Our html/php code shows the name from the database but the rest of them doesn't show. Although we used the same syntax for every part but we changed the data name same with the database column name.
Why is that so?
Here is our code:
<?php
$servername = "localhost";
$username = "";
$password = "1234";
$dbname = "straypaws";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Dog_name, Dog_loc, Dog_desc1, Dog_desc2, Dog_info1 FROM dog_info";
$result = $conn->query($sql);
?>
<div class="no-name-C61RwL helveticaneue-regular-normal-black-48px"><?php
if($row = mysqli_fetch_array($result)) {
echo $row["Dog_name"];
} ?>
</div>
<div class="dog-profile-C61RwL helveticaneue-regular-normal-black-72px">DOG PROFILE</div>
<img class="group-49-C61RwL animate-enter7" src="img/group-49#2x.svg" />
<div class="group-51-C61RwL">
<div class="group-49-Hjsg7h">
<div class="burnham-park-baguio-oKkELk helvetica55roman-regular-normal-black-24px"><?php
if($row = mysqli_fetch_array($result)) {
echo $row["Dog_loc"];
} ?>
You can't fetch same result multiple times.
Just store you fesult in array and then show it trought foreach
$dogs = [];
while($row = $result->fetch_assoc){
$dogs[] = $row;
}
And then in you html:
foreach($dogs as $dog){
echo $dog['Dog_name'];
}
Related
There is a drop-down list on the PHP website that contains names taken from the database,
after selecting a name from the list, e.g. "Aprilia", I would like all records to be displayed from the database
where mark = Aprilia
I know I should add in the code below just in a SELECT WHERE x = y query
but I just don't know how to do it;
it should look like this in my opinion:
$result = mysqli_query($conn, "SELECT * FROM motorcycles WHERE mark = X");
And I just can't find this X (X should be the user pick from the dropdown list)
How to write it?
Photos:
https://imgur.com/a/PMu4At7
<?php
require_once 'header.php';
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "projectinz";
// Create connection
//$conn = new mysqli($servername, $username, $password, $dbname);
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>
<select name="mark" id="mark">
<?php
$query = $conn->query("SELECT mark FROM motocykle");
while($kategoria = mysqli_fetch_array($query))
{
echo '<option>'.$kategoria['mark'].'</option>';
}
?>
</select>
<?php
$wynik = mysqli_query($conn,"SELECT * FROM motocykle");
while($row = mysqli_fetch_array($wynik))
{
echo "<br>".$row['mark']." ".$row['model']." ".$row['capacity']." ".$row['power']."<br>";
}
mysqli_close($conn);
?>
</body>
</html>
EDIT !!!
find1.php
<?php
require_once 'header.php';
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "projectinz";
// Create connection
//$conn = new mysqli($servername, $username, $password, $dbname);
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>
<form action="../includes/find.inc.php" method="post">
<select name="mark" id="mark">
<?php
$query = $conn->query("SELECT mark FROM motocykle");
while ($kategoria = mysqli_fetch_array($query)) {
echo '<option value="id">'.$kategoria['mark'].'</option>';
}
?>
</select>
<button type="submit" name="findmoto">Find</button>
</form>
<?php
$wynik = mysqli_query($conn,"SELECT * FROM motocykle ");
while ($row = mysqli_fetch_array($wynik)) {
echo "<br>".$row['mark']." ".$row['model']." ".$row['capacity']." ".$row['power']."<br>";
}
mysqli_close($conn);
?>
</body>
</html>
find.inc.php
<?php
session_start();
if (isset($_POST['findmoto'])) {
require 'dbh.inc.php';
$id = $_POST["id"];
$marka = $_POST["mark"];
echo $id;
echo $marka;
$display = "SELECT * FROM motocykle WHERE id='$id';";
$run = mysqli_query($conn, $display);
if ($run) {
echo $display;
} else {
echo "not";
}
}
BUT:
https://imgur.com/a/RA5wuus
Where is the problem?
<option value="1">Name</option> has attribute value witch is sent to POST when form is submitted. You need to set it with probably ID from table and then after POST you filter by that ID in WHERE part of your SQL.
First change
<select name="mark" id="mark">
<?php
$query = $conn->query("SELECT id, mark FROM motocykle");
while($kategoria = mysqli_fetch_array($query))
{
echo '<option value="'.$kategoria['id'].'">'.$kategoria['mark'].'</option>';
}
?>
</select>
Second change
if(isset($_POST['findmoto'])) {
require 'dbh.inc.php';
$selectedId = intval($_POST["mark"]);
echo $selectedId;
$display = "SELECT * FROM motocykle WHERE id='$selectedId';";
$run = mysqli_query($conn, $display);
if($run)
{
echo $display;
}
else
{
echo "not";
}
}
I made a simple HTML web page with a list of emplyees (only two atm).
