Here's the code:
$coinno = $_POST["CoinNo"];
$week = $_POST["week"];
$payer = $_POST["payer"];
$payee = $_POST["payee"];
$servername = "localhost";
$username = "username"; // Edited from original
$password = "password";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO `Transactions`(`Week`,`Coin`,`Payer`,`Payee`)
VALUES ($week,$coinno,'$payer','$payee')";
if($conn->query(sql)===TRUE)
{
echo "Success";
}
else
{
echo "Error ".$sql."<br>".$conn->error;
}
$conn->close;
?>
And my error message:
Error INSERT INTO Transactions(Week,Coin,Payer,Payee) VALUE (1,2,'Bob','Carol')
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'sql' at line 1
I've tried using backticks, single quotes, etc. Even copying the generated query from phpMyAdmin did not help.
Another notable issue is that my Transactions table has a single row but when I try to select it I get 0 results. Could there possibly be a connection?
if($conn->query(sql)===TRUE)
-->
if($conn->query($sql)===TRUE)
Also, get into the habit of flipping the if around -- some functions return either FALSE or something useful. That is, they may never return TRUE. That is:
if($conn->query($sql) !== FALSE)
Related
I have looked at several examples on how to call a MySQL stored procedure from PHP but none have helped me. The stored procedure works when run inside PHPMyAdmin but I am having trouble calling it from the web.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$result = mysqli_query($conn,"CALL standings_build()");
if (mysqli_query($conn,$sql))
header('refresh:1; url=schedule_main_scores.php');
else
echo "failed";
?>
There's 2 problems here.
You're querying twice and using the wrong variable, being $sql instead of $result.
$result = mysqli_query($conn,"CALL standings_build()");
if (mysqli_query($conn,$sql))
^^^^^^^^^^^^ calling the query twice
^^^^ wrong variable, undefined
all that needs to be done is this:
if ($result)
and an else to handle the (possible) errors.
Error reporting and mysqli_error($conn) would have been your true friends.
http://php.net/manual/en/function.error-reporting.php
http://php.net/mysqli_error
Side note: You really should use proper bracing techniques though, such as:
if ($result){
echo "Success";
}
else {
echo "The query failed because of: " . mysqli_error($conn);
}
It helps during coding also and with an editor for pair matching.
Hello I am new at php and mysql and I don't know what is wrong.
I cant show the results from query and the connection with mysql is successfully connected.
I don't use wampserver I just install php,mysql and Apache separately.
Thanks in advance.
Code
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$servername = "localhost";
$username = "root";
$password = "";
// Create connection
$conn = mysqli_connect($servername, $username, $password);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
$sql="select * from `books`;";
$result=mysqli_query($conn,$sql);
if (!$result){
echo "query cannot execute";
};
?>
its only show me "query cannot execute"
You need to pass fourth parameter database name in mysqli_connect()
It would be
$conn = mysqli_connect($servername, $username, $password,"YOUR_DATABASE");
Read http://php.net/manual/en/mysqli.error.php to check error in query.
Read http://php.net/manual/en/mysqli-result.fetch-array.php
To fetch data from query result
I am new in PHP and would need some explanation. Here is a code where we connect to MySQL with PHP. Can you please explain me where is the statement that makes the connection? I can see only that we define what the value of $conn is, but does it mean execution as well? The other thing is: where do we create the database? I can see that we give the string "CREATE DATABASE myDB" as a value to $sql and we have an if statement, but does the expression ($conn->query($sql) === TRUE) also evaluated? It is strange for me, can somebody please explain it to me?! :) Thanks!
<?php
$servername = "localhost";
$username = "username";
$password = "password";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Create database
$sql = "CREATE DATABASE myDB";
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
} else {
echo "Error creating database: " . $conn->error;
}
$conn->close();
?>
Here is a simple explanation of which lines do what. If you would like to know specifically what the individual parts of these mean, then please say which ones so they can be further explained to you. Or the correct links pointed to.
I notice that you are using the W3Schools example, as an almost exact copy and paste. Have you installed MySQL on your machine and created a username and password?
<?php
$servername = "localhost"; // This is the location of your server running MySQL
$username = "username"; // This is the username for MySQL
$password = "password"; // This is the password for MySQL
// Create connection
$conn = new mysqli($servername, $username, $password); // This is where you create a connection
// Check connection
if ($conn->connect_error) { // This checks if the connection happened
die("Connection failed: " . $conn->connect_error); // and produces an error message if not
} // otherwise we move on
// Create database
$sql = "CREATE DATABASE myDB"; // This is the SQL query which is sent to the MySQL server
if ($conn->query($sql) === TRUE) { // When the if statement begins here, it executes the query and test if it returns true
echo "Database created successfully"; // If it returns true then here is the message is returns
}
else {
echo "Error creating database: " . $conn->error; // Or if there was error with the query this is returned
}
$conn->close(); // Close the connection when it is no longer in use
?>
Although, your question does not belong here (This place is to help with your coding issues), but I will give you a bit explanation.
