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PHP - Help On Connecting To Db
Hi there,
I'm working on a project already started, and using WAMP+WINDOWS7+SQL2008R2, At this point it's connecting to a MSSQL DATABSE LIKE THIS: $dns = "TempGes2"; $con = odbc_connect($dns, $user, $pwd); This works fine, but i can't find where is the TempGes2 configured, it isn't in the odbcad32.exe, Any ideas, i need to point this to another DATABASE, Thank you in advance Similar TutorialsHi, my school say not to use XAMPP, and I already have PHP installed, MySQL installed, Apache I am not sure, but assuming if I have it installed, how do I connect to localhost b/c beforer when I use XAMPP, the default for where to get php/mysql running was: C:\xampp\htdocs\ Please any help appreciated! So I'm currently watching this tutorial : http://youtu.be/9E0s4gsUeU0 And am trying to build the sample application shown there. However I do seem to run into some problems, so if someone could take the time to help me out; I would be very grateful! --- The application is a simple HTML form, which takes the data input and stores it into a MySQL database table. My project is organized as this : The actual form is stored on the insert.php file. Code: [Select] <?php include_once('resources/init.php'); if (isset($_POST['title'], $_POST['post'])) { // functions go here! } ?> <html> <head> <title> Post something! </title> </head> <body> <div id="form"> <form method="POST" action= " "> <label for="title">Title:</label><br> <input type="text" name="title" id="title" /><br> <label for="post">Post:</label><br> <textarea name="post" id="post" rows="15" cols="50"></textarea><br> </form> </div> </body> </html> The config.php and init.php are the configuration and initialize files. config.php Code: [Select] <?php $config['DB_HOST'] = 'localhost'; $config['DB_USER'] = 'root'; $config['DB_PASS'] = ''; $config['DB_NAME'] = 'form'; //foreach ( $config as $k => $v ) { // define(strtouper($k), $v); //} ?> init.php Code: [Select] <?php include_once('config.php'); mysql_connect(DB_HOST, DB_USER, DB_PASS); mysql_select_db(DB_NAME); ?> --- When I try this in my browser I get the error : Code: [Select] Warning: mysql_connect() [function.mysql-connect]: Unknown MySQL server host 'DB_HOST' (11004) in C:\xampp\htdocs\form\resources\init.php on line 5 Warning: mysql_select_db() [function.mysql-select-db]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\form\resources\init.php on line 6 Warning: mysql_select_db() [function.mysql-select-db]: A link to the server could not be established in C:\xampp\htdocs\form\resources\init.php on line 6 Could someone tell me what is wrong? Thanks in advance! I have the following code created, however when i run it it doesn't come back as saying connected successfully?
<?php
$host = "localhost";
$my_user = "root";
$my_password = "";
$my_db = "notes_db"
$con = mysqli_connect("localhost","my_user","my_password","my_db");
echo ("Connected Successfully");
?>
Can someone tell me why in the following code, in the 'try' section, this code works: (first line after 'try')
$conn = new PDO("mysql:host=$servername;dbname=$dbasename", $username, $password); even though I have not declared a variable named $dbname,
but if I change it to "$databasename", which I have declared, it doesn't work. I don't understand that.
<code>
<?php Hey all
I'm working on a script which needs to connect to the Google Calendar API, but when I run the code I get
Fatal error: Uncaught exception 'Google_Auth_Exception' with message 'Error refreshing the OAuth2 token, message: '{ "error" : "invalid_grant" }'' in /home/public_html/cal_test/google-api-php/src/Google/Auth/OAuth2.php:341 Stack trace: #0 /home/public_html/cal_test/google-api-php/src/Google/Auth/OAuth2.php(300): Google_Auth_OAuth2->refreshTokenRequest(Array) #1 /home/public_html/cal_test/google-api-php/src/Google/Auth/OAuth2.php(230): Google_Auth_OAuth2->refreshTokenWithAssertion() #2 /home/public_html/cal_test/google-api-php/src/Google/Service/Resource.php(171): Google_Auth_OAuth2->sign(Object(Google_Http_Request)) #3 /home/public_html/cal_test/google-api-php/src/Google/Service/Calendar.php(1133): Google_Service_Resource->call('list', Array, 'Google_Service_...') #4 /home/public_html/cal_test/index.php(34): Google_Service_Calendar_CalendarList_Resource->listCalendarList() #5 {main} thrown in /home/public_html/cal_test/google-api-php/src/Google/Auth/OAuth2.php on line 341
The code i'm using is
<?php
error_reporting(-1);
ini_set('display_errors', 'On');
require_once "google-api-php/src/Google/Client.php";
require_once "google-api-php/src/Google/Service/Calendar.php";
// Service Account info
$client_id = 'CLIENT_ID';
$service_account_name = 'EMAIL';
$key_file_location = 'FILE_LOC';
// Calendar id
$calName = 'CAL_ID';
$client = new Google_Client();
$client->setApplicationName("Calendar test");
$service = new Google_Service_Calendar($client);
$key = file_get_contents($key_file_location);
$cred = new Google_Auth_AssertionCredentials(
$service_account_name,
array('https://www.googleapis.com/auth/calendar.readonly'),
$key
);
$client->setAssertionCredentials($cred);
$cals = $service->calendarList->listCalendarList();
//print_r($cals);
//exit;
$events = $service->events->listEvents($calName);
foreach ($events->getItems() as $event) {
echo $event->getSummary();
}
?>
i've spent serveral hours trying to locate a fix on Google, but so far i'm stuggling
Can anyone help me?
