My user ranking system php sql help me - php

see screenshot development
as seen in the picture, where is the red circle should the number of user rankig appear. example: 1ts, 2th, 3rd...
I have tried everything but not me.
My table is users
ID Name username password Wins Loses
I leave my php code file profile.php
<?
require_once('values.inc.php');
require_once('func.inc.php');
if(!($_GET[user])) exit('Error.');
open_conn();
$res=mysql_query("SELECT ID, Name, username, Thumbnail, LinkText, LinkUrl, country FROM users WHERE Status='1' ORDER BY RAND() LIMIT 2");
while($row=mysql_fetch_assoc($res)) {
$username[] = $row[username];
$ID[] = $row[ID];
$Thumbnail[] = $row[Thumbnail];
$Name[] = $row[Name];
$LinkText[] = $row[LinkText];
$Model=mysql_query("SELECT ID, Name, username, Wins, Loses, Draws, Thumbnail, LinkText, LinkUrl, country, (Wins) / (Wins+Loses) AS WinPercent, (Wins+Loses+Draws) AS Matches, DATE_FORMAT(DateTime, '%d %M %Y') AS Date FROM users WHERE username='$_GET[user]' LIMIT 1");
$Model=mysql_fetch_assoc($Model);
$WinPercent=#round(($Model['Wins'] / ($Model['Wins']+$Model['Loses'])) * 100, 2);
$LosePercent=#round(($Model['Loses'] / ($Model['Wins']+$Model['Loses'])) * 100, 2);
if($Model['LinkText'] != '' && $Model['LinkUrl'] != '') {$Url="<br><br>$Model[LinkText]";}
if($row['LinkText'] != '' && $row['LinkUrl'] != '') {$Url[]="<br><br>$row[LinkText]<br><br>";}else{$Url[]='';}
}
close_conn();
?>
<?php if (isset($Model['username'])) { ?>
<div class="content-holder">
<div class="profile-holder">
<div class="profile-image">
<img src="images/uploads/<?php echo ucwords($Model['Thumbnail'])?>" class="photo">
</div>
<div id="photo-upload-holder">
<label for="pic" id="photo-upload-label">
<h1 class="txtShadow"><?php echo ucwords($Model['Name'])?></h1>
</label>
</div>
<!--<p class=\"userName\">#dada</p>-->
</div>
<div class="dashboard-Details overFlow">
<div class="leftColumn pull">
<div class="displayBlock"><br>
<i class="heart-icon"></i>
<span><?php echo ucwords($Model['Wins'])?></span>
</div><br>
<div class="displayBlock">
<i class="star-icon"></i>
<span><?php echo ($WinPercent)?>%</span>
</div>
</div>
<div class="rightColumn push">
<h2><?php echo ucwords($Model['username'])?></h2>
<?php if($Model['country']):?>
<h4> <?php echo $countrys; ?> <?php echo ucwords($Model['country'])?></h4>
<?php endif?>
<?php if($Model['LinkText']):?>
<h4> Instagram: #<?php echo ucwords($Model['LinkText'])?></h4>
<?php endif?>
</div>
<?php } else { ?>
<?php echo $profile14; ?>

use quotes here
$username[] = $row['username'];
$ID[] = $row['ID'];
$Thumbnail[] = $row['Thumbnail'];
$Name[] = $row['Name'];
$LinkText[] = $row['LinkText'];

