mysql-connect

PHP Select SQL view no output

由 ▼魔方 西西 提交于 2019-12-11 08:21:35
问题 I creating a php file that pulls a view from an SQL database. Can someone let me know why this isn't working? It seems to be timing out. I am not getting a connection error, either. Thank you in advance. <?php require ('mysqli_connect.php'); $sql = "SELECT * FROM testview ;"; $result = mysqli_query($dbc,$sql); // Check connection if ($dbc->connect_error) { die("Connection failed: " . $dbc->connect_error); } $result=mysqli_query($sql); if ($result->num_rows > 0) { echo "<table><tr><th>userID<

fork + PHP CodeIgniter: 'MySQL server has gone away' - How not to lose connection when forking with CI?

由 社会主义新天地 提交于 2019-12-11 04:44:32
问题 I installed fork on my Ubuntu Server (using PHP-Apache-Codeigniter). I have this code: <?php public function add_keyword() { $keyword_p = $this->input->post('key_word'); $prod = $this->input->post('prod_name'); $prod = $this->kas_model->search_prod_name($prod); $prod = $prod[0]->prod_id; $country = $this->input->post('key_country'); $keyword = explode(", ", $keyword_p); var_dump($keyword); $keyword_count = count($keyword); echo "the keyword count: $keyword_count"; // Create fork $pid = pcntl

mysql_connect doesn't work second time

由 不羁岁月 提交于 2019-12-11 02:49:14
问题 duplicate of unanswered: How to reconnect in php adodb after exceptions: Mysql server gone away or Lost connection to MySQL server during query mysql_connect works the first time, but never works after that... $connectDb = mysql_connect(secret, secret, secret); mysql_select_db("secret", $connectDb); $sleepPeriod = 1800; sleep($sleepPeriod); while (true) { $result = mysql_query("good query", $connectDb); if (!$result) { if (mysql_error()=='MySQL server has gone away') { echo "MySql connection

keeping db connection active across pages

由 好久不见. 提交于 2019-12-10 16:37:27
问题 I'm a learner. Is there a way to stay connected in to the mysql database as the user is taken to the next page. For example, the db connection is made, the user is logged in, and then goes to the next page to access a table in the database. Instead of having to make the db connection again, is there a way to keep the previous connection active? Or does it matter at all in a low-traffic site? I read a post yesterday about something related to sessions, and the responder talked about sending a

Windows 7 PHP MySQL Connection Issues [duplicate]

由 泄露秘密 提交于 2019-12-07 08:59:14
问题 This question already has answers here : Cannot connect to MySQL 4.1+ using old authentication (7 answers) Closed 6 years ago . When I run $conn = mysql_connect($host, $user, $pass); I am able to get a connection to the database but when i do a var_dump($conn) I get back bool(true) which is limiting me from having multiple connection to multiple servers. The original setup on this computer was XAMPP with a version of PHP 5.2 that was upgraded to PHP 5.3.4 via the PHP installer. It could

MySQL database config in a separate class

由 Deadly 提交于 2019-12-06 08:35:32
Is it possible to keep all my database related configuration (hostnames, usernames, passwords, and databases) as well as the function to connect to and select the correct database in a separate class? I tried something like this: class Database { var $config = array( 'username' => 'someuser', 'password' => 'somepassword', 'hostname' => 'some_remote_host', 'database' => 'a_database' ); function __construct() { $this->connect(); } function connect() { $db = $this->config; $conn = mysql_connect($db['hostname'], $db['username'], $db['password']); if(!$conn) { die("Cannot connect to database server

php mysql_connect resource is allways the same

由 邮差的信 提交于 2019-12-05 19:21:36
I'm having a problem my php script witch you can see here: http://codepad.org/F0qhElRC Is not opening a new resource for each connection from a child. Already I tryed using 127.0.0.1 or my local network IP but nothing worked, also max_user_connections in mysql is set to 0 opening a connection from web gives me a different resource for the web. But in the command line it uses the same. In mysql_connect I specified new_link to be true, I placed a __destruct in the bd class to close the connection if the child get's destroyed. Closing the connection does not make php open a new one even after I

Windows 7 PHP MySQL Connection Issues [duplicate]

由 可紊 提交于 2019-12-05 13:57:27
This question already has answers here : Cannot connect to MySQL 4.1+ using old authentication (7 answers) Closed 6 years ago . When I run $conn = mysql_connect($host, $user, $pass); I am able to get a connection to the database but when i do a var_dump($conn) I get back bool(true) which is limiting me from having multiple connection to multiple servers. The original setup on this computer was XAMPP with a version of PHP 5.2 that was upgraded to PHP 5.3.4 via the PHP installer. It could connect to its local database with no problem (other than returning a boolean instead of the resource link

php MySql connectivity error

由 情到浓时终转凉″ 提交于 2019-12-04 04:53:52
问题 I have the following php code to connect to my mysql DB. mysql_connect( "myserver.com" , "root", "redhat","datastore") or die(mysql_error()); when I ran this code I got the error saying: Warning: mysql_connect(): Host '10.21.21.10' is not allowed to connect to this MySQL server in C:\xampp\htdocs\inventory\net.php on line 20 Host '10.21.21.10' is not allowed to connect to this MySQL server But the IP of myserver.com when I did ping myserver.com in command prompt it gave 10.25.15.95 ,so I

How to insert data in two different tables of two different database in php

由 ▼魔方 西西 提交于 2019-12-04 04:28:19
问题 I have to insert data in two different database's table. I have created database1 and table1 for database1, also i have created database2 and table2 for database2. For inserting data i have written code, $connect = mysql_connect("localhost","root",""); //database connection mysql_select_db("database1",$connect); // select database1 mysql_select_db("database2",$connect); // select database2 $sql = mysql_query("INSERT INTO database1.table1 (contact_first, contact_last, contact_email) VALUES(
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