How can i echo sql attributes to an HTML table using PHP - php

<div class="col-md-6">
<?php
include 'includes/connection.php';
$query = "SELECT * FROM customer";
$result = mysql_query($query);
while ($person = mysql_fetch_array($result)) {
//echo "<p>" . $person['customerID'] . "</p>";
//echo "<p>" . $person['firstName'] . "</p>";
//echo "<p>" . $person['lastName'] . "</p>";
//echo "<p>" . $person['address'] . "</p>";
$customerID = $person['customerID'];
$firstName = $person['firstName'];
$lastName = $person['lastName'];
$address = $person['address'];
echo $customerID; //A test echo that is working just fine. But i need to echo this to the table
}
?>
<table class="table table-striped">
<tr>
<th>Customer ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Address</th>
</tr>
</th>
<tr>
<td>
<? echo $customerID; ?>
</td>
<td>
<? echo $firstName; ?>
</td>
<td>
<? echo $lastName; ?>
</td>
<td>
<? echo $address; ?>
</td>
</tr>
</table>
</div>
</div>
I am learning PHP and i need help with this as soon as possible.
When i run the web page nothing is being shown to the table apart from the table headers. Please help for i am learning PHP. I am using xampp on which i created a database.

<? echo $customerID; ?>
should be
<?php echo $customerID; ?>

Firstly:
Please, don't use mysql_* functions, They are no longer maintained and are officially deprecated. Learn about prepared statements instead, and use PDO or MySQLi. This article will help you decide.
But to answer your question, you need to output your table data inside of your while loop like this:
<table class="table table-striped">
<tr>
<th>Customer ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Address</th>
</tr>
</th>
<?php while ($person = mysql_fetch_array($result)) { ?>
<tr>
<td>
<?php echo $person['customerID']; ?>
</td>
<td>
<?php echo $person['firstName']; ?>
</td>
<td>
<?php echo $person['lastName']; ?>
</td>
<td>
<?php echo $person['address']; ?>
</td>
</tr>
<?php } ?>
</table>

Related

How to fix a table causing "502: Bad Gateway" error in PHP

I am creating a simple database that can do basic CRUD operations (create, read, update, delete) using php. I am able to complete the create, and able to see the results if I directly query the mySQL DB in the back end. But, I am having trouble getting the table to display on the webpage. It is instead displaying a "Bad Gateway" error if I attempt to display the database entries.
I tried removing the reference to the table, specifically
<?php while ($row = mysqli_fetch_array($results)) { ?>
<tr>
<td><?php echo $row['name']; ?></td>...
and the web page on front end works fine. Albeit can only see the data if I query the backend.
<?php include('php_code.php'); ?>
...
...
...
<?php $results = mysqli_query($db, "SELECT * FROM info"); ?>
<table>
<thead>
<tr>
<th>Name</th>
<th>Address</th>
<th>City</th>
<th colspan="2">Action</th>
</tr>
</thead>
<?php while ($row = mysqli_fetch_array($results)) { ?>
<tr>
<td><?php echo $row['name']; ?></td>
<td><?php echo $row['address']; ?></td>
<td><?php echo $row['city']; ?></td>
<td>
<a href="test.php?edit=<?php echo $row['id']; ?>"
class="edit_btn" >Edit</a>
</td>
<td>
<a href="php_code.php?del=<?php echo $row['id']; ?>"
class="del_btn">Delete</a>
</td>
</tr>
<?php } ?>
</table>
<!--in php_code.php-->
//to retrieve records
$select_query = "SELECT * FROM info";
$result = mysqli_query($db, $select_query);
I should be able to see the table with data containing name, address and city. But I am getting a 502 error instead.
Try this
<?php
include('php_code.php');
$results = mysqli_query($db, "SELECT * FROM `info` ");
$return = <<<HTML
<table>
<thead>
<tr>
<th>Name</th>
<th>Address</th>
<th>City</th>
<th colspan="2">Action</th>
</tr>
</thead>
<tbody>
HTML;
while ($row = mysqli_fetch_array($results)) {
$return .= <<<HTML
<tr>
<td>{$row['name']}</td>
<td>{$row['address']}</td>
<td>{$row['city']}</td>
<td><a href="test.php?edit={$row['id']}" class="edit_btn" >Edit</a></td>
<td>Delete</td>
</tr>
HTML;
}
$return .= <<<HTML
</tbody>
</table>
HTML;
echo $return;
?>
your php_code.php should really only have the database config...
you are closing the php statements so you cannot retrieve the result of your query. Don't split php parts and just echo html like this
<?php
echo " <table>
<thead>
<tr>
<th>Name</th>
<th>Address</th>
<th>City</th>
<th colspan='2'>Action</th>
</tr>
</thead> ";
while ($row = mysqli_fetch_array($results)) {
echo "<tr>
<td>$row['name']</td>
<td>$row['address']</td>
<td> $row['city']</td>
<td>
<a href='test.php?edit=$row['id']'
class='edit_btn' >Edit</a>
</td>
<td>
<a href='php_code.php?del=$row['id']'
class='del_btn'>Delete</a>
</td>
</tr>";
}
echo "</table>";
?>