Related

How to avoid blank first result when searching from databse

I have search webpage, but when the search is run, there is a blank first result.
<?php include "headnav.php";
$count = 0;
$sql = "SELECT * FROM lost_property";
if (!empty($_POST)) {
$name = mysqli_real_escape_string($dbconnect, htmlspecialchars($_POST['name']));
$item = mysqli_real_escape_string($dbconnect, htmlspecialchars($_POST['item']));
$area = mysqli_real_escape_string($dbconnect, htmlspecialchars($_POST['area']));
$sql = "
SELECT *
FROM lost_property
JOIN item
ON lost_property.itemID = item.itemID
JOIN area
ON lost_property.areaID = area.areaID
WHERE name LIKE '%$name%'
AND item LIKE '%$item%'
AND area LIKE '%$area%'
ORDER
BY lost_property.name ASC
";
$search_query = mysqli_query($dbconnect, $sql);
$count = mysqli_num_rows($search_query);
}
$result = $dbconnect->query($sql);
?>
<body>
<div class="form">
<h1>Search for lost property here:</h1>
<form action="" method="POST">
Name:
<input type="text" placeholder="Name" name="name">
Item type:
<select name="item" class="dropdown">
<option value="" disabled selected>Item</option>
<?php
$item_sql = "SELECT DISTINCT item FROM `lost_property`
JOIN item ON (lost_property.itemID = item.itemID)
ORDER BY item ASC
";
$item_query = mysqli_query($dbconnect, $item_sql);
$item_rs = mysqli_fetch_assoc($item_query);
do {
?>
<option value="<?php echo $item_rs['item']; ?>"><?php echo $item_rs['item']; ?></option>
<?php
} while ($item_rs = mysqli_fetch_assoc($item_query));
?>
</select>
Area (where it was found):
<select name="area" class="dropdown">
<option value="" disabled selected>Area</option>
<?php
$area_sql = "SELECT DISTINCT area FROM `lost_property`
JOIN area ON (lost_property.areaID = area.areaID)
ORDER BY area ASC
";
$area_query = mysqli_query($dbconnect, $area_sql);
$area_rs = mysqli_fetch_assoc($area_query);
do {
?>
<option value="<?php echo $area_rs['area']; ?>"><?php echo $area_rs['area']; ?></option>
<?php
} while ($area_rs = mysqli_fetch_assoc($area_query));
?>
</select>
<input type="submit" value="Search" name="btn">
</form>
</div>
<div class="gallery">
<h2>Search results:</h2>
<?php
//check for results. If there are none display error
if ($count < 1) {
?>
<div class="error">
<h1>No results were found.</h1>
</div>
<?php
} //end if
else {
do {
?>
<!-- display image and information from database and show in gallery -->
<div class="results">
<h3><?php echo $search_rs['name']; ?></h3>
<h3><?php echo $search_rs['item']; ?></h3>
<p><?php echo $search_rs['area']; ?></p>
</div>
<?php
} // end of do
while ($search_rs = mysqli_fetch_assoc($search_query));
} //end else
//if there are any display
?>
</div>
</table>
</body>
</html>
It's hard to see without any CSS, but no matter what is searched, there is always one result, but the h3 and p fields don't have any content. If there wasn't the no results error message it would pop up there too. What is causing this first result?
Use while (){}, if you use do instead it will run once first (credit to Magnus Eriksson).
It ends up like
else {
while ($search_rs = mysqli_fetch_assoc($search_query)) {
?>
<!-- display image and information from database and show in gallery -->
<div class="results">
<h3><?php echo $search_rs['name']; ?></h3>
<h3><?php echo $search_rs['item']; ?></h3>
<p><?php echo $search_rs['area']; ?></p>
</div>
<?php
} // end of do
} //end else
//if there are any display
?>
instead of
else {
do {
?>
<!-- display image and information from database and show in gallery -->
<div class="results">
<h3><?php echo $search_rs['name']; ?></h3>
<h3><?php echo $search_rs['item']; ?></h3>
<p><?php echo $search_rs['area']; ?></p>
</div>
<?php
} // end of do
while ($search_rs = mysqli_fetch_assoc($search_query));
} //end else
//if there are any display
?>

Liking system, how do I fix it?