<form method="post" action = "del.php">
<table border = "1">
<tr>
<th>Employee Name</th>
</tr>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "lavoratori";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT nome, id FROM operai";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo"<tr>";
echo'<td><input type = "checkbox" name = checkbox[]" value = '.$row['id']."<td>".$row['nome']."</td>";
echo"</tr>";
}
echo"</table>";
}
$conn->close();
?>
<input type = "submit" name = "delete" id = "delete" value = "Delete Records">
</form>
This is del.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "lavoratori";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['delete'])){
$chkarr = $_POST['checkbox'];
foreach($chkarr as $id){
$sql = "DELETE FROM operai WHERE id = .$id.";
$result = $conn->query($sql);
}
header("Location: /test_purpose/home.php");
}
$conn->close();
?>
Can you guys tell me what's going on? I'm new to PHP.
What i'm trying to do is to delete a specific a row from selecting with a checkbox. And it does not work, it redirects me back to the main page, without deleting anything obviously.
This line is wrong:
$sql = "DELETE FROM operai WHERE id = .$id.";
Replace with:
$sql = "DELETE FROM operai WHERE id = {$id}";
You also need to consider SQL Injection: https://stackoverflow.com/a/60496/1403785
I have been trying to echo out database data through a while loop now for awhile but it doesn't work, I can't find where the issue is. I have tried echoing out data manually and that works just fine.
<?php $results = mysqli_query($con,"SELECT * FROM guestbook ORDER BY id DESC"); ?>
<?php while ($row = mysqli_fetch_assoc($results)) : ?>
<li>
<?php echo $row['message']; ?>
</li>
<?php endwhile ?>
First of all make sure you are connected to the database. I don't know if you omitted that on purpose or not. Also, check if the connection is established.
<?php
//Insert your server info here
$servername = "servername";
$username = "root";
$password = "root";
$dbname = "test_database";
// Create and check your connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM guestbook ORDER BY id DESC";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo '<li>' . row["message"] . '</li>';
}
} else {
// Your query produced 0 results
echo "Empty result!";
}
// Remember to close the connection
mysqli_close($conn);
?>
What I wan't to do is create buttons that are automatically generated from the database. So when I add a new record in the database the button is created Is this possible with a loop? So yes how do I create the button.
This is what I have so far:
<?php
$servername = "localhost";
$username = "root";
$password = "Iamthebest1009";
$dbname = "dktp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM theme";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "". $row["theme_name"]. "<br>";
}
} else {
echo "no results";
}
$conn->close();
?>
Yes it is possible. you need to echo html
<?php
$servername = "localhost";
$username = "root";
$password = "Iamthebest1009";
$dbname = "dktp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM theme";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$your_url ="https://www.google.com";
echo "". $row["theme_name"]. "<br>";
echo '<input type="button" name="' . $row["theme_name"]. '" value="'. $row["theme_name"].'">';
}
} else {
echo "no results";
}
$conn->close();
?>
I have a database which has a table called 'propImages' and there are two columns.- 'pid' and 'location'.
And i have data in the database where multiple images can contained by single pid.
image contains database data
now i want to retrieve images from database according to given pid. there can be more than one image.
All i know it there should be an iteration to retrieve images.
I want to display images in HTML .
can you please show me the way to do it in php?
Thanks in advance guys
This may help you
<?php
include 'inc/database.php';
$conn = new mysqli($servername, $username, $password, $database);
$propid = $_GET['propid'];
$sql = "SELECT * FROM propImages WHERE propid='" . $propid . "';";
$result = $conn->query($sql);
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<img src=" . $row['image'] . ">";
}
}
else {
echo "No results";
}
?>
in the inc/database.php :
<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "database";
?>
To see how it works try visiting : file.php?propid=22
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "databasename";
// Create connection
$con = mysqli_connect($servername, $username, $password, $dbname);
//create sql
$sql = "SELECT * FROM `propImages` where pid='$YOUR_PID'";
$result = mysqli_query($con, $sql);
$row = mysqli_num_rows($result);
//retrive data print here
if($row > 0){
while($col = mysqli_fetch_assoc($result))
{
echo $col['location'];
}
} else {
echo 'no result found.';
}
?>
wish it helps