PHP mysql_query error_问答_开发者_运维开发者技术经验分享

开发者 https://www.devze.com 2022-12-16 08:46 出处：网络

I\'m trying to learn and I\'m stuck.I don\'t understand why this doesn\'t work.If I just leave the include and remove the function call and don\'t wrap the database connection in a function it works p

相关专题：php

I'm trying to learn and I'm stuck. I don't understand why this doesn't work. If I just leave the include and remove the function call and don't wrap the database connection in a function it works properly.

What is it that I'm missing here?

Error Message:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home3/badamsne/public_html/views/dogs.php on line 24 Database query failed:

Web page code:

    <?php
    include("../model/db_conn.php");

    db_conn();
?>

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
    <title></title>
</head>
<body>
  <?php



    // 3. Perform database query
    $result = mysql_query("SELECT * FROM dogs", $connection);
    if(!$result) {
        die("Database query failed: " . mysql_error());
        }

     // 4. Use returned data
           while ($row = m开发者_如何学JAVAysql_fetch_array($result)) {
                echo $row[0]." ".$row[1]."<br />";
           }
  ?>
</body>
</html>
<?php
// 5. Close connection
   mysql_close($connection);
?>

PHP Function in separate file:

<?php

function db_conn() {

    // 1. Create database connection
    $connection = mysql_connect("localhost","website_admin","p@ssw0rd");
    if(!$connection) {
        die("Database connection failed: " . mysql_error());
        }

    // 2. Select a database to use
    $db_select = mysql_select_db("website_db", $connection);
    if(!$db_select) {
        die("Database selection failed: " . mysql_error());
       }
}
?>

Thanks! Tom

you never return $connection from your function, nor do you capture it as a variable in the top file.

Change this:

<?php
include("../model/db_conn.php");

db_conn();

To This:

 <?php
include("../model/db_conn.php");

$connection = db_conn();

And add

return $connection;

to your function.

Your problem here is, that the $connection variable isn't global. Which means it's only "visible" in the context of function db_conn() and not in the main program.

There are three simple solutions to your problem:

Add 'global $connection;' as the first command to your function.
You don't have to provide the MySQL connection resource to the mysql_query function if you've just opened one connection. So 'mysql_query( "SELECT * FROM dogs" );' would work too.
You could also return the $connection variable from your function.

It looks like $connection is created as a local variable to the db_conn() function. You could declare it global so that other code has access to it:

function db_conn()
{
    global $connection;

    $connection = mysql_connect( ... );

}

// now $connection is available everywhere

note: Most programming languages have caveats regarding misusing global variables. See Global Variables Considered Harmful by W.A. Wulf and M. Shaw (ACM SIGPLAN, 1973).