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.
Related
This question already has answers here:
How to include a PHP variable inside a MySQL statement
(5 answers)
Closed 2 years ago.
I want to select an email from the DB which needs to return the related Account_ID
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "connected accounts details";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT acc_id FROM users
WHERE
email = $user_email";
$result = $conn->query($sql);
echo "$result";
$conn->close();
This code returns nothing, just blank.
While using the same connection/db data is being inserted in db successfully.
Try this code
// Php Sql Select From DB....
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "connected accounts details";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT `acc_id` FROM `users` WHERE `email` = ?";
$result->$conn->prepare($sql);
$result->bind_param('s',$user_email);
$result->execute();
$data = $result->get_result();
// Fetch all
$data->fetch_all(MYSQLI_ASSOC);
print_r($data);
$mysqli -> close();
// #link http://php.net/manual/en/mysqli-result.fetch-all.php
$sql = "SELECT acc_id FROM users
WHERE
email LIKE '%$user_email%' ";
$result = mysqli_query($conn, $sql);
$row = $result->fetch_assoc();
$id = $row['acc_id'];
echo "$id";
$conn->close();
Got the wanted Record, trying this way.
Thanks everyone answering on this post.
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' ";
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");
}
?>
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. ;)
whats the solution for this problem Number of elements in type definition string doesn't match number of bind variables
session_start();
$regValue1 = $_GET['un'];
$regValue2 = $_GET['pass'];
$regValue3 = $_GET['fn'] ;
$regValue4 = $_GET['ln'];
$regValue5 = $_GET['age'] ;
$regValue6 = $_GET['sex'];
$regValue7 = $_GET['em'] ;
echo "hello: ".$regValue3.".";
$servername = "localhost";
username = "root";
$password = "b4sonic";
$dbname = "blog";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$stmt = $conn->prepare("INSERT INTO register(un,pass,fn,ln,age,sex,email) VALUES (?,?,?,?,?,?,?)");
bind_param("sss",$regValue1,
$regValue2,$regValue3,$regValue4,$regValue5,$regValue6,$regValue7);
$stmt->execute();
echo "New records created successfully";
$stmt->close();
$conn->close();
?>
I have this code in php , i am trying to insert data to mysql but i face this problem Number of elements in type definition string doesn't match number of bind variables
Your type string in your statement doesn't have enough type specifiers in it.
bind_param("sss",$regValue1,$regValue2,$regValue3,$regValue4,$regValue5,$regValue6,$regValue7);
says that you have type "sss" which only corresponds to 3 of the 7 variables you specified. You need to add types for the rest.
From the documentation:
var1
The number of variables and length of string types must match the parameters in the statement.
The commands you are using are for using PDO to connect to an sql database and you are using mysqli. I gave an example using mysqli below that should work. The other option would be to change the connection to PDO type rather than mysqli.
<?php
session_start();
$regValue1 = $_GET['un'];
$regValue2 = $_GET['pass'];
$regValue3 = $_GET['fn'] ;
$regValue4 = $_GET['ln'];
$regValue5 = $_GET['age'] ;
$regValue6 = $_GET['sex'];
$regValue7 = $_GET['em'] ;
echo "hello: ".$regValue3.".";
$servername = "localhost";
$username = "root";
$password = "b4sonic";
$dbname = "blog";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO register(un,pass,fn,ln,age,sex,email)".
"VALUES (`$regValue1`,`$regValue2`,`$regValue3`,`$regValue4`,`$regValue5`,`$regValue6`,`$regValue7`)";
$conn->query($sql);
echo $conn->affected_rows. " new records created successfully";
$conn->close();
?>