I wrote this code and I want to style it with css. I added style to it but same result appears can any body help please .
<p>
<phpcode>
<?php
$term = mysql_real_escape_string($_REQUEST['term']);
$servername = "localhost";
$username = "***********";
$password = "***********";
$dbname = "*************";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Install , power , RPM FROM SN WHERE serial = '$term'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "serial number: " . $term.'<br>';
echo "installation: " . $row["Install"].'<br>';
echo "power: " . $row["power"].'<br>';
echo "RPM: " . $row["RPM"].'<br>';
}
} else {
echo "no result";
}
$conn->close();
?>
</phpcode>
</p>
<style>
.results div:nth-of-type(1){
color:red;
background:blue;
font-size:3rem;
}
.results div:nth-of-type(2){
color:blue;
background:yellow;
font-size:2rem;
}
.results div:nth-of-type(3){
color:green;
background:orange;
font-size:1rem;
}
.results div:nth-of-type(4){
color:orange;
background:black;
font-size:0.5rem;
}
</style>
while( $row = $result->fetch_assoc() ) {
echo "
<div class='results'>
<div>serial number:{$term}</div>
<div>installation:{$row['install']}</div>
<div>power:{$row['power']}</div>
<div>rpm:{$row['rpm']}</div>
</div>";
}
Related
When I click on artikle It needs to diretct me to page.php where I display whole article.
Problem is im not sure how to with $_GET superglobal var properly take information. I have to get ID with $_GET.
I already have included database in my index.php where I displayed several articles.
Im sending id like this:
echo '<a href="clanak.php? class="form-field-textual" id='.$row['id'].'">';
//article.php
<?php
$servername = "localhost:3306";
$user = "root";
$pass = "";
$dbo = "projekt";
// Create connection
$conn = new mysqli($servername, $user, $pass, $dbo);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM projekt";
if ($conn->query($sql)===TRUE) {
$specID=$conn->insert_id;
echo "id: " . $row["id"]. " - Name: " .
$row["kategorija"]. " " . $row["naslov"]. "<br>";
} else {
echo "0 results";
}
$conn->close();
?>
$row_id = $row['id'];
echo 'some text';
<?php
$servername = "localhost:3306";
$user = "root";
$pass = "";
$dbo = "projekt";
// Create connection
$conn = new mysqli($servername, $user, $pass, $dbo);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$id = isset($_GET['id']) ? (int)$_GET['id'] : '';
$sql = "SELECT * FROM projekt WHERE id = $id";
if ($conn->query($sql)===TRUE) {
$specID=$conn->insert_id;
echo "id: " . $row["id"]. " - Name: " .
$row["kategorija"]. " " . $row["naslov"]. "<br>";
} else {
echo "0 results";
}
$conn->close();
Can anyone help please?
This is my code.I want to add an alt option for images but everything I try is throwing errors.
I have tried studying the php handbook and have tried copying code from other questions but so far no luck.
<?php
$servername = "localhost";
$username = "logosewe_5";
$password = "password";
$dbname = "logosewe_5";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM mk2";
$sql = "SELECT * FROM mk2 WHERE brand='2786'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<img src='$row['ming']'>';
}
} else {
echo "0 results";
}
$conn->close();
?>
You have a string concatenation problem :
echo '<img src="' . $row['ming'] . '">';
Will output : <img src="myimage.jpg">.
To add a alt attribute, you could do :
echo '<img src="' . $row['ming'] . '" alt="' . $row['name'] . '">';
Will output something like
<img src="myimage.jpg" alt="image name">
change your code like this:
<?php
$servername = "localhost";
$username = "logosewe_5";
$password = "password";
$dbname = "logosewe_5";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM mk2"
;
$sql = "SELECT * FROM mk2 WHERE brand='2786'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<img src="'.$row['ming'].'" alt="'.$altText.'">'; //change here
}
} else {
echo "0 results";
}
$conn->close();
?>
The problem was that you did not have an alt attribute in your echo statement and the second was your concatenation was a little off with misplaced quotes.
while($row = $result->fetch_assoc()) {
echo '<img src=' . $row['ming'] . 'alt="Your alt message">';
}
So I want to echo a specific value from my database, but it doesn't wok. I'd like to echo that value into a text form something like this:
<h6 class="brand-before" align="center"><small>Konyhakész fa</small></h6><br>
<input type="text" name="konyha" size="18" align="center" value="<?php echo $konyha; ?>" />
And this is how I connect to MySQL:
<?php
mysql_connect("localhost", "koristuzep", "***") or die("Kapcsolódás az adatbázishoz sikertelen.");
mysql_select_db("koristuzep")or die("Kapcsolódás az adatbázishoz sikertelen.");
$konyha = mysql_query("SELECT * FROM $arak WHERE ID=3")
?>
you should fetch the results first from $konyha as given in this example
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " .$row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
I have this php code to connect database it works fine.
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cv";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
mysqli_set_charset($conn,"utf8");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM services";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["service_name"]. " " . $row["service_desc"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
My question is: how to create function getServices() and give argument to show this results in another php file using foreach or while, like this:
<?php foreach ($results as $key=>$result) : ?>
<?php echo $result['service_name']; ?>
<?php endforeach; ?>
Okay. So you have to do the following things:
db.php :
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cv";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
mysqli_set_charset($conn,"utf8");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function yourFunctionName() {
$sql = "SELECT * FROM services";
$result = $conn->query($sql);
return $result;
}
$conn->close();
otherfile.php :
<?php
include 'db.php';
$data = yourFunctionName();
if ($data->num_rows > 0) {
// output data of each row
while($row = $data->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["service_name"]. " " . $row["service_desc"]. "<br>";
}
} else {
echo "0 results";
}
?>
Hope this works, haven't tested yet.
<?php
$servername = "localhost";
$username = "root";
$password = "william";
$dbname = "camping";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT fnavn,enavn,epost,tlf FROM knr ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["fnavn"]. " - Name: " . $row["enavn"]. " " . $row["epost"]. "ire: " . $row["tlf"]."<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Tried it, and it doesnt work, anyone who knows why? It's something wrong with the second if statement I think
You would need to join the tables:
SELECT * FROM produkt
LEFT JOIN produkttype ON (produkt.ptype = producttype.ptype)
WHERE utleid="No"