I'm running this code on GoDaddy webhosting and I'm getting 'The database could not be found' echoed.
Obviously the database in question can't be selected, even though I've privileged the user and checked the db name.
I don't get anything out of here mysqli_error()
$db= 'test2' ;
$con = mysqli_connect('whatever','whatever','whatever') or die ('The connection to the database could not be established.');
mysqli_select_db($db , $con) or die ('The database could not be found' . mysqli_error());
As per the mysqli_select_db documentation, it expects the parameters this way:
mysqli_select_db ( mysqli $link , string $dbname ) : bool
So your parameters are put in backwards, change it to this:
mysqli_select_db($con, $db) ...
Or, alternatively, just select the database inside mysqli_connect().
$con = mysqli_connect('whatever','whatever','whatever', $db) ...
Side note, your die() isn't really doing anything, you won't get an actual error code out of that. To use mysqli_error(), you need to pass your database handle:
die('There was an error: ' . mysqli_error($con));
For the die() that is attached to mysqli_connect(), you should do this:
die('There was an error: ' . mysqli_connect_error());
Related
I have a php script which is supposed to insert a new row to a remote database. Something is not quite working as it should, but when I try to debug with mysqli_error it doesn't return anyting. Code as follows:
$connect = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or die("Unable to Connect to '$dbhost'");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = "INSERT INTO #DBname.TableName# (Section,Gender,WinningTeam,LosingTeam,FixtureD,FixtureT,Venue,Court,Texts,SetsWon,SetsLost,Winner_Score,Loser_Score) VALUES ($Section,$Gender,$WinningTeam,$LosingTeam,$FixtureD,$FixtureT,$Venue,$Court,$Texts,$SetsWon,$SetsLost,$Winner_Score,$Loser_Score)";
$result = mysqli_query($connect, $query) or die (mysqli_error());
The error message says mysqli_error expects exactly 1 paramater, 0 was given. However when I replace it with mysqli_error($connect) I get Error 404 - php script not found on server.
Pretty sure I'm missing something obvious, can anyone help out?
Kind regards
function mysqli_error() needs one parameter according to php.net
So in your case:
$result = mysqli_query($connect, $query) or die (mysqli_error($connect));
I have written that seems to not be working, but MySQL does not return any error. It is supposed to get data from database1.table to update database2.table.column
<?php
$dbh1 = mysql_connect('localhost', 'tendesig_zink', 'password') or die("Unable to connect to MySQL");
$dbh2 = mysql_connect('localhost', 'tendesig_zink', 'password', true) or die("Unable to connect to MySQL");
mysql_select_db('tendesig_zink_dev', $dbh1);
mysql_select_db('tendesig_zink_production', $dbh2);
$query = " UPDATE
tendesig_zink_dev.euid0_hikashop_product,
tendeig_zink_production.euid0_hikashop_product
SET
tendesig_zink_dev.euid0_hikashop_product.product_quantity = tendesig_zink_production.euid0_hikashop_product.product_quantity
WHERE
tendesig_zink_dev.euid0_hikashop_product.product_id = tendesig_zink_production.euid0_hikashop_product.product_id";
if (mysql_query($query, $dbh1 ))
{
echo "Record inserted";
}
else
{
echo "Error inserting record: " . mysql_error();
}
?>
The manual page for mysql_error() mentions this about the optional parameter you're omitting:
link_identifier
The MySQL connection. If the link identifier is not
specified, the last link opened by mysql_connect() is assumed. If no
such link is found, it will try to create one as if mysql_connect()
was called with no arguments. If no connection is found or
established, an E_WARNING level error is generated.
So it's reading errors from $dbh2, which is the last connection you've opened. However, you never run any query on $dbh2:
mysql_query($query, $dbh1 )
Thus you get no errors because you are reading errors from the wrong connection.
The solution is to be explicit:
mysql_error($dbh1)
As about what you're trying to accomplish, while you can open as many connections as you want, those connections won't merge as you seem to expect: they're independent sessions to all effects.
All your tables are on the same server and you connect with the same users, there's absolutely no need to even use two connections anyway.
You can't just issue a cross-database update statement from PHP like that!
You will need to execute a query to read data from the source db (execute that on the source database connection: $dbh2 in your example) and then separately write and execute a query to insert/update the target database (execute the insert/update query on the target database connection: $dbh1 in your example).
Essentially, you'll end up with a loop that reads data from the source, and executes the update query on each iteration, for each value you're reading from the source.
I appreciate everyone's help/banter, here is what finally worked for me.
<?php
$dba = mysqli_connect('localhost', 'tendesig_zink', 'pswd', 'tendesig_zink_production') or die("Unable to connect to MySQL");
$query = " UPDATE
tendesig_zink_dev.euid0_hikashop_product, tendesig_zink_production.euid0_hikashop_product
SET
tendesig_zink_dev.euid0_hikashop_product.product_quantity = tendesig_zink_production.euid0_hikashop_product.product_quantity
WHERE
tendesig_zink_dev.euid0_hikashop_product.product_id = tendesig_zink_production.euid0_hikashop_product.product_id";
if (mysqli_query($dba, $query))
{
echo "Records inserted";
}
else
{
echo "Error inserting records: " . mysqli_error($dba);
}
?>
For some reason, the following code inside the query works in my MySQL command console, yet when I try to run it as a Query in PHP, something keeps going wrong and I'm not sure what. Here is the code I've done so far.
