11-09-2009, 11:51 PM
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#13 (permalink)
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The Addict
Join Date: May 2009
Posts: 287
Thanks: 5
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PHP Code:
// Take this out when you're ready to make the code live.
error_reporting(E_ALL);
// Connect to server and select database.
$hostname = 'CHANGE ME';
$username = 'CHANGE ME';
$password = 'CHANGE ME';
$database = 'CHANGE ME';
$link = mysql_connect($host, $username, $password) || die(mysql_error());
// mysql_select_db($database, $link)or die(mysql_error());
// Set some default data.
$name = 'John';
$lastname = 'Doe';
$email = 'user@example.com';
$id = 123145;
// update data in mysql database
$sql = "UPDATE " . $database .
" SET name='" . mysql_real_escape_string($name) .
"', lastname='" . mysql_real_escape_string($lastname) .
"', email='" . mysql_real_escape_string($email) .
"' WHERE id='" . mysql_real_escape_string($id) . "'";
$result = mysql_query($sql, $link);
Ok, wow, I was stupid and escaped $db_name. This will cause MySQL to try to update the data into a nonexistence/new database. Sorry!!
I would recommend the code in the post above me ( sketchMedia's), or use this code, it's just easier for me to read. I was going through the code trying to fix any mundane bug and I found that I was escaping the table name. |
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