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PHP - Mysql_connect Question
Hi, this is more of a question than asking for help so if its in the wrong section could someone please move.
my question is could you connect to a database by using an array that has been generated from getting contents from a file, for example... Code: [Select] $file_c = file_get_contents("example.txt"); $colms = explode(",",trim($file_c)); mysql_connect("$colms[0]","$colms[1]","$colms[2]") or die(mysql_error()); mysql_select_db("$colms[3]") or die(mysql_error()); if not how else would someone go about doing this. Similar TutorialsHi, I am really plugging into how to write functions in PHP. But I was going to delve into a user management program and try to create it, but dont want to be the older version of what I was before if you like, where I just type in for the sake of typing in code so I thought I would question what their doing. But a peice of code, a very small snippet, this came up: @mysql_connect What does this actually mean with the @ sign infront of the mysql_connect function? Seen this a few times but just never appreciated what the at sign means, any help is wonderfully appreciated of course. Thanks, Jeremy i think this has to do something with me dropping my table and trying to reinstall the blog but now i get "can't connect to mysql5.000webhost.com" so then i tried using localhost and i get... "can't connect to localhost using password YES" so i assume my password is right and for some reason it isnt connecting yet all the variables match up correct password, database,user i double and triple checked spelling mysql_connect("$host", "$username", "$password")or die("cannot connect to server"); i create a connection and let say there is a problem with the server so that the connection can't established my question is how can i avoid the default error to be displayed which is WARNING: mysql_connect() blablabla as it is annoying to see such error message. thanks in advance Quote Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'nyhungry_builder'@'localhost' (using password: YES) in /home3/nyhungry/public_html/musicmakerstore/storescript/connect_to_mysql.php on line 19 could not connect to mysql I am getting this error from the code below how come? Code: [Select] <! <?php // Place db host name. Sometimes "localhost" but // sometimes looks like this: >> ???mysql??.someserver.net $db_host = "localhost"; // Place the username for the MySQL database here $db_username = "nyhungry_builder"; // Place the password for the MySQL database here $db_pass = "blablabla"; // Place the name for the MySQL database here $db_name = "nyhungry_musicbeatmaker"; // Run the actual connection here mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); mysql_select_db("$db_name") or die ("no database"); ?> I am working on a cpanel kind of host and you know how the databases are setup like blablab_databasename etc.. What is going on here i have check the database name, the table, password and the host thousands of times and nothing.. Help I am trying to extend my knowledge from using basic mysql_connect etc. into something better. I am searching on Google, and looking up as much info as possible but find that I learn best with a brief overview that allows me to channel down my searches. (1) I know PDO v mysql_connect v mysqli isn;t right - I think PDO is a class, mysql_connect is a command and mysqli is something else. Am I on the right track? (2) What is the best solution? I am using a full OOP coding style and want to learn the best - I am a novice at this but would consider myself an 8/10 programmer at other parts of PHP. An very brief overview to get me started and channel my Google searches would be much appreciated. - I have heard PDO is slower. - I don't need support for different databases (only MySQL). Thanks in advance. Hi, I currently have an HTML form that posts data (method:POST) to a PHP script. The PHP script then retrieves the data and displays it on screen so the user can do a final check. I then want to have a button, so that when pressed, the data gets sent to a MySQL database. What I have so far is along these lines: Code: [Select] <?php if(isset($_POST['var1'])) {$var1 = $_POST['var1'];} else {$var1 = "";} if(isset($_POST['var2'])) {$var2 = $_POST['var2'];} else {$var2 = "";} echo "<b>You have entered the following:</b><p></p>", "$var1 <br>","$var2<p></p>", "If this is correct, please press continue."; print "<html><body><button name=Button1 onclick="">Continue</button></body></html>"; and onclick I want the following to run: Code: [Select] $link = mysql_connect ('localhost', 'root'); if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db ('DB1', $link); mysql_query ("INSERT INTO table1 (field1, field2) VALUES ('$var1','$var2')"); mysql_close($link); How can I do this?! Is it neccesary to use another "submit" button that submits $var1 & $var2 to a different PHP script, which then runs the mysql_connect part? Or can I enclose the mysql_connect part in an IF statement that runs only if Button1 has been pressed (and I enclose the rest of the PHP script in an IF statement that runs only if Button1 has NOT been pressed) and get