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.
Related
I've looked all over here. Please be patient as I am new to php and mysql.
I got WAMPP installed & seems to be working OK. I created a simple "test" database from phpMyAdmin and "firsttable" in that. I can do a simple connect using example from w3schools, but trying to select & display data I entered only throws back errors.
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Connect
$conn = mysqli_connect($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT reference, firstname, lastname, room FROM firsttable";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["reference"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "room:" . $row["room"]. "<br>";
}
} else {
echo "0 results";
}
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$conn->close();
?>
First off, I get a parse error on line 17. The one that reads:
if ($result->num_rows > 0) {
The error says: Trying to get property of non-object.
I tried wrapping the whole php code in tags and saving it as html, but then it appeared that no row data was ever found.
I am able to use very simple code that connects successfully. I can confirm the database is in there, so is the table, and the contents I added to it.
Please, what am I doing wrong?
You need to specify the database when you connect:
$database = 'test';
$conn = mysqli_connect($servername, $username, $password, $database);
where $database is the name of your database (test in this case). MySQL doesn't know which database your table resides in without you telling it.
In addition, you should always include error checking for your database connection (you have two of these, you don't need the last one) as well as any queries. Sans this, you can check your error logs for more information when something fails.
I have read a bunch of the other posts about similar issues but I am very new to this so I am likely missing something. Most of the other questions I found were a lot more complicated than mine. I am trying to follow the w3schools tutorial for this and am testing locally using XAMPP. Right now I am just trying to get this to work successfully and eventually I am planning to submit data into a mySQL database from a web form.
<?php
$servername="localhost";
$dbname="mysql";
// Create connection
$conn = mysqli_connect($servername, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO owner_table (ownerFirst, ownerLast, mobile)
VALUES ('John', 'Doe', '1111111111')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
NOTE:
The connection is not failing from what I can tell. Removing everything after the first if statement, and adding else { echo "success" } outputs success. Again though, I am new to PHP.
If someone knows that the answer has been given in another post PLEASE PLEASE comment here and ill close this out.
http://www.w3schools.com/php/php_mysql_insert.asp
You should follow the proper syntax:
$conn = mysqli_connect(<host Name>, <username>, <password>, <database name>);
Reference: http://php.net/manual/en/function.mysqli-connect.php#refsect1-function.mysqli-connect-examples
Add two more parameters- username and password here:
$username = "root"; // Default values
$password = ""; // Default values
$conn = mysqli_connect($servername, $username, $password, $dbname);
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 have a website in PHP. I try to store the session variable $_SESSION['user_name'] to a mysql database when a logged in user visits a specific webpage on my site.
<?php
$servername = "localhost";
$username = "user1";
$password = "user1";
$dbname = "payment";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = 'INSERT INTO users
VALUES ('.$_SESSION['user_name'].')';
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Error message:
Notice: Undefined variable: _SESSION in /opt/lampp/htdocs/succes.php on line 16
Tried a bunch of things but can't figure it out. What is wrong here?
You need to call session_start() at the beginning of your script (before using any $_SESSION variables). Also, you need quotes around the variable in you query:
$sql = 'INSERT INTO users
VALUES ("'.$_SESSION['user_name'].'")';
Please note that this is not safe; you are wide open to SQL injection. Instead, you should use prepared statements:
<?php
$servername = "localhost";
$username = "user1";
$password = "user1";
$dbname = "payment";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = 'INSERT INTO users
VALUES (?)';
$stmt = $conn->prepare($sql);
$stmt->bind_param('s', $_SESSION['user_name']);
if ($stmt->execute()) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Before you use any $_SESSION variables you need to call session_start().
Of topic a bit though, something to look into PDO. It can be a bit a tad slower than mysqli() however supports many more Database types. Here is a good article on Tuts+ explaining some of the differences as well as explaining essential security steps.
If I could be a bit biased I have created a PHP Class for PDO Connections which can be found on GitHub
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.