I want to delete multiple images from source folder through PHP but the code for deleting the multiple images from the folder is not working

<?php include("header.php"); ?>
<?php
if (#$_POST['delete']=="Delete"){
$count=count($_POST['delbx']);
for($i=0;$i<$count;$i++){
$delete = "DELETE FROM admin WHERE a_id='".$_POST['delbx'][$i]."'";
$resulty = mysqli_query($conn, $delete) or die(mysql_error());
$select_delete = "SELECT `a_image` FROM admin WHERE a_id='".$_POST['delbx'][$i]."'";
$resultrowdy = $conn->query($select_delete);
$rowdy = $resultrowdy->fetch_assoc();
$path="admin/".$rowdy['a_image'];
echo $path;
unlink($path);
echo '<script>window.location="view_user.php"</script>';
}
} ?>
<div class="table-responsive">
<table class="table">
<caption>All Users</caption>
<?php
$sql = "SELECT a_id, a_name, a_phone, a_password, a_role, a_mail, a_image FROM admin";
$result = $conn->query($sql);
if ($result->num_rows > 0) {?>
<thead>
<tr>
<th><form action="view_user.php" method="post"><input name="delete" type="submit" id="delete" value="Delete"></th><th>S. No.</th> <th>Name</th> <th>Phone No.</th> <th>Mail Id</th> <th>Role</th> <th>Password</th> <th>Image</th>
</tr>
</thead>
<?php
while($row = $result->fetch_assoc()) { ?>
<tbody>
<tr>
<th scope="row">
<?php echo $row["a_id"]; ?>
</th>
<td align="center" bgcolor="#FFFFFF">
<input name="delbx[]" type="checkbox" id="delbx[]" value="<?php echo $row["a_id"]; ?>" />
</td>
<td>
<?php echo $row["a_name"]; ?>
</td>
<td>
<?php echo $row["a_phone"]; ?>
</td>
<td>
<?php echo $row["a_mail"]; ?>
</td>
<td>
<?php echo $row["a_role"]; ?>
</td>
<td>
<?php echo $row["a_password"]; ?>
</td>
<td>
<img src="admin/<?php echo $row["a_image"]; ?>" width="60" height="40">
</td>
<th>
Edit
</th>
</tr>
</tbody>
<?php
}
} else {
echo "0 results";
}?>
</table>
</form>
</div>
<?php include("footer.php"); ?>
The code I mention is not deleting the multiple images from the source folder but deleting the multiple data from database whereas I am trying to delete images from the source folder along with data please help thanks in advance
One of the problem is you are deleting the row and trying to select image column from the deleted row.. dont use user supplied variables directly in your query
your code should be
for($i=0;$i<$count;$i++){
$select_delete = "SELECT `a_image` FROM admin WHERE a_id='".$_POST['delbx'][$i]."'";
$resultrowdy = $conn->query($select_delete);
$rowdy = $resultrowdy->fetch_assoc();
$delete = "DELETE FROM admin WHERE a_id='".$_POST['delbx'][$i]."'";
if(mysqli_query($conn, $delete)){
$path="admin/".$rowdy['a_image'];
unlink($path);
echo '<script>window.location="view_user.php"</script>';
}
}

