I am connected to the database, a page has lots of content so I'll only share the part that doesnt return a value in php, but it returns a value in MySQL
Here is the code;
$query = "SELECT firstname FROM users WHERE id = '17'";
$query_run = mysql_query($query);
$row = mysql_fetch_row($query_run);
echo $row[0];
Changing the code I shared first to this, solved the problem. Thanks to anyone who tried to help.
$query = "SELECT firstname FROM users WHERE id = '17'";
$query_run = mysqli_query($conn, $query);
$row = mysqli_fetch_row($query_run);
echo $row[0];
And made bit changes to the connect.inc.php which I also shared in comment.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "notsitesi";
$conn = new mysqli($servername, $username, $password, $dbname);
if($conn->connect_error) {
die("Connection failed");
}
?>
Related
I'm trying to echo the query made to my database. So far the code runs but nothing happens. I'm very new at coding so help or an explanation about what I'm doing wrong would be awesome! Thanks a lot!!!
<?php
$servername = "servername";
// REPLACE with your Database name
$dbname = "dbsname";
// REPLACE with Database user
$username = "username";
// REPLACE with Database user password
$password = "password";
$ID = 1;
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
echo "Error de conexion.";
}
$sql = "SELECT * FROM registro WHERE ID = '1' ";
$result = $conn->query($sql);
while ($data = $result->fetch_assoc()){
$sensor_data[] = $data;
}
echo $data;
$conn->close();
Thanks everyone who replied, #AbraCadaver was right, I can't echo an array so I used print_r() instead and printed the variable $data. Now the php file prints the array with the query results. Thank you all!
$sql = "SELECT * FROM registro WHERE ID = $ID";
while ($data = $result->fetch_assoc()){
$sensor_data[] = $data;
print_r($data);
}
I have a test MySQL with a field which is unique as e-mailadres. Now I want to check before adding a record if the email already exists. I have the following code
<?php
$servername = "localhost";
$username = "xxxxxx";
$password = "xxxxxx";
$dbname = "xxxxxxx";
$servername = "localhost";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//test met voorbeeld
$emailtest="adres#me.com ";
$sql = "SELECT * FROM salonkeuze WHERE emailadres='$emailtest' ";
$data = mysqli_query($connect, $sql);
$row = mysqli_fetch_assoc($data);
$id = $row['voorkeur_id'];
Echo $id;
?>
When I use an email address which is in de table. I don’t get any output. What do I wrong?
Here $emailtest="adres#me.com "; you can extra space after com, try to remove it -> $emailtest="adres#me.com";
Firstly, you use wrong variable in this line:
$data = mysqli_query($connect, $sql);
Correct:
$data = mysqli_query($conn, $sql);
And it's optional, but it will be good, if you use trim() function for your $email. Because if your data comes from inputs and if $email have spaces, you can't show correct results:
$emailtest = "adres#me.com ";
$emailtest = trim($emailtest);
$sql = "SELECT * FROM salonkeuze WHERE emailadres='$emailtest' ";
<?php
$uname = $_POST["username"]$uname =
stripslashes(trim($uname));
$pward = $_POST["password"];
$pward = stripcslashes(trim($pward));
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "index_data";
$conn = new mysqli($servername, $username,
$password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM 00_user_details WHERE
username = '$uname' ";
$result = mysqli_query($conn, $sql);
$rows = mysqli_fetch_row($result);
echo "_".$rows["first_name"]."_";
?>
I am a bigger.
It's my code but it outputs nothing .
Any if you respected sir/ma'm help me kindly...
Please put semicolon after $_POST["username"] and use mysqli_fetch_assoc in place of mysqli_fetch_row.
mysqli_fetch_assoc return a row as an associative array where the column names will be the keys storing corresponding value.
Please find actual code as below:
$uname = $_POST["username"];
$uname = stripslashes(trim($uname));
$pward = $_POST["password"];
$pward = stripcslashes(trim($pward));
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "index_data";
$conn = new mysqli($servername, $username,$password, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM 00_user_details WHERE username = '$uname'";
$result = mysqli_query($conn, $sql);
$rows = mysqli_fetch_assoc($result);
echo "_".$rows["first_name"]."_";
This program is meant to delete a record when given the id.
php:
if ($_GET['type']=="file"){
$servername = "localhost";
$username = "****";
$password = "****";
$dbname = "****";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (mysqli_connect_error($conn)) {
die("Connection failed: " . mysqli_connect_error($conn));
}
$sql = "SELECT id,user, FROM CreationsAndFiles WHERE id =".$_GET['id']." LIMIT 1";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($result);
if ($row['user'] == $login_session){
$sql = "DELETE FROM CreationsAndFiles WHERE id=".$_GET['id'];
if(mysqli_query($conn, $sql)){echo "deleted";}
}
mysqli_close($conn);
//header("location: index.php?page=CreationsAndFiles");
}
the header is type=file&id=9
there is a record where id=9
It for no apparent reason will not work.
Your SQL syntax is wrong;
SELECT id,user, FROM CreationsAndFiles...
^ extra comma
should be simply
SELECT id,user FROM CreationsAndFiles...
You may want to sanitize your input though, for example simply entering type=file&id=id will most likely do bad things.
I've created an app were you can register as a user. You can sign up and then you're in the database "myAppDataBase" in "firsttable". A second table contains a list of lets say other important users that I manually created in the PHPmyAdmin-Website/"App". This table is called "secondtable".
My code to get the data is as follows:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "mydatabas";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}else{
//Print ("successfully connected");
}
$query = "SELECT * FROM firsttable";
$result = mysqli_query($conn, $query) or die("Error: " . mysqli_error($query));
$num = mysqli_num_rows($result);
$rows = array();
while ($r = mysqli_fetch_assoc($result))
{
$rows[] = $r;
Print ("sf");
}
Print json_encode($rows);
mysqli_close($conn);
?>
The only thing i changed was this line: THIS WORKS
$query = "SELECT * FROM firsttable";
But when I change it to this it won't work anymore.
$query = "SELECT * FROM secondtable";
Any help?
Change this:
mysqli_error($query)
With this:
mysqli_error($conn) // with your connection
Explanation:
mysqli_error() function needs connection link identifier not your query as param.
Mysqli_error PHP Manual
I SOLVED IT! Somehow, my second wasn't encoded the right way. I simply added this coder and it worked:
mysqli_set_charset($conn, 'utf8mb4');
Thanks for all you help though. ;)