge-danger">Male</span>
</article>
EDIT: Here is the HTML spot I'm talking about
Ok, It`s easy to solve. First your string has only one word (because doesn´t have spaces) an second, when you have a paragraph, the tag that wrap content has an autoalignement.
Maybe you can try adding some css rules for your parent div?
You may check this : https://stackoverflow.com/a/3367759/8186765
I have 2 tables:
TABLE users (id, name, image, user_group_id)
TABLE user_groups (id, name, menu)
column menu value contains something like this:
<img src="../images/employees/'.$row_UserDetails['Image'].'" alt="">
Now at img src its creating a problem.
'.$row_UserDetails['Image'].'
This doesn't display the value.
Why is that and how do we fix this?
I want to display results from recordset UserGroup like this:
Now the value coming from php mysql is: <img src="../images/employees/'.$row_UserDetails['Image'].'" alt="">
you see .$row_UserDetails['Image'].
This value must be abc.jpg which comes from Recordest UserDetails:
On this php page if i want to display user name, i use php echo like this: <?php echo $row_UserDetails['Name'];?>
When using echo in php you will need to escape the quotes as they cause problems.
<header id=\"header-navbar\" class=\"content-mini content-mini-full\"> <ul class=\"nav-header pull-right\"> <li> <div class=\"btn-group\"> <button class=\"btn btn-default btn-image dropdown-toggle\" data-toggle=\"dropdown\" type=\"button\"> <img src=\"../images/employees/\"" .$row_UserDetails['Image']. "\" alt=\"\"> <span class=\"caret\"></span> </button> <ul class=\"dropdown-menu dropdown-menu-right\"> <li> <a tabindex=\"-1\" href=\"my-profile.php\"> <i class=\"si si-user pull-right\"></i> My Profile </a> </li> <li> <a tabindex=\"-1\" href=\"../logout.php\"> <i class=\"si si-logout pull-right\"></i>Log out </a> </li> </ul> </div> </li> </ul>
</header>
You say you echo $row_UserDetails['Image'], but I don't see an echo statement anywhere. Also, if that PHP code is really in your database like that, I believe you are on the wrong track.
Much better would be to use some sort of placeholder, and replace that by your desired database value when rendering that menu.
Something like this:
// assume something like "... <img src="../images/employees/%{image}"> ..."
// in the menu column, where '%{image}' serves as a placeholder
echo str_replace('%{image}', $row['image'], $row['menu']);
You need to echo your tag or echo your data inside the src attribute to use the value return by $row_UserDetails['Image'].
If you have done so, this may be the problem.
You have a issue with cocatenation. Your code should be like this
<?php echo "<img src='../images/employees/".$row_UserDetails['Image']."' alt='test'/>"; ?>
This article mat help you : https://teamtreehouse.com/community/why-do-we-use-concatenation-when-outputting-this-img-tag-with-php
Best regards !
I have the following html code:
<div class="media row-fluid">
<div class="span3">
<div class="widget">
<div class="well">
<div class="view">
<img src="img/demo/media/1.png" alt="" />
</div>
<div class="item-info">
Title 1
<p>Info.</p>
<p class="item-buttons">
<i class="icon-pencil"></i>
<i class="icon-trash"></i>
</p>
</div>
</div>
</div>
<div class="widget">
<div class="well">
<div class="view">
<img src="img/demo/media/2.png" alt="" />
</div>
<div class="item-info">
This is another title
<p>Some info and details go here.</p>
<p class="item-buttons">
<i class="icon-pencil"></i>
<i class="icon-trash"></i>
</p>
</div>
</div>
</div>
</div>
Which basically alternates between a span class with the widget class, and then the widget class without the span3 class.
What I wanted to know was if there was a way to have php "echo" or populate the details for and details under the "item-info" class. Would I need to use a foreach statement to get this done? I would be storing the information in a mysql database, and while I can get it to fill in the info one by one (repeatedly entering the and echoing out each image and item title) it's not practical when the content needed to be displayed is over 15 different items. I'm not well versed in foreach statements so I could definitely use some help on it.
If someone could help me perhaps structure a php script so that it can automatically output the html based on the number individual items in the database, that'd be greatly appreciated!
I'm wondering if the html + php (not including the foreach) would look like this:
<div class="span3">
<div class="widget">
<div class="well">
<div class="view">
<img src="img/<? $file ?>" alt="" />
</div>
<div class="item-info">
<?$title?>
<p>Info.</p>
<p class="item-buttons">
<i class="icon-pencil"></i>
<i class="icon-trash"></i>
</p>
</div>
</div>
</div>
EDIT:
I wanted to add some more information. The items populated would be based on a type of subscription - which will be managed by a group id.