Outputting Data into an HTML table using php/html

I am trying to output data onto my webpage from a database. I am able to so successfully, but in a very untidy way. I have tried to put the data in a table on the website with no success.
Below, data is retrieved from the db, but just echoed out crudely. It looks untidy but it outputs successfully onto the webpage
<?php
$query = "SELECT name, email, address FROM SHHowners";
$result = mysqli_query($conn, $query);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while ($row = mysqli_fetch_assoc($result)) {
echo "Name: " . $row["name"] . " - Email: " . $row["email"] . " - Address: " . $row["address"] . "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
I try to put the data into an HTML table(on the same page, by combining PHP and HTML with no success. How can put this data into an organised table successfully?
<div class="container">
<h2>Bordered Table</h2>
<p>The .table-bordered class adds borders to a table:</p>
<table class="table table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Address</th>
</tr>
</thead>
<tbody>
<tr>
<td><?php $row["name"] ?></td>
<td><?php $row["email"] ?></td>
<td><?php $row["address"]?></td>
</tr>
</tbody>
</table>
</div>
Below is roughly how I would like the data to be structured, how can achieve this?
Name | Email | Address
Jo |J#o.com|12 Street
Ben |B#e.com|23 street
try this:
<div class="container">
<h2>Bordered Table</h2>
<p>The .table-bordered class adds borders to a table:</p>
<table class="table table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Address</th>
</tr>
</thead>
<tbody>
<?php
while ($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><?php $row["name"] ?></td>
<td><?php $row["email"] ?></td>
<td><?php $row["address"]?></td>
</tr>
<?php } ?>
</tbody>
</table>
</div>
Try below code :
<div class="container">
<h2>Bordered Table</h2>
<p>The .table-bordered class adds borders to a table:</p>
<table class="table table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Address</th>
</tr>
</thead>
<tbody>
<tr>
<?php
$query = "SELECT name, email, address FROM SHHowners";
$result = mysqli_query($conn, $query);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while ($row = mysqli_fetch_assoc($result)) { ?>
<td><?php $row["name"]?></td>
<td><?php $row["email"]?></td>
<td><?php $row["address"]?></td>
<?php }
} ?>
</tr>
</tbody>
</table>
</div>
<?php
$query = "SELECT name, email, address FROM SHHowners";
$result = mysqli_query($conn, $query);
?>
<div class="container">
<h2>Bordered Table</h2>
<p>The .table-bordered class adds borders to a table:</p>
<table class="table table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Address</th>
</tr>
</thead> <tbody>
<?php
if (mysqli_num_rows($result) > 0) {
// output data of each row
while ($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><?php $row["name"] ?></td>
<td><?php $row["email"] ?></td>
<td><?php $row["address"]?></td>
</tr>
<?php }
} else {
echo "<tr><td colspan=3>0 results</td></tr>";
}
mysqli_close($conn);
?>
</tbody>
</table>
</div>
You can always print the results from the select like this:
while ($row = mysqli_fetch_assoc($result)) {
echo "<tr>
<td>".$row["name"]."</td>
<td>".$row["email"]."</td>
<td>".$row["address"] . "</td></tr>";
}
It's not a very clean way, but for each row should print another row, below the three table headers.
There isn't really a very neat way to do this beyond the other answers.
But what you can do is keep the HTML and PHP separate as much as possible in the same file.
The PHP at the top of the file can be as follows:
$query = "SELECT name, email, address FROM SHHowners";
$result = mysqli_query($conn, $query);
$htmlToDisplay = ''; //this variable will contain table rows
if (mysqli_num_rows($result) > 0) {
// output data of each row
while ($row = mysqli_fetch_assoc($result)) {
//create the HTML row - we'll use this later
$htmlToDisplay .= "<tr><td>".$row['name']."</td><td>". $row['email']."</td><td>".$row['address']."</td></tr>";
}
} else {
$htmlToDisplay = "<tr><td>0 results</td><tr>";
}
Then you can have your HTML after your closing PHP tag (?>) as follows:
<table class="table table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Address</th>
</tr>
</thead>
<tbody>
<!-- echo your php row(s) here in a slightly neater way since it's one line -->
<?php echo $htmlToDisplay; ?>
</tbody>
</table>
This will make the the different parts of the file (the PHP and HTML) more readable.
You could also look at using a PHP template engine like smarty.

