How to display information form a database about gender user? - php

In the database I have a gender table where male is M and female is F. Then I want to display full information about the gender of the user, so if M is to be male, and if F is to be female. The problem is that it always displays Female. Can someone see the code and give a hint where the error lies?
public function getData()
{
$id = $_GET['id'];
$sql = $this->database->connect()->prepare("SELECT user.id, first_name, last_name, avatar, birth_date, gender, age, school, work, city, martial_status, number_phone FROM user JOIN user_data ON user.id = user_data.user_id where user.id = :abc and user_id = :abc");
$sql->bindParam(':abc',$id, PDO::PARAM_INT);
$sql->execute();
if($sql->rowcount())
{
$row = $sql->fetch();
}
$this->userData = $row;
}
public function display(string $colName)
{
echo (isset($this->userData[$colName])) ? $this->userData[$colName] : '';
}
<main>
<div class='container'>
<div class="row">
<div class="col-xl-12 d-flex justify-content-center">
<div id="name"><span><?php echo $profil->display('first_name') ?></span> <span><?php echo $profil->display('last_name') ?></span></div>
</div>
</div>
<div class="row" id="margin">
<div class="col-xl-12 d-flex justify-content-center">
<div>Płeć: <?php if($profil->display('gender')=='M'){ echo "Mężczyzna";} else{ echo "Kobieta";} ?></div>
</div>
</div>
</div>
</main>

Related

PHP Output (PDO & CRUD)

I need help with my PHP script.
I have 3 Tables in my Database (its German so sorry if you dont understand ^^):
-firma (in english company)
-produkt (in english product)
-firmaprodukt (in english companyproduct)
If I click on a company, the product image should be output (from the "Produkt" database) and the appropriate URL (from the "firmaprodukt").
So far I've managed that when you click on a company, the ID is displayed (example: "localhost/TESTING//index?1=") but I'm struggling with the output.
That's my index.php script for the output:
<div class="col-xl-6">
<div id="login_content">
<div class="scrollP">
<?php
$fpresults = $fpCrud->getPicUrl($FirmenID);
?>
<div class="col-12 d-flex justify-content-center heading-div">
<h3> <?= $result['firmenname']?> </h3>
</div>
<table border="0">
<tbody>
<?php
foreach ($fpresults as $fpresult){
?>
<tr>
<div class="box customer-box" data-parent="#login_card" id="project_reg_box">
<img src="dashboard/TESTING/<?= $fpresult['ProduktLogo']?>" height="100%" >
</div>
</tr>
<?php
}
?>
</tbody>
</table>
</div>
</div>
</div>
And that's in my crud.php:
public function getPicUrl($FirmenID) {
$stmt = $this->conn->query(
"SELECT fp.FirmenID, f.firmenname, p.ProduktLogo, fp.url
FROM firma f
JOIN firmaprodukt fp ON fp.FirmenID = f.FirmenID
JOIN produkt p ON fp.ProduktID = p.ProduktID
WHERE fp.FirmenID = :FirmenID");
$stmt->execute(array(':FirmenID' => $FirmenID));
$data = $stmt->fetch();
return $data;
}
In my formprocess.php I don't have anything for that.
My solution:
index.php
<?php
if (isset($_GET['firmaprodukte'])) {
$FirmenID = (int) $_GET['firmaprodukte'];
$firmaprodukte = $fpCrud->getPicUrl($_GET['firmaprodukte']);
$firmenresults = $firmaCrud->getFirma($_GET['firmaprodukte']);
?>
<div class="col-xl-6">
<div class="scrollP">
<div id="login_content">
<div id="login_card" class="container">
<?php
$fpresults = $fpCrud->getPicUrl($FirmenID);
?>
<div class="col-12 d-flex justify-content-center heading-div">
<h3> <?= $firmenresults['firmenname']?> </h3>
</div>
<table border="0">
<tbody>
<?php
foreach ($fpresults as $fpresult){
?>
<tr>
<div class="box customer-box" data-parent="#login_card" id="project_reg_box">
<img src="dashboard/TESTING/<?= $fpresult['ProduktLogo']?>" height="80vh" max-width="200vh">
</div>
</tr>
<?php } ?>
</tbody>
</table>
</div>
</div>
</div>
</div>
<?php } ?>
Crud.php
public function getPicUrl($FirmenID) {
$stmt = $this->conn->prepare("SELECT fp.FirmenID, f.firmenname, p.ProduktLogo, fp.url FROM firma f JOIN firmaprodukt fp ON fp.FirmenID = f.FirmenID JOIN produkt p ON fp.ProduktID = p.ProduktID WHERE fp.FirmenID = :FirmenID");
$stmt->execute(array(':FirmenID' => $FirmenID));
$data = $stmt->fetchAll();
return $data;
}
formprocess.php
if(isset($_POST['firmaprodukte'])) {
$FirmenID = $_POST['FirmenID'];
if($crud->getPicUrl($FirmenID)) {
header('location: FirmaProdukt.php');
exit;
} else{
header('location: FirmaProdukt.php');
exit;
}
}