Button1 to reload the same PHP script? Either way, how do I define an onclick event inside echo? This prob a noobish question, but I seem to get errors if I try to put (single) quotes "" ' ' inside the quotes of the echo: echo " onclick"" "; Thanks! When I try to connect to my database I am getting this error: Quote Warning: mysql_connect() [function.mysql-connect]: php_network_getaddresses: getaddrinfo failed: No such host is known. in C:\wamp\www\Envne\envne.com\mysqlAdmin\classes\Database.php on line 9 Warning: mysql_connect() [function.mysql-connect]: [2002] php_network_getaddresses: getaddrinfo failed: No such host is kn (trying to connect via tcp://root:3306) in C:\wamp\www\Envne\envne.com\mysqlAdmin\classes\Database.php on line 9 Warning: mysql_connect() [function.mysql-connect]: php_network_getaddresses: getaddrinfo failed: No such host is known. in C:\wamp\www\Envne\envne.com\mysqlAdmin\classes\Database.php on line 9 Here is the EXACT code on line 9: mysql_connect('root', '', 'localhost'); I have never gotten this error before... how do I fix it? Hi, I get this error when I load the page, but after hitting ctrl f5 to refresh.. the error goes away and the page works perfectly. I have edited out my details. Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'user'@'localhost' (using password: YES) Access denied for user 'user'@'localhost' (using password: YES) I know everything is set right.. I just don't get it.. I am going crazy of this. Ive cleared my browser cache.. even tried it on my android phone and it does the same thing. Thanks, Deceevn I am using a script that gets connected to PHP table and returns the visitors country name from: Database MyProjectA Table: ip2country and inside the script it uses the mysql_connect() function and it works fine BUT the code used in other pages to get data from other tables does not work. I mean, I have a combo box for example, that populates data from: Database MyProjectA Table: mybookings But it doesn't populate as long as the code for the 'user's country script' is active. The moment I comment out the 'user's country script', the combo box works. Perhaps I am using 2 database pooling (or connection pooling) so what I did, I closed the previous connection after I get the visitor's country name but still the combo box has no values in it. Do you get the picture. Please read the question again. Thank you I don't know how else to explain this. I have two php pages in the same domain, same folder, with exact same line to connect to mysql, and one works, and the other doesn't. I have no clue. This is the line $con = mysql_connect("localhost", "myusername", "mypassword"); Hi all I have been trying to do a query but I get Call to undefined function mysql_connect() What other way can I perform a query? Or do I need to get something enabled? Thank in advance Dave httpd-2.2.17-win32-x86-openssl-0.9.8o php-5.3.6-Win32-VC9-x86 mysql-5.5.11-win32 PHP code: $dbcnx = mysql_connect('localhost','root','admin'); Error message: No connection could be made because the target machine actively refused it. I used mysql.exe -u root -p to verify that mysql is working and using the right username and password. Any ideas how to fix it? Thanks, Richard We have a 10 year old system that has 100's of php files that all connect to a central database using a single shared php include file that does a mysql_connect before any other code executes
include_once("db_connect.php");
mysql_query(...);
We never used the returned connection id (link identifier) from the mysql_connect function call as it was always implied in every subsequent mysql_query statement etc.
Now we've moved some of our bigger tables to a separate server, so we have to update small areas of some files that need to connect to this other system and send the old existing query to the new server instead of to the main server and process the results
So, going back through hundreds of files and converting mysql_query($string) to mysql_query($string, $connection1) is not practical
Possible solution one: anytime I need to use the new server I do this
------------------------------------------------------------------------------
mysql_connect(NEW_SERVER_IP,"...",""");
mysql_query($string);
...
// get back to normal
mysql_connect(OLD_SERVER_IP,"...","");
Possible solution two: modify the master include file
------------------------------------------------------------------------------
//add a new line ABOVE the existing mysql_connect statement...
$new_server=mysql_connect(NEW_SERVER_IP,"...",""");
mysql_connect(OLD_SERVER_IP,"...","");
and then anytime I need to use the new database I have to go back and modify all of the mysql_query,mysql_affected_rows, etc to reference the $new_server link identifier
solution one seems much simpler, but all of the mysql_connect reconnections I assume will be extremely inefficient
solution two seems cleaner but if I miss one mysql_ statement that needs the new connection object as a reference, the code will completely break or worse yet it will produce unexpected results
Example: If I miss converting a mysql_errno() function that really needs to now be mysql_errno($new_server) it would cause major issues.