Thanks
Edited by dweb, 09 October 2014 - 06:58 AM. Hi all.
My database just stopped connecting? I have tried everything i could but no way! What could have happened and what can i do? I even went as far as creating the database connection inside the page to see if it will work but still no way!
Thanks Windows 2012 R2 Web Farm IIs 8.5 PHP 7.2
I am trying to create a logon page using PHP create a database on my SQL 2014 server. Installed the PHP SQL Server Drivers for PHP 7.2
<?php
// Connect to MSSQL
if (!$link) { I get an HTTP 500 internal server error cant get past this one Any thing else I can provide to help us fix this let me know
Thank you
Tom Hello again guys! What if i have 2 databases, db1 and db2. i want to read raw data from db1 and match it's content with data from db2, then produce some information. How can i do it at the same time? I know that to be able to access mysql data, we have to use the mysql_connect() and mysql_select_db(). mysql_select_db() only allows to connect to single database. what should i do? thanks for your help in advance. For some reason Im not able to connect to mysql database, when i fill in the form and select search, it just basically refreshes the page but does not come up with no error messages or any results from my database. any help is appreciated. HTML Code Code: [Select] <table id="tb1"> <tr> <td><p class="LOC">Location:</p></td> <td><div id="LC"> <form action="insert.php" method="post"> <select multiple="multiple" size="5" style="width: 150px;" > <option>Armley</option> <option>Chapel Allerton</option> <option>Harehills</option> <option>Headingley</option> <option>Hyde Park</option> <option>Moortown</option> <option>Roundhay</option> </select> </form> </div> </td> <td><p class="PT">Property type:</p></td> <td><div id="PS"> <form action="insert.php" method="post"> <select name="property type" style="width: 170px;"> <option value="none" selected="selected">Any</option> <option value="Houses">Houses</option> <option value="Flats / Apartments">Flats / Apartments</option> </select> </form> </div> </td><td> <div id="ptype"> <form action="insert.php" method="post"> <input type="radio" class="styled" name="ptype" value="forsale"/> For Sale <p class="increase"> <input type="radio" class="styled" name="ptype" value="forrent"/> To Rent </p> <p class="increase"> <input type="radio" class="styled" name="ptype" value="any"/> Any </p> </form> </div> </td> </tr> </table> <div id="table2"> <table id="NBtable"> <tr> <td><p class="NBS">Number of bedrooms:</p></td> <td><div id="NB"> <form action="insert.php" method="post"> <select name="number of bedrooms"> <option value="none" selected="selected">No Min</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> to <select name="number of bedrooms"> <option value="none" selected="selected">No Max</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> </form> </div> </td> <td><p class="PR">Price range:</p></td> <td><div id="PR"> <form action="insert.php" method="post"> <select name="price range"> <option value="none" selected="selected">No Min</option> <option value="50,000">50,000</option> <option value="60,000">60,000</option> <option value="70,000">70,000</option> <option value="80,000">80,000</option> <option value="90,000">90,000</option> <option value="100,000">100,000</option> <option value="110,000">110,000</option> <option value="120,000">120,000</option> <option value="130,000">130,000</option> <option value="140,000">140,000</option> <option value="150,000">150,000</option> <option value="160,000">160,000</option> <option value="170,000">170,000</option> <option value="180,000">180,000</option> <option value="190,000">190,000</option> <option value="200,000">200,000</option> <option value="210,000">210,000</option> <option value="220,000">220,000</option> <option value="230,000">230,000</option> <option value="240,000">240,000</option> <option value="250,000">250,000</option> <option value="260,000">260,000</option> <option value="270,000">270,000</option> <option value="280,000">280,000</option> <option value="290,000">290,000</option> <option value="300,000">300,000</option> <option value="310,000">310,000</option> <option value="320,000">320,000</option> <option value="330,000">330,000</option> <option value="340,000">340,000</option> <option value="350,000">350,000</option> </select> to <select name="price range"> <option value="none" selected="selected">No Max</option> <option value="50,000">50,000</option> <option value="60,000">60,000</option> <option value="70,000">70,000</option> <option value="80,000">80,000</option> <option value="90,000">90,000</option> <option value="100,000">100,000</option> <option