Cannot show html tables in php

Below I'll post my code and I am going to explain my problem: The fact is that this code should show the value of every record of the query that I established, the problem is that the query is right (cause I checked it on phpMyAdmin) but the page don't show me the <tbody>, it show me nothing, I tried to put a single echo output with the <td> tag but nothing, can someone solve my problem and see the error that I have done?
<div class="container-fluid col-lg-6">
<table class="table">
<thead class="thead-default">
<tr>
<th>ID Pratica</th>
<th>ID Utente</th>
<th>Data Inizio</th>
<th>Descrizione</th>
<th>Data Stimata</th>
<th>Stato</th>
</tr>
</thead>
<tbody>
<?php
$query_search_all="SELECT * FROM `pratiche`";
$result = mysql_query($query_search_all);
while ($row=mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<td>".$row['id_pratica']."</td>";
echo "<td>".$row['id_utente']."</td>";
echo "<td>".$row['data_inizio']."</td>";
echo "<td>".$row['descrzione']."</td>";
echo "<td>".$row['data_stimata']."</td>";
echo "<td>".$row['stato']."</td>";
}
?>
</tbody>
</table>
</div>
You forgot the <tr> tag.
<div class="container-fluid col-lg-6">
<table class="table">
<thead class="thead-default">
<tr>
<th>ID Pratica</th>
<th>ID Utente</th>
<th>Data Inizio</th>
<th>Descrizione</th>
<th>Data Stimata</th>
<th>Stato</th>
</tr>
</thead>
<tbody>
<?php
$query_search_all="SELECT * FROM `pratiche`";
$result = mysql_query($query_search_all);
while ($row=mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<tr>";
echo "<td>".$row['id_pratica']."</td>";
echo "<td>".$row['id_utente']."</td>";
echo "<td>".$row['data_inizio']."</td>";
echo "<td>".$row['descrzione']."</td>";
echo "<td>".$row['data_stimata']."</td>";
echo "<td>".$row['stato']."</td>";
echo "</tr>";
}
?>
</tbody>
</table>
</div>
Can you please check the following code???
<?php
$query_search_all="SELECT * FROM pratiche ";
$result = mysql_query($query_search_all);
while ($row=mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<tr>";
echo "<td>".$row['id_pratica']."</td>";
echo "<td>".$row['id_utente']."</td>";
echo "<td>".$row['data_inizio']."</td>";
echo "<td>".$row['descrzione']."</td>";
echo "<td>".$row['data_stimata']."</td>";
echo "<td>".$row['stato']."</td>";
echo "</tr>";
}
?>

Why is my table being rendered empty?

I'm writing some php that renders a table from a database, this should be simple, but for some reason, it is rendering extra cells and all cells are empty. Here is my code:
<?php
$db= new PDO("mysql:host=localhost;dbname=mydb", "user", "password");
$query= $db->query("SELECT yarnId, yarnName, longYarnDescription, sale_price, cost, contents, onSale, yarnImage, activeFlag FROM yarn");
$result= $query->fetchAll();
?>
<table border="1">
<tr>
<th>yarnId</th>
<th>yarnName</th>
<th>description</th>
<th>sale price</th>
<th>cost</th>
<th>contents</th>
<th>onSale</th>
<th>yarnImage</th>
<th>activeFlag</th>
<th>edit</th>
</tr>
<?php for($r=0; $r<count($result); $r++){?>
<tr>
<?php for($c=0; $c<count($result[0]); $c++){?>
<td><?php echo $result[r][c];?></td>
<?php }?>
<td><button name=edit>edit</button></td>
</tr>
<?php }?>
</table>
If anyone can tell me why it's empty and why there are extra cells, it would be greatly appreciated.
The following code uses while()loop instead of for()
<table border="1">
<tr>
<th>yarnId</th>
<th>yarnName</th>
<th>description</th>
<th>sale price</th>
<th>cost</th>
<th>contents</th>
<th>onSale</th>
<th>yarnImage</th>
<th>activeFlag</th>
<th>edit</th>
</tr>
<?php
while($row = $query->fetch()) {
echo "<tr>";
for ($x=0;$x<= 8; $x++) {
echo "<td>" . $row[$x] . "</td>";
}
echo "<td><button name=\"edit\">edit</button></td></tr>\n";
}
?>
</table>

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