I have the following script for showing posts and liking them, but if I like one post it likes all the posts on the page, I can't think of another way to do it, can anyone give me some advice?
<?php
if ($sort == 1){
$result = $conn->query("SELECT * FROM posts ORDER BY date DESC LIMIT 4 ");
}
elseif($sort == 2)
{
$result = $conn->query("SELECT * FROM posts WHERE date > NOW() - INTERVAL 24 HOUR ORDER BY likes DESC");
}
elseif($sort == 3)
{
$result = $conn->query("SELECT * FROM posts ORDER BY likes DESC");
}
if ($result->num_rows > 0) :
while($row = mysqli_fetch_assoc($result)) : ?>
<div class="card mb-4">
<img class="card-img-top" src="<?php echo $row['image1'] ?>" alt="Card image cap">
<div class="card-body">
<h2 class="card-title"><?php print title; ?></h2>
<p class="card-text"><?php print text; ?></p>
Read More →
</div>
<div class="card-footer text-muted">
Posted on <?php print $row['date'] ?> by
<?php print $row['author']; ?>
<?php
$id=$row['id'];
if($_POST['like']) {
$update = "UPDATE posts set `likes` = `likes`+1 where `id` ='$id'";
if ($conn->query($update) === TRUE) {
} else {
echo "Error updating record: " . $conn->error;
}
} ?>
<form action="" method="POST">
<button type = "submit" value = "like" name='like'style="font-size:24px"><?php echo $row['likes']; ?><i class="fa fa-thumbs-o-up"></i>
</form>
</div>
</div>
<?php endwhile; endif; ?>
Your while loop contains the update query so your code should be change like this.
in order to get the id to like you just need to use a hidden field to post that id like in this code
<?php
if($_POST['like']) {
$id=$POST['id'];
$update = "UPDATE posts set `likes` = `likes`+1 where `id` ='$id'";
if ($conn->query($update) === TRUE) {
} else {
echo "Error updating record: " . $conn->error;
}
} ?>
<?php
if ($sort == 1){
$result = $conn->query("SELECT * FROM posts ORDER BY date DESC LIMIT 4 ");
}
elseif($sort == 2)
{
$result = $conn->query("SELECT * FROM posts WHERE date > NOW() - INTERVAL 24 HOUR ORDER BY likes DESC");
}
elseif($sort == 3)
{
$result = $conn->query("SELECT * FROM posts ORDER BY likes DESC");
}
if ($result->num_rows > 0) :
while($row = mysqli_fetch_assoc($result)) : ?>
<div class="card mb-4">
<img class="card-img-top" src="<?php echo $row['image1'] ?>" alt="Card image cap">
<div class="card-body">
<h2 class="card-title"><?php print title; ?></h2>
<p class="card-text"><?php print text; ?></p>
Read More →
</div>
<div class="card-footer text-muted">
Posted on <?php print $row['date'] ?> by
<?php print $row['author']; ?>
<form action="" method="POST">
<input name="id" type="hidden" value="<?php echo $row['id']; ?>">
<button type = "submit" value = "like" name='like'style="font-size:24px"><?php echo $row['likes']; ?><i class="fa fa-thumbs-o-up"></i>
</form>
</div>
</div>
<?php endwhile; endif; ?>

How to select two rows each time from a table in SQL and display them in PHP?

I have a table with fields like id, date, heading, news. I want to display the fields date, heading and news in an Owl Carousel slider. Can anyone suggest how to pick two rows at a time from an SQL table to display two entries at a time in the Owl Carousel slider. I have used an SQL query like this:
<?php
$sql3 = "SELECT date, heading, news FROM news ORDER BY news.date LIMIT 0, 1";
$result3 = mysql_query($sql3) or die(mysql_error());
while($row3 = mysql_fetch_array($result3)) {
?>
<div class="item ">
<div class="l_blk">
<div class="news_container">
<div class="col-md-1 date_b"><p>Mar<br>09</p></div>
<div class="col-md-11 cont_b">
<p>
<span class="news_t">"<?php echo $row3['heading']; ?>"</span><br>
<?php echo $row3['news']; ?>
</p>
</div>
</div>
</div>
<div class="l_blk">
<div class="news_container">
<div class="col-md-1 date_b"><p>Mar<br>09</p></div>
<div class="col-md-11 cont_b">
<p>
<span class="news_t">"<?php echo $row3['heading']; ?>"</span><br>
<?php echo $row3['news']; ?>
</p>
</div>
</div>
</div>
</div>
<?php
}
?>
Can anyone suggest how to do this?
$sql3 = "SELECT date, heading, news FROM news ORDER BY news.date LIMIT 0, 1";
Don't you change the LIMIT 0, 2
Which means select 2 starting with the value at index 0?
Once you've done that, take out the duplicate code inside the loop
Try this pls.
<?php
$sql3 = "SELECT date, heading, news FROM news ORDER BY news.date"; // No limit
$result3= mysql_query($sql3) or die(mysql_error());
$counter = 0;
while($row3 = mysql_fetch_array($result3, MYSQL_ASSOC))
{
if($counter % 2 == 0){
echo '<div class="item ">'.PHP_EOL;
}
?>
<div class="l_blk">
<div class="news_container">
<div class="col-md-1 date_b"><?php $row3['date'] ?></div>
<div class="col-md-11 cont_b">
<p>
<span class="news_t">"<?php echo $row3['heading']; ?>"</span><br>
<?php echo $row3['news']; ?>
</p>
</div>
</div>
</div>
<?php
if($counter % 2 == 1){
echo '</div>'.PHP_EOL;
}
$counter++;
}
if($counter % 2 == 1){
echo '</div>'.PHP_EOL;
}
?>

How can I include an image that is tied to a user's database information?

