I usually am immersed in the Microsoft Stack but dabble in PHP from time to time. A long standing question I've had with PHP that I've never seem to be able to find the answer to is how do you apply your already declared require("dbConnect.php") database connection to your mysql_query()? For clarification please see my code example below:
require("dbConnect.php");
$db_host = 'localhost';
$db_user = 'UserName';
$db_pwd = 'Password';
$database = 'DbName';
$table = 'tblQuote';
if (!mysql_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
// sending query
$result = mysql_query("SELECT QuoteID, FirstName, LastName, PhoneNumber, Email, QuoteDate FROM tblQuote ORDER BY QuoteDate DESC");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
So in looking at this you can see the standard require() declaration at the top... which already holds my connection info. But every single MySQL Query example I've ever found always creates it's own connection... which I get for demonstration purposes... but I've never been able to figure out how I can use my already existing connection thereby bypassing rewriting the exact same connection info over and over again when it comes to writing queries. I know for you PHP developers this question is like 101 but I've not been able to find an answer to this seemingly basic question... admittedly I may be asking the question wrong so any help would be appreciated!
From the PHP documentation: http://php.net/manual/en/function.mysql-query.php
mixed mysql_query ( string $query [, resource $link_identifier = NULL ] )
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.
So since you've already created one in your dbConnect.php, the one you just made will be used (It won't create a new one for every query). To pass it explicitly into your mysql_query function call, you can return the MySQL resource that was returned from your mysql_connect call like so:
dbConnect.php
return mysql_connect(....);
Then in the code you pasted above:
$mysql_conn = require('dbConnect.php');
...
$result = mysql_query('...', $mysql_conn);
Then you will explicitly have the connection and pass it to your query - there will be no mistaking it, regardless of how large your codebase becomes. When you require the file, you'll have access to the connection variable, but in the above example, how you get the connection is more semantically clear.
Also, notice that this function has been deprecated in PHP>=5.5, so you'll want to use PDOs or MySQLi which have future support.
Hope this helps!
Related
I am attempting to create new tables every time I post to this method, but for some reason I can not figure out why it dies.
<?php
$host = "127.0.0.1";
$username = 'cotten3128';
$pwd = 'pwd';
$database = "student_cotten3128";
$pin = $_REQUEST['pinSent'];
$words = $_REQUEST['resultSent'];
$tableName = $pin;
$db = new mysqli($host, $username, $pwd, $database);
if ($sql = $db->prepare("CREATE TABLE $pin (id INT(11) AUTO_INCREMENT);")) {
$sql->execute();
$sql->close();
}else{
echo $mysql->error;
die('Could not create table');
}
for($i=0;$i<count($words);$i++){
if($sql = $db->prepare("INSERT INTO ".$pin.$words[$i].";")) {
$sql->execute();
$sql->close();
}else{
echo $mysql->error;
die("Could not add data to table");
}
}
mysqli_close();
?>
Any help or insight would be greatly appreciated.
The intention of my post is to help you finding the issue by yourself. As you did not added much information I assume my post is helpful for you.
Based on the code you have shared I guess you mean one of your called die() functions is executed.
Wrong function call
As Jay Blancherd mentioned mysql_close is the wrong function. You rather have to use mysqli_close as you created a mysqli instance.
Beside of that mysql_* is deprecated and should not be used anymore.
Debugging Steps
Not only for this case but in general you should ask yourself:
Is there an error message available? (Frontend output, error log file, ...)
YES:
What's the message about?
Is it an error you can search for? E.g. via a search engine or the corresponding documentation?
Look up in the bug tracker (if available), by the software developer of the software you are using, and if it has not been reported yet report the issue.
NO: (if none error message available OR you cannot search for it as it is a custom error message)
Search in the files of the software you are using for the error message and start a core-debugging.
STILL NO SOLUTION?:
Ask on stackoverflow.com e.g. and tell your issue and the steps you have performed to find and fix the bug. Post only as much code as necessary plus use a proper format.
Debugging in your case:
In order to narrow down the scope. Which of the die() is executed? Depending on that echo the query to execute just before it actually is executed. Then copy the SQL query to an SQL editor and look at it syntax. After that you probably know the problem already.
i am having difficulty with connecting my database to mysql, i have tried many different ways of trying to get it to connect to my database, but none these methods seem to work. I am only new to php and i could be missing something, but i dont know what it is.
This is for a search engine, i have the form created.
I would think that my problem is coming from this line of code, mysql_connect("localhost", "root", "") or die("could not connect");?!
I was wondering if i could be told what is the problem and how to fix it, thank you so much.
here is my code below.
