This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Reference - What does this error mean in PHP?
(38 answers)
Closed 3 years ago.
I get the error message: Connection failed: Unknown database 'first'
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "first";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name, age FROM table";
$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["name"]. "age: " . $row["age"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The database I'd like to select is there, but I get the error over and over
Related
This question already has answers here:
MySQL Error: : 'Access denied for user 'root'#'localhost'
(28 answers)
Closed 3 years ago.
I want to display data from a database from instead of showing the data in html it tells me "Connection failed: Access denied for user ''#'localhost' (using password: NO)"
I've tried searching the problem online and when other people had the same issue it was the case where they either had 'root'#'localhost' or 'user'#'localhost'. But in my case it doesn't even show any kind of username, it's blank.
<?php
$servername = "localhost";
$username = "john";
$password = "johndoe";
$dbname = "login";
// Create connection
$conn = new mysqli($localhost, $john, $johndoe, $login);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT firstname, lastname FROM cards";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["id"]. " - Name: ". $row["firstname"]. " " . $row["lastname"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Where did you declare these variables?
$conn = new mysqli($localhost, $john, $johndoe, $login);
In your code, the real variable names are: $servername, $username, $password, $dbname and that's what you should use to retrieve their values, which are "localhost, "john", "johndoe" and "login".
Try:
$conn = new mysqli($servername, $username, $password, $dbname);
This question already has an answer here:
PHP: 500 Error to error page
(1 answer)
Closed 5 years ago.
I want to select data from a MySQL database with PHP. Problem is, that when I try to echo out the $result, I get a 500 Error. When I leave out the echo $result;, I get a 200 OK return.
You guys gut any ideas?
Here's the PHP:
$q = $_GET['q'];
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "test";
//establish connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check Conncetion
if(!$conn) {
die("Connection failes: " . mysqli_connect_error());
}
$sql = "SELECT * FROM phptesting";
$result = mysqli_query($conn, $sql);
echo $result;
/*
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"]. "- Name: " . $row["first_name"]. " " . $row["last_name"]";
}
*/
mysqli_close($conn);
For your info, $q is just an integer with an id for testing it out.
echo is used to output primitive data type such as String, Integers.
$result holds the metadata of your SQL query result.
In ur code ur trying to echo the metadata. And This caused PHP fatal error.
This question already exists:
PHP's white screen of death [duplicate]
Closed 5 years ago.
I am always getting blank page when using Php and mysql connections
this is my php file
this is my html file
Try this code instead:
<?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();
?>
This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 6 years ago.
I am facing trouble printing the details of a username from MYSQL. The quotes in WHERE name = "xxx" is the cause.
This is the code:
$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);
}
$sql = "SELECT name, email FROM MyTable WHERE name=$name";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["name"]. " - email: " . $row["email"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
How do I replace the WHERE name = $name?
you are missing an ''
should be:
WHERE name = '$name'
notice the quotes
I think you shoudl use this notation (previously be sure of a proper sanitize of $name for preventing potential SQL injection)
"SELECT name, email FROM MyTable WHERE name='$name'";
This question already has an answer here:
Mysqli update throwing Call to a member function bind_param() error [duplicate]
(1 answer)
Closed 7 years ago.
After migrating from localhost to webserver I'm getting this error. Maybe it could be because of the PHP version on webserver (PHP 5.2). Any ideas?
<?php
$servername ="";
$username ="uran";
$password ="";
$dbname ="";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM answer WHERE topic_key='$id'";
$result = $conn->query($sql);
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<strong>Príspevok č.:</strong> " . $row["id"]. "<br>"." <strong>Napísal:</strong> "
. $row["name"]. "<br>". $row["topic"]."<br>" . $row["reg_date"]."<br>"." <br><br>";
}
$conn->close();
?>
I think that your query fails to that $conn->query($sql) will return FALSE (see mysqli::query).
Please add the following code to your code to get the error message:
if (!$result) {
printf("Error: %s\n", $conn->error);
}