//2. Perform database query
$query = "SELECT skills.element_id, content_model_reference.element_id, element_name FROM skills, content_model_reference WHERE (skills.element_id = content_model_reference.element_id)";
$result = mysql_query($query);
//Tests if there was a query error
if(!$result){
die("Database query failed.");
}
Is there something preventing the code that worked in MySQL (The line with SELECT) from working, or is my syntax somehow wrong?
EDIT: So it's saying I didn't select a database. Yet I thought I had. Here is the code above it:
//1. Create a database connection
$dbhost = "host"; //Host: Can be either an IP address, or a domain (like google.com).
$dbuser = "user";//User: The user that is connecting to the database.
$dbpass = "pass";//Password: This is the password that the user is using.
$dbname = "db";//Name: This is the name of the database.
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);//The value, 'handle,' is the connection.
//Test if connection occurred. Die ends the program/php, and in this case, also prints a message
if(mysqli_connect_errno()){
die("Database connection failed: ".
mysqli_connect_error().
" (". mysqli_connect_errno() . ")"
);
}
Like I said, the error message I am getting is pertaining only to the query, the server is fine with my database connection.
You're using mysqli_* for the connection, but you're using mysql_* for the QUERY... don't think you can do that, has to be one or the other (MYSQLI_ preffered). Also the query should be:
$result = mysqli_query($connection,$query);
Here is the code:
<?php
$con=mysql_connect('localhost', 'itorras', 'passwordhere');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db( "my_db" ) or die( 'Error'. mysql_error() );
$sql="INSERT INTO Brackets(have things here)
VALUES(and here)";
if (!mysqli_query($con,$sql))
{
die('Error with adding row: ' . mysqli_error($con));
}
echo "Thank you, your bracket has been submited.";
echo "<a href='index.html'>Click here to go back to home page</a>";
?>
I am using the username and password that is given the rights to mess with the database.
So when i try to submit something into the file none of the top errors run but then the bottom error prints error with adding row and nothing more. This file worked fine on my local server but has not worked on the web host. I am using godaddyweb hosting, so phpmyadmin and cpanelx.
If you need any more info let me know. Have been at this for a couple hours.
You are using mysql_connect but mysqli_query. You are mixing mysql and mysqli. Use only mysqli_*.
It appears you are using two different libraires at the same time
The original mysql API is deprecated for various reasons and you should not use it in new code
Instead, use PDO or mysqli
In your code, you are using the original mysql for the db connect, and then you use mysqli for the query. Instead you should only use mysqli
Replace the following code :
$con=mysql_connect('localhost', 'itorras', 'passwordhere');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
With :
$db = new mysqli('localhost', 'user', 'pass', 'demo');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
Then you can query the database with something similar :
$sql = <<<SQL
SELECT *
FROM `users`
WHERE `live` = 1
SQL;
if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}
For a complete overview, you can take a look at the documentation
http://ca2.php.net/mysqli
I have this simple code:
<?php
//Open the mySQL connection
$conn_string = "'localhost', 'Vale', 'test'";
$dbh = mysql_connect($conn_string);
//Check that a connection with the DB has been established
if (!$dbh)
{
die("Error in MySQL connection: " . mysql_error());
}
...
And I get the error: Error in MySQL connection: php_network_getaddresses: getaddrinfo failed: The requested name is valid, but no data of the requested type was found.
I cannot figure out what the problem is, I have been google-ing but all the suggestions have failed (tried 127.0.0.1 instead of localhost, 127.0.0.1:3306, etc.)
I have code that works with postgre, but I need to use mysql, and I am trying to modify it, but I cannot pass the first line and get a connection. Any suggestion, please? Thank you!
mysql_connect doesn't take a comma seperated string. It takes 3 individual strings.
Change it to this:
$dbh = mysql_connect($server, $mysql_user, $password);
If you absolutely have a comma separated string, and can't get around this, you can split the string like this:
$config = str_replace("'", '', $conn_string); // replace the quotes.
$config = preg_split('/,/', $conn_string); // split string on ,
if($config != $conn_string) { // make sure this returned an array
count($config) === 3 OR die("Invalid database configuration string");
$dbh = mysql_connect($config[0], $config[1], $config[2]);
if(FALSE === $dbh) {
die("Coult not connect: " . mysql_error());
}
}
You might want to consider MySQLi, instead of MySQL.
<?php
$dbh=mysqli_connect("localhost","Vale","Password","test"); /* Vale is your username? And the name of your database if test, right? And your Username's Password is blank? */
if(mysqli_connect_errno()){
echo "Error".mysqli_connect_error();
}
?>
As other users have pointed out mysql_connect expects the database, username, and password as separate arguments rather than a single string.
I think another highly important issue to point out is that this particular extension is deprecated.
Please see: http://uk1.php.net/function.mysql-connect
A better solution would be to use mysqli_connect: http://uk1.php.net/manual/en/function.mysqli-connect.php
$db = mysqli_connect( 'localhost', 'Vale', 'test', 'yourDatabaseName' );
mysql_connect requires three argument not single string
$dbh = mysql_connect('localhost', 'Vale', 'test');