How to display the datas from two tables using search query?

I need to display datas from particular based on the results.
I have two tables movies_info => has all movies info & tv_shows_info => has TV shows info. I have search box if i enter the search it should search from both the table and display the datas but the condition in the results if it has movie_id, it should show movie details and if the result has tv_show_id it should show tv-show details.
as of now i have used basic search to fetch datas from one table and i have tried of using two but its not working.
<?php
// sql query for retrieving data from database
$sql_query = "SELECT * FROM `movies_info`";
$result = mysqli_query($connection, $sql_query);
// SQL Query for filter
if(isset($_POST['search_button']))
{
$value_to_search = $_POST['value_to_search'];
// search in all table columns
// using concat mysql function
$search_query = "SELECT * FROM `movies_info` WHERE CONCAT(`movie_name`, `movie_original_name`, `release_year`, `movie_genre`, `movie_country`, `movie_stars`, `movie_director`) LIKE '%".$value_to_search."%'";
//$search_query = "SELECT * FROM `movies_info`,`tv_shows_info` WHERE CONCAT(`movie_name`, `release_year`, `movie_genre`, `movie_country`, `tv_show_name`) LIKE '%".$value_to_search."%' GROUP BY `movie_id`";
//$search_query = "SELECT * FROM `movies_info` UNION ALL `tv_shows_info` WHERE CONCAT(`movie_name`, `release_year`, `movie_genre`, `movie_country`, `tv_show_name`) LIKE '%".$value_to_search."%'";
//$search_query2 = "SELECT * FROM `tv_shows_info` WHERE CONCAT(`tv_show_name`, `tv_show_start_year`, `tv_show_end_year`, `tv_show_genre`, `tv_show_country`) LIKE '%".$value_to_search."%'";
//$search_query .= $search_query2;
$search_result = filterTable($search_query);
}
else {
$search_query = "SELECT * FROM `movies_info`";
$search_result = filterTable($search_query);
//echo 'No Search Results Found';
}
?>
<!-- /w3l-medile-movies-grids -->
<div class="general-agileits-w3l">
<div class="w3l-medile-movies-grids">
<!-- /movie-browse-agile -->
<div class="movie-browse-agile">
<!--/browse-agile-w3ls -->
<div class="browse-agile-w3ls general-w3ls">
<div class="tittle-head">
<h4 class="latest-text">Search Results for : "<?php echo $value_to_search ?>"</h4>
<div class="container">
<div class="agileits-single-top">
<ol class="breadcrumb">
<li>Home</li>
<li class="active" style="text-transform:Capitalize;">Search Results </li>
</ol>
</div>
</div>
</div>
<div class="container">
<div class="browse-inner">
<?php
echo $search_result;
$rowcount = mysqli_num_rows($search_result);
for($i=1;$i<=$rowcount;$i++){
$row=mysqli_fetch_array($search_result);
?>
<div class="col-md-2 w3l-movie-gride-agile">
<a href="movie.php?movie_id=<?php echo $row['movie_id']; ?>" class="hvr-shutter-out-horizontal"><img src="<?php echo $row['movie_image']; ?>" title="<?php echo $row['movie_name']; ?>" alt=" " />
<div class="w3l-action-icon"><i class="fa fa-play-circle" aria-hidden="true"></i></div>
</a>
<div class="mid-1">
<div class="w3l-movie-text">
<h6><?php echo $row['movie_name']; ?></h6>
</div>
<div class="mid-2">
<p><?php echo $row['release_year']; ?></p>
<div class="block-stars">
<ul class="w3l-ratings">
<li>
<span>IMDB <i class="fa fa-star" aria-hidden="true"></i> <?php echo $row['movie_rating']; ?> </span>
</li>
</ul>
</div>
<div class="clearfix"></div>
</div>
</div>
<div class="ribben two">
<p>NEW</p>
</div>
</div> <?php } ?>
</div>
as of now i can get the values from one table.,
My exact need is it can be Movie or TV-Show but i should get the datas from both the table if it is a movie it should show some particular info and if thats a TV-show it should show someother info.
First, you need to use prepared statements which you can use even if you are using the LIKE function in MySQL. Then you need to add spaces between the column names to prevent the values from being blended together like "star warsstar wars". This will cause a movie like "star wars to be a result when a user searches for "ss" which would be inaccurate.
$search_result = array();
$search_query = "SELECT * FROM `movies_info` WHERE CONCAT(`movie_name`,' ', `movie_original_name`,' ',`release_year`,' ',`movie_genre`,' ',`movie_country`,' ',`movie_stars`,' ',`movie_director`) LIKE CONCAT('%',?,'%')";
if($stmt = $db->prepare($search_query)){
$stmt->bind_param('ii',$_SESSION['iidn_reference'],$o['parent_ref_id']);
if($stmt and $stmt->execute()){
$result = $stmt->get_result();
while($row = $result->fetch_array(MYSQLI_ASSOC)){
$search_result[] = $row;
}
}
}
After that, you could combine these tables with MySQL UNION but I strongly recommend just searching both tables and aliasing the table names to match.
SELECT 'movie' as `type`, `movie_id` as `id`, `movie_name` as `name`...
SELECT 'show' as `type`, `show_id` as `id`, `show_name` as `name`...