I've been coding PHP for 2 weeks (it's not pretty) and I have not been able to find the answer to my question. I want an admin type user to be able to fill a form and post it to a page where base level users can view the content. I've gotten all of this to work like a charm, but my dilemma is to allow the admin user to include an image as well. Maybe I just don't know what to search for.
Here is the php code and the form for the admin user page:
<?php
error_reporting(E_ALL & ~E_NOTICE);
session_start();
include_once("connection.php");
if (isset($_SESSION['adminid'])) {
$adminid = $_SESSION['adminid'];
$adminemail = $_SESSION['adminemail'];
if ($_POST['submit']) {
$title = $_POST['title'];
$deadline = $_POST['deadline'];
$content = $_POST['content'];
$sql_blog = "INSERT INTO blog (title, deadline, content, logoname) VALUES ('$title', '$deadline', '$content', '$logoname')";
$query_blog = mysqli_query($dbcon, $sql_blog);
echo "<script>alert('Your inquiry has been posted')</script>";
}
} else {
header('Location: index.php');
die();
}
$sql = "SELECT adminid, adminemail, adminpassword, adminname FROM admin WHERE adminemail = '$adminemail' LIMIT 1";
$query = mysqli_query($dbcon, $sql);
if ($query) {
$row = mysqli_fetch_row($query);
$adminname = $row[3];
}
?>
and here is the code for the base level user page: (i commented out the image block where I want the admin's image to be shown.
<main>
<div class="container">
<div class="row topbuffpost">
<h1>business inquiries</h1>
<hr>
<?php
include_once('connection.php');
$sql = "SELECT * FROM blog ORDER BY id DESC";
$result = mysqli_query($dbcon, $sql);
while ($row = mysqli_fetch_array($result)) {
$title = $row['title'];
$content = $row['content'];
$date = strtotime($row['deadline']);
?>
<div class="col-md-4 col-lg-3">
<div class="card hoverable">
<!-- <div class="card-image">
<div class="view overlay hm-white-slight z-depth-1">
<img src="">
<a href="#">
<div class="mask waves-effect">
</div>
</a>
</div>
</div> -->
<div class="card-content">
<h5> <?php echo $title; ?> <br/> <h6>Deadline |<small> <?php echo date("j M, Y", $date); ?> </small> </h6></h5> <br/>
<p> <?php echo $content; ?> </p>
<div class="card-btn text-center">
Read more
<i class="fa fa-lightbulb-o"></i>&nbsp propose a plan
</div>
</div>
</div>
</div>
<?php
}
?>
</div>
</div>
</main>
All of this works perfectly, I just can't figure out how to have an image display in the same way as the title, deadline, and content. Youtube wont help either, too much outdated php + I haven't been coding long enough to really work things out on my own.
You can save all user images under a folder (let's call /images/user) and record the file name into database.
if ($_POST['submit']) {
$title = $_POST['title'];
$deadline = $_POST['deadline'];
$content = $_POST['content'];
$logoname = basename($_FILES["fileToUpload"]["logoname"]; // <-- Make sure your form is ready to submit an file
// Update below as per your need.
$target = 'images/users/' . $logoname;
move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target);
$sql_blog = "INSERT INTO blog (title, deadline, content, logoname) VALUES ('$title', '$deadline', '$content', '$logoname')";
$query_blog = mysqli_query($dbcon, $sql_blog);
echo "<script>alert('Your inquiry has been posted')</script>";
}
You can then display the image your page
<main>
<div class="container">
<div class="row topbuffpost">
<h1>business inquiries</h1>
<hr>
<?php
include_once('connection.php');
$sql = "SELECT * FROM blog ORDER BY id DESC";
$result = mysqli_query($dbcon, $sql);
while ($row = mysqli_fetch_array($result)) {
$title = $row['title'];
$content = $row['content'];
$date = strtotime($row['deadline']);
$logoname = 'images/user/' . $row['logoname'];
?>
<div class="col-md-4 col-lg-3">
<div class="card hoverable">
<div class="card-image">
<div class="view overlay hm-white-slight z-depth-1">
<img src="<?php echo $logoname; ?>">
<a href="#">
<div class="mask waves-effect">
</div>
</a>
</div>
</div>
<div class="card-content">
<h5> <?php echo $title; ?> <br/> <h6>Deadline |<small> <?php echo date("j M, Y", $date); ?> </small> </h6></h5> <br/>
<p> <?php echo $content; ?> </p>
<div class="card-btn text-center">
Read more
<i class="fa fa-lightbulb-o"></i>&nbsp propose a plan
</div>
</div>
</div>
</div>
<?php
}
?>
</div>
</div>
</main>