I was initially going to use <? (if $_SESSION['group_id']==1)>
echo <div class="item-info">
$title
<p>$info</p>
</div>
so that only the subscribed items would populate. But, I would need it to iterate through all the items for group1 table and list it. Currently I know that I can do
<? (if $_SESSION['group_id']==1)
while ($row=mysql_fetch_assoc($sqlItem))
{
$itemInfo = $row['info'];
$image = $row['image'];
$title = $row['title'];
$url = $row['url'];
};
>
$sqlItem for now can only be assigned one thing (manually - as in: $sqlItem = '123'), unless I iterate through which is what I'm trying to figure out.
Just read that 'mysql_fetch_assoc' is being depreciated with 5.5, here is the new way and looks better, easier I think.. Hope this helps, was updated today.
I hope this helps http://php.net/manual/en/mysqli-stmt.fetch.php
replace the printf with echo '//then your html stuff
This will iterate through the rows in your database until their are no more matching records.
shouldn't a while be enough? It depends on the structure of your database and website (we didn't need so much HTML I think. Some more PHP maybe). Hope this helps.
I have this PHP function, which I use to display my userdata. If run once it will basicly display a div with the userdata in it.
function listings($fornavn, $efternavn, $email, $adresse, $tlf, $postnr, $city, $fodselsdag, $brugerid,$bartype,$idbar)
{
?>
<div class="container">
<span class="records" id="id-<?php echo $brugerid;?>">
<div class="customer bidDiv clearfix">
<?php if ($bartype=='temp_vip') { ?>
<ul>
<li>
<span class="actionSpan" id="<?php echo $brugerid;?>" value="<?php echo $idbar;?>">
<a class="edit-opt" id="godkend" href="#">GODKEND</a>
</span>
</li>
<li>
<span class="actionSpan" id="delete-<?php echo $brugerid;?>" value="<?php echo $bartype;?>">
<a class="delete-opt" id="delete" href="#">Afvis</a>
</span>
</li>
<?php }else{ ?>
<ul>
<li>
<span class="actionSpan" id="delete-<?php echo $brugerid;?>" value="<?php echo $bartype;?>">
<a class="delete-opt" id="delete" href="#">Slet</a>
</span>
</li>
<li>
<a class="edit-opt" href="editform.php?id=<?php echo $brugerid."&bartype=".$bartype;?>" rel="facebox">Rediger</a>
</li>
<?php if ($bartype =='vip'){?>
<li>
<a class="print-opt" href="print.php?id=<?php echo $brugerid;?>" rel="facebox">Print</a>
</li>
<?php }else{
// Dont render vip link
}}?>
</ul>
<p class="contact-data">
<?php echo $email;?><br>
Tlf.: <?php echo $tlf;?>
</p>
<div class="base-data">
<h4><?php echo ucwords($fornavn)." ".ucwords($efternavn);?></h4>
<p>Fødselsdag <?php echo $fodselsdag;?></p>
<address><?php echo ucwords($adresse) ." ". $postnr ." ". ucwords($city);?></address>
</div>
</div><!-- end customer -->
</div>
</span>
<?php
I mostly use this function in a loop, to display all users from my MySQL database, who has a specific relation to something else.
This is all working great. But currently it would pull all results at once. Right now I only have 5 members, so its no problem, but say I get 800 or 2000 members, the list of users would get very long.
So I want to implement some kind of limit on how many users it displays, but still be able to browse through all the users. Many sites use something like pages and split up the results that way. Links like:
[1] [2] [3] [Last page>>]
I can't figure out how to start doing this? How would I proceed?
Also looked into this Jquery plugin:
http://andersonferminiano.com/jqueryscrollpagination/
But it keeps reloading my PHP file which results in the Loops being restarted, and it simply displays the results over and over again endlessly.
Can someone help me with the logic behind creating this? Or better, point me in a direction where I could use the jquery pluging above - where it only loads the loop one time, and renders the results as I scroll.
Any help or pointers will be greatly appreciated!
Thanks
Jquery pagination by default will only break-up the given table into several tabs on the page, that is all the data is actually loaded. There are some that let you do an AJAX request to fetch the next page, to work in conjunction with your PHP code.
Ideally, you want to limit the query using LIMIT (number of rows in the result object) and OFFSET (Start from row X ) which will give you number of records starting from your offset row up until the limit, and then use logic on the PHP side to determine the maximum number of pages.
$page = $_POST["page"] * 25; // make sure that page 0 is the starting page or calculate
$SQL = "SomeQueryText LIMIT 25 OFFSET '$page'" ;
$result = query ($SQL);