issue displaying multiple users info by id

I'm trying to get it where it will get the friends id from the friends table, and then get the friends information from the users table. I've tried using a foreach but had no luck.
Here's what I have right now where it's only echoing one friend and not the three that I have in the table. Any ideas on how I can fix this issue? Maybe I wasn't using the for each properly? Thank You In Advance!
<?php
//Gets users information from users.
$stmt = $DB_con->prepare('SELECT friendsid FROM friends WHERE userid='.$_SESSION['user']['id']);
$stmt->execute();
if($stmt->rowCount() > 0) {
$row=$stmt->fetch(PDO::FETCH_ASSOC);
extract($row);
} else {
$friendsid = $row['friendsid'];
}
$stmt = $DB_con->prepare('SELECT username,userprofile,status FROM users WHERE id='.$friendsid);
$stmt->execute();
if($stmt->rowCount() > 0) {
while($row=$stmt->fetch(PDO::FETCH_ASSOC)) {
extract($row);
?>
<div class="row">
<div class="col-sm-3">
<div class="well">
<h4><strong><?php echo $row['username'];?></strong></h4>
<img src="images/profile/<?php echo $userprofile;?>" class="img-circle" height="70" width="70" alt="Avatar">
</div>
</div>
<div class="col-sm-9">
<div class="well">
<h3 class="text-left"><?php echo $row['status'];?></h3>
</div>
</div>
</div>
<?php
}
} else {
?>
<?php
}
?>
Prepared statements use parameters and you should use a JOIN.
$stmt = $DB_con->prepare('SELECT u.* FROM users AS u JOIN friends AS f ON u.id = f.friendsid WHERE f.userid = :user_id');
$stmt->bindParam(':user_id', $_SESSION['user']['id']);
$stmt->execute();
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
// Do whatever you want with $row['id'], $row['username'], $row['userprofile']
}

My user ranking system php sql help me

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'];

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>

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