<?php
//connect
mysql_connect("localhost", "root", "") or die("could not connect");
mysql_select_db("search") or die("could not find database");
$output = '';
if (isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i","", $searchq);
$query = mysql_query("SELECT * FROM gp management system WHERE Title LIKE '%$searchq%' OR Description LIKE '%$searchq%' OR Keyword LIKE '%$searchq%'") or die("could not search");
$count = mysql_num_rows($query);
if($count == 0) {
$output = 'There was no search results!';
}
else{
while($row = mysql_fetch_array($query)) {
$Title = $row['Title'];
$Description = $row['Description'];
$Keyword = $row['Keyword'];
$output .= '<div> ' .$Keyword. ' </div> ';
}
}
}
?>
SQL errors:
SELECT * FROM gp management system WHERE Title...
^^-- table name
^^^^^^^^^^ aliasing 'gp' as 'management'
^^^^^^-- extra unknown garbage
Table names shouldn't have spaces to begin with, but if you insist on having them, then you need proper quoting:
SELECT * FROM `gp management system` etc...
^--------------------^
Never output a fixed/useless "could not search" error. Have the DB tell you what went wrong:
$result = mysql_query($sql) or die(mysql_error());
^^^^^^^^^^^^^^^^^^^^^^^
As your only new, you really should stop using mysql_ and change to mysqli_ or (my preference) PDO(). Your message in your title seems to be coming from your selectdb line, so you're actually connecting to the database fine, it's just not able to locate the schema that you are trying to use (i.e. "search" does not exist as a schema name in your DB environment). Unless that's not the error that you are getting, in which case it's a flaw in your actual SQL query. Without an accurate statement of what you are getting back on the screen when you try to run the script then not much can be done to help.
I would rather do somehow like I did ages ago, see below:
function db_connect($host, $username, $passwd, $db_name) {
$db = mysql_connect($host, $username, $passwd);
if (!$db) {
echo "Login to database failed: ";
echo mysql_error();
return false;
}
$result=mysql_select_db($db_name); //Select the database
if (!$result) {
echo "Cannot find the database: ".$db_name;
echo mysql_error();
return false;
}
return $db; //Database descriptor
}
However you shouldnt rely on the old MYSQL extension, but use MYSQLi.
EDIT: as #tadman pointed out there were a # sign infront the mysql_connect call, however there is an if statement for checking and a call to error output.
As others have said, your code is something of a mess - you seem to be new to programming, not just in PHP. But everyone has to start somewhere, so....
You might want to start by learning a bit about how to ask questions - specifically you have included a lot of code which is not relevant to the problem you are asking about - which is about connecting to MySQL. OTOH you have omitted lots of important information like the operating system this is running on and what diagnostics you have attempted. See here for a better (though far from exemplary example).
Do read your question carefully - the title implies an issue with mysql_select_db() while in the body of the content you say you think the problerm is with mysql_connect(). If you had provided the actual output of the script we would have been able to tell for ourselves.
You should be checking that MySQL is actually running - how you do that depends on your OS. You should try connecting using a different client (the mysql command line client, sqldeveloper, Tora, PHPMyAdmin....there are lots to choose from).
You should also check (again this is operating system specific) that your MySQL instance is running on the same port as is configured in your php.ini (or overridden in your mysql_connect() call).
MySQL treats 'localhost' as something different from 127.0.0.1 on Unix/Linux systems. If this applies to you try 127.0.0.1 in place of localhost.
or die("could not connect");
Does not help in diagnosing the problem. If instead you...
or die(mysql_error());
You will get a more meaningful response.
BTW: while writing code using the mysql extension rather than mysqli or PDO is forgivable, I am somewhat alarmed to see what appears to be medical information managed with no real security.
I am getting this error:
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource
Here's my Connection.php :
$userDB_server = "";
$userDB_user = "";
$userDB_password = "";
$userDB_database = "";
$connection = mysql_connect("$userDB_server","$userDB_user","$userDB_password") or die ("Unable to establish a DB connection");
$userDB = mysql_select_db("$userDB_database", $connection) or die ("Unable to establish a DB connection");
$gameDB_server = "";
$gameDB_user = "";
$gameDB_password = "";
$gameDB_database = "";
$gameDB_connection = mysql_connect("$gameDB_server","$gameDB_user","$gameDB_password", true) or die ("Unable to establish a DB connection");
$gameDB = mysql_select_db("$gameDB_database", $gameDB_connection) or die ("Unable to establish a DB connection");
Here's my function :
require_once('Connection.php');
$findQuery = sprintf("SELECT * FROM `Keys` WHERE `ID` = '$gID'");
$findResult = mysql_query($findQuery, $connection) or die(mysql_error());
$resultRow = mysql_fetch_assoc($findResult) or die(mysql_error());
The error is on "$findResult = mysql_query($findQuery, $connection) or die(mysql_error());"
But I don't see a problem anywhere.