So, I'm wondering if others have faced this dilema of migrating single mysql_connect code to multi mysql_connect mode and if there is a better way of going about implementing it.
Thanks in advance for any help
Steve
Im making a site for my family to leave messages for everyone to see but every now and then the message wont load and i get theses errors, Warning: mysql_connect() [function.mysql-connect]: Too many connections in /home/#####/public_html/#####.com/messages.php on line 17 Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/#####/public_html/#####.com/messages.php on line 18 Warning: mysql_query() [function.mysql-query]: Access denied for user '#####'@'localhost' (using password: NO) in /home/#####/public_html/#####.com/messages.php on line 21 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/#####/public_html/#####.com/messages.php on line 21 Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/#####/public_html/#####.com/messages.php on line 24 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/#####/public_html/#####.com/messages.php on line 25 Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in /home/#####/public_html/#####.com/messages.php on line 54 i also recieved this error once Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' (2) The hosting support team have been no help at all. MySQL Client Version: 5.1.50-community PHP Version: 5.3.5 Apache Version: 2.2.17 Okay, I am trying to get MySQL to cooperate with PHP. All of my attempts have been unsuccessful. Before you post a reply, please note I have taken these steps before I posted this topic: 1. My libmySQL.dll is in both my php and C:\Windows folder 2a. (Mental note: my installation directory is C:\Program Files\Apache 2.2\php\ext) 2b. The extension_dir in my php.ini file is as follows: Code: [Select] extension_dir = "C:\Program Files\Apache 2.2\php\ext"3. The php.ini file is also in the C:\Windows folder. 4. The extensions that are enabled for php are as follows: Code: [Select] [PHP_MYSQL] extension=php_mysql.dll [PHP_MYSQLI] extension=php_mysqli.dll [PHP_PDO_MYSQL] extension=php_pdo_mysql.dll5. Any section referencing to MySQL in my php info is not showing up. 6. I have tried replacing my php_mysql.dll, didn't work either. Okay, so when I try to install my forums for my website, it returns with this error: Fatal error: Call to undefined function mysql_connect() in C:\Program Files\Apache 2.2\htdocs\forum\upload\includes\class_core.php on line 312 Please notify me of anything that has or will work(ed). Please help me to solve this error Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\shop-script-free\includes\database\mysql.php on line 13 Warning: mysql_get_server_info() expects parameter 1 to be resource, boolean given in C:\wamp\www\shop-script-free\includes\database\mysql.php on line 14 Access denied for user 'SYSTEM'@'localhost' (using password: NO) Thanks you very much when in a form, I wish to build a conditional that if the response to a radio button is a value of 2 (female), it will display an input requesting for users maiden name. If not 2 goes to the next input statement. Here is code I was experimenting with: <html> <body> <form action="" name="test" method='POST'> <input type="radio" id="sex" value=1 checked><label>Male</label> <input type="radio" id="sex" value=2><label>Femaleale</label> <?php $result = "value"; if ($result == 2) echo "<input type='int' id='gradYear' size='3' required>"; else echo "Not a female!" ?> <input type="submit" value="GO"> </form> </body> </html> The code passes debug, however, Not a Female is displayed. My question is - Can I do this and if so, what value do I test against id='sex' or value. I tried each one but gave the same results. I realize that $_POST[sex] would be used after the submit button is clicked. But this has me stumped. Thanks for the assis in advance. Hi, I am looking into a way of adding addons to a class I made.. I thought something like this would work, but not sure how to implement it: Code: [Select] <?php class foo { function foobar($text) { $text = $text . 'b'; return $text; } } class bar extends foo { function foobar($text) { $text = $text . 'c'; return $text; } } $var = new bar(); $var->foobar('a'); // this would then return abc ?> Now i want to be able to call up foo, and it calls the foobar of foo, and the foobar of bar. is that possible? hosh I'm considering getting a VPS, but I'm not entirely sure what this would be able the handle. The specs doesn't look that good to me, but my friend swears it will handle running apache/php/mysql handling large websites and databases without a problem. I want to run multiple databases that just store statistics and I'm a bit worried about the RAM and the company doesn't even mention a cpu. Here's the specs: 256MB RAM 300GB HDD 10Mbps unmetered 1 IP Address My friend runs two counter-strike: source servers off his VPS (same stats).. so I'm actually considering this, but it just seems like it's not powerful enough. Oh.. forgot the most important part.. it's only $9. |
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