PHP reads each line and EXECUTES It. the create connection part opens a new connection using the "new" object and save it a variable ($conn),
($conn->connect_error) checks if the connection was successful with connect_error property. if it was connected, continue, or else through and error and stop.
If connection was successful, then create the database based on connection opened in variable ($conn).
I have been developing a CRUD application using PHP & MySQL database.
I was succeeded by creating, displaying, updation parts. But I stuck at the deletion part of a row from a database table.
I tried my best solving all the PHP shown errors but now in final it is now showing a message which I wrote to echo in case of failure.
I request someone to please help me with this problem.
Thankyou in advance.
Code I wrote for deletion:
//include database connection
include 'db_connect.php';
//$mysqli->real_escape_string() function helps us prevent attacks such as SQL injection
$query = "DELETE
FROM `grocery`
WHERE `GrocerID` ='".$mysqli->real_escape_string($_GET['id'])."'
limit 0,1";
//execute query
if( $mysqli->query($query) ){
//if successful deletion
echo "User was deleted.";
}else{
//if there's a database problem
echo "Database Error: Unable to delete record.";
}
$mysqli->close();
?>
Code I wrote for delete link in display table:
//just preparing the delete link to delete the record
echo "<a href='delete.php?id={$GrocerID}'>Delete</a>";
Code I wrote for db config:
<?php
//set connection variables
$host = "localhost";
$username = "root";
$password = "secret";
$db_name = "crud"; //database name
//connect to mysql server
$mysqli = new mysqli($host, $username, $password, $db_name);
//check if any connection error was encountered
if(mysqli_connect_errno()) {
die("Connection failed: " . $conn->connect_error);
exit;
}
?>
I tried this and got working, can you update the code and see if this works?
$host = "localhost";
$username = "root";
$password = "secret";
$db_name = "crud"; //database name
//connect to mysql server
$mysqli = new mysqli($host, $username, $password, $db_name);
//check if any connection error was encountered
if(mysqli_connect_errno()) {
die("Connection failed: " . $conn->connect_error);
exit;
}
// Delete row
if ($mysqli->query (sprintf( "DELETE FROM grocery WHERE email = '".$mysqli->real_escape_string($_GET['id'])."' LIMIT 1") )) {
printf ( "Affected Rows %d rows.\n", $mysqli->affected_rows );
}
I hope this helps.
Provide a connection :
if( $mysqli->query($con, $query) ){
I'm an iOS developer trying to update a bit 'o php to work with the mysqli_* stuff instead of the deprecated mysql_* stuff.
I know next to nothing about php/mysql, and I'm stuck. I'm using a script found at http://www.w3schools.com/php/php_mysql_select.asp, with changes to reflect my field names etc
When I call the following from a browser I get an access error (Connection failed: Access denied for user 'whoisit7'#'localhost' (using password: YES)). The server name, password and username etc I'm using all work with an existing script in a browser but not with this new one.
Can anyone point me at what I'm getting wrong?
<?php
$servername = "localhost";
$userName = "whoisit7_ios";
$password = "blahblahblah";
$dbName = "whoisit7_scathing";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT name, latitude, longitude FROM location where status = 1";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "name: " . $row["name"]. " - latlong: " . $row["latitude"]. " " . $row["longitude"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
In PHP, variable names are case-sensitive. So, Change your connection variable names to:
$servername = "localhost";
$username = "whoisit7_ios";
$password = "blahblahblah";
$dbname = "whoisit7_scathing";
Read it if you are new to PHP: http://php.net/manual/en/language.variables.basics.php
Php is case sensitive, so change the below line from:
$conn = mysqli_connect($servername, $username, $password, $dbname);
to
$conn = mysqli_connect($servername, $userName, $password, $dbName);
You have made error in $username which is $userName and $dbname which should be $dbName as you declared above.
Just change -->
localhost to 127.0.0.1 //you need to change params in mysql config if u want to use it as localhost
And
$username to $userName //As u need to keep these variables similar :)
this Problem seems to be coming from database privileges as i have also have encountered this before. Give this fix a try.
GRANT ALL PRIVILEGES ON database_name TO usernamehere#host IDENTIFIED BY 'yourpassword';
FLUSH PRIVILEGES;
Just change the lower case letters to your own credentials.