What I've tried :
I've tried with and without the "true" on the second connection,
didn't seem to make a difference anywhere.
Echoing the $connection and $gameDB_connection shows nothing,
Using var_dump on $connection shows "resource(9) of type (mysql link)"
Removing the $connection from the mysql_query has it connect to the
other DB (gameDB_connection) and I get an error that the table
doesn't exist (its not on that DB).
Adding / changing / removing the backquote ( ` ) from the query seems
to have no effect on the error
The variable $gID echo's correctly, so it's not null (its 1001 in
this case)
If I run the SELECT part in the actual sql form (instead of via php),
it lists them all correctly
The Connection.php is used in other places (one page reads
from both databases at the same time) successfully. No errors anywhere else
Anyone have any idea what's wrong?
Based on the comments, it sounds like the problem is caused by using require_once() inside a function.
One of two thing is happening. Either:
You've already included Connection.php somewhere else, so when you get to the function, it's not actually included.. because of the once part of require_once.
or...
It is working the first time you call the function, but the second time you call it, the file has already been included and does not get included again.
The problem is that when the file is first included (assuming that's from within this function), the $connection variable is created in the function scope, and like any other function variable, is gone at the end of the function. When you call the function a second time, the include doesn't happen because you're using require_once.
You could probably fix this by calling require() instead of require_once(), but that will end up reconnecting to the database every time you call the function - which is a lot of unnecessary overhead. It's much cleaner to just move the include outside of the function, and either pass the connection into the function, or use it as a global variable.
That would look like this:
require_once('Connection.php');
function getResult() {
global $connection;
$findQuery = "SELECT * FROM `Keys` WHERE `ID` = '$gID'";
$findResult = mysql_query($findQuery, $connection) or die(mysql_error());
$resultRow = mysql_fetch_assoc($findResult) or die(mysql_error());
}
All that being said, there's 2 major problems with this code.
You're using the mysql_* functions which are deprecated and will soon be removed from new versions of PHP. See this question for more details: Why shouldn't I use mysql_* functions in PHP?
It's not actually that hard to switch to something like the mysqli_* functions instead - there's a non-object set of functions that are almost identical to what you're using now.
You're including a variable in your query without properly escaping it. At the very least you should be calling mysql_real_escape_string() (or mysqli_real_escape_string()), but a better solution is to look into prepared statements. You can find more information on prepared statements here: How can I prevent SQL injection in PHP?
pg_query and pg_query_params in PHP can be executed without specifying a connection resource in the arguments' list, so PHP will use the last connection opene by pg_connect.
Is there a way to retrieve the default connection string that PHP will use intead of trying to figure out where it is coming from across a couple of PHP files? Something like get_default_dbConnection()?
Thanks.
That doesnt mean that there is a default connection string, but that you dont have to provide a param that references the connection. Check out the example on php: pg_query
$conn = pg_pconnect("dbname=publisher");
if (!$conn) {
echo "An error occured.\n";
exit;
}
$result = pg_query($conn, "SELECT author, email FROM authors");
if (!$result) {
echo "An error occured.\n";
exit;
}
In this example there is $conn = pg_pconnect("dbname=publisher") and in the call to pg_query they pass a long $conn. That is the connection reference. Now you could also just use
$conn = pg_pconnect("dbname=publisher");
$result = pg_query("SELECT author, email FROM authors");
pg_query will now automatically use the last opened connection (which is $conn)
Update:
You cannot get the resource, PHP uses an internal method php_pgsql_get_default_link to get the default connection but will not return it anywhere (See php-5.4.9\ext\pgsql\pgsql.c in the source).
You can get information on the database name or user using pg_dbname or pg_parameter_status
I have been writing php and mySQL functions all day and as I was writing the simplest part of my project I have hit a wall.
The function should simply count how many entries are in the database and return that number (If there is a more simple way please let me know, this is my first php + mysql project)
Here is the code:
function quoteCount(){
global $db;
$totalQuoteNum = array();
$query = "SELECT * FROM Quotes";
$result_set = mysqli_query($db, $query)
or die ("Query $query failed ".mysqli_error($db)); //fails here
$totalQuoteNum = mysql_num_rows($result_set)
or die ('couldnt count rows'.mysqli_error($db));
echo 'COUNTED EVERYTHING!!!';
return $totalQuoteNum;
};
Now when the die statement prints I get the string but not the mysqli error.
Things I have tried and ruled out:
$db is correct
query works in mysql
I wasnt sure if the database was connected, so I added the connect inside this function and stil nothing.
Any ideas? From what I see it should work and its not giving me any error to work from. Please help!
Based on the comments, it seems as though $db is the database name.
Functions such as mysqli_query() expect a database link (resource), not simply the database name.
This resource is created by constructing a new mysqli object. Following your procedural style, use mysqli_connect().