Recommended user video Like youtube

I am trying to make a youtube like main page. With the code below I want to make videos that are recommended for my users.
The following code shows only a user's video.
<?php $query = "SELECT
user.uid,
user.user_name,
user.user_avatar,
user_posts.uid_dk,
user_posts.post_id,
user_posts.post_name,
user_posts.post_info,
user_posts.post_time,
user_posts.post_ext,
user_posts.post_num,
user_posts.post_views
FROM user
JOIN user_posts
ON user_posts.uid_dk = user.uid
WHERE user_name='$user_name' LIMIT 5";
$run_query = mysql_query($query);
while($data=mysql_fetch_assoc($run_query)){
$post_name=$data['post_name'];
$post_time = $data['post_time'];
$post_views = $data['post_views'];
$post_numid = $data['post_num'];
$post_id = $data['post_id'];
$user_name = $data['user_name'];
$user_avatar = $data['user_avatar'];
?>
<div class="onerilent"><img src="<?php echo $user_avatar;?>"><?php echo $user_name ;?> Recommended for you</div>
<div class="onmnwrp">
<div class="onmn">
<div class="onmn_img"><img src="<?php echo $base_url.'user_uploads/'.$post_num;?>.png"></div>
<div class="onmg_tit"><?php echo $post_name;?></div>
<div class="onm_snm">gönderen: <?php echo $user_name;?></div>
<div class="onm_tim"><?php echo $post_views;?> views</div>
</div>
</div>
<?php } ?>
I want to show this section only one time.
<div class="onerilent"><img src="<?php echo $user_avatar;?>"><?php echo $user_name ;?> Recommended for you</div>
Anyone can help me in this regard ?
The easiest way would be with a counter, like this:
<?php
$query = "SELECT
user.uid,
user.user_name,
user.user_avatar,
user_posts.uid_dk,
user_posts.post_id,
user_posts.post_name,
user_posts.post_info,
user_posts.post_time,
user_posts.post_ext,
user_posts.post_num,
user_posts.post_views
FROM user
JOIN user_posts
ON user_posts.uid_dk = user.uid
WHERE user_name='$user_name' LIMIT 5";
$run_query = mysql_query($query);
$counter = 1;
while($data=mysql_fetch_assoc($run_query)){
$post_name=$data['post_name'];
$post_time = $data['post_time'];
$post_views = $data['post_views'];
$post_numid = $data['post_num'];
$post_id = $data['post_id'];
$user_name = $data['user_name'];
$user_avatar = $data['user_avatar'];
if($counter == 1){
$counter++;
echo '<div class="onerilent"><img src="'.$user_avatar.'">'.$user_name.' Recommended for you</div>';
}
?>
<div class="onmnwrp">
<div class="onmn">
<div class="onmn_img"><img src="<?php echo $base_url.'user_uploads/'.$post_num;?>.png"></div>
<div class="onmg_tit"><?php echo $post_name;?></div>
<div class="onm_snm">gönderen: <?php echo $user_name;?></div>
<div class="onm_tim"><?php echo $post_views;?> views</div>
</div>
</div>
<?php
}
?>
note the $counter variable is set to 1 before the loop, and inside the loop there is a condition to check if it is set to the value 1, and if it is, then it echo's your html and increments the $counter, so that it is no longer set to 1

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