value="110,000">110,000</option> <option value="120,000">120,000</option> <option value="130,000">130,000</option> <option value="140,000">140,000</option> <option value="150,000">150,000</option> <option value="160,000">160,000</option> <option value="170,000">170,000</option> <option value="180,000">180,000</option> <option value="190,000">190,000</option> <option value="200,000">200,000</option> <option value="210,000">210,000</option> <option value="220,000">220,000</option> <option value="230,000">230,000</option> <option value="240,000">240,000</option> <option value="250,000">250,000</option> <option value="260,000">260,000</option> <option value="270,000">270,000</option> <option value="280,000">280,000</option> <option value="290,000">290,000</option> <option value="300,000">300,000</option> <option value="310,000">310,000</option> <option value="320,000">320,000</option> <option value="330,000">330,000</option> <option value="340,000">340,000</option> <option value="350,000">350,000</option> </select> </form> </div> </td> </tr> </table> PHP Code Code: [Select] <?php $server = ""; // Enter your MYSQL server name/address between quotes $username = ""; // Your MYSQL username between quotes $password = ""; // Your MYSQL password between quotes $database = ""; // Your MYSQL database between quotes $con = mysql_connect($server, $username, $password); // Connect to the database if(!$con) { die('Could not connect: ' . mysql_error()); } // If connection failed, stop and display error mysql_select_db($database, $con); // Select database to use $result = mysql_query("SELECT * FROM tablename"); // Query database while($row = mysql_fetch_array($result)) { // Loop through results echo "<b>" . $row['id'] . "</b><br>\n"; // Where 'id' is the column/field title in the database echo $row['location'] . "<br>\n"; // Where 'location' is the column/field title in the database echo $row['property_type'] . "<br>\n"; // as above echo $row['number_of_bedrooms'] . "<br>\n"; // .. echo $row['purchase_type'] . "<br>\n"; // .. echo $row['price_range'] . "<br>\n"; // .. } mysql_close($con); // Close the connection to the database after results, not before. ?> To connect to my database I an entering mysql database details just at the first top lines server,username,password and database. is thats correct? Thank You for your help in adavance. Hi I am struggling to connect to my database and was wondering if someone could please help me. I am not sure what I need to put on line 10 in this code
<?php
define('DB_HOST', 'localhost');
define('DB_USER', 'root');
define('DB_PASS', '');
define('DB_NAME', 'tonyruttle');
define("PERPAGE", 10);
$mydb = (DB_HOST, DB_USER, DB_PASS, DB_NAME);
/***********************************************
** SEARCH FOR MATCHING PRODUCTS
************************************************/
$showResults = 0;
$search = '';
if (isset($_GET['srch'])) {
$search = $_GET['srch'];
$showResults = 1;
$srch = array_filter(array_unique(explode(' ', trim($_GET['srch']))));
array_walk($srch, function(&$v, $k) { // add the +'s to the search tags
$v = '+'.$v;
});
$tagparam = join(' ', $srch);
//
// FINDTOTAL RECORDS IN SEARCH RESULTS
//
$params[] = $tagparam;
$res = $db->prepare("SELECT COUNT(*) as tot
FROM television p
WHERE MATCH(title,keywords) AGAINST(? IN BOOLEAN MODE)
");
$res->execute( $params );
$total = $res->fetchColumn();
$page = $_GET['page'] ?? 1;
$params[] = ($page-1)*PERPAGE; // append parameters offset
$params[] = PERPAGE; // and number of records for limit clause
//
// GET A PAGEFUL OF RECORDS
//
$sql = "SELECT
id
, title
, active
FROM television p
WHERE MATCH(title,keywords) AGAINST(? IN BOOLEAN MODE)
ORDER BY TITLE
LIMIT ?,? ";
$stmt = $db->prepare($sql);
$stmt->execute($params);
if ($stmt->rowCount()==0) {
$results = "<h3>No matching records</h3>";
} else {
$results = "<tr><th>Product Id</th><th>Title</th><th>Active</th><th colspan='2'>Action</th></tr>\n";
foreach ($stmt as $rec) {
$cls = $rec['active']==0 ? "class='inactive'" : '';
$results .= "<tr $cls><td>{$rec['id']}</td><td>{$rec['title']}</td><td class='ca'>{$rec['active']}</td>
<td class='ca'><a href='?action=edit&id={$rec['id']}'><img src='images/edit-icon.png' alt='edit'></a></td>
</tr>\n";
}
}
}
?>
<div id='title'>Product List</div>
<form id='form1' ">
<fieldset>
<legend>Search titles and tags</legend>
<input type='text' name='srch' id='srch' size='50' value='<?=$search?>' placeholder='Search for...' >
<input type="hidden" name="page" id="page" value="1">
<input type="hidden" name="action" value="search">
<input type="submit" name="btnSub" value="Search">
</fieldset>
<div id='formfoot'></div>
</form>
<?php if ($showResults) { ?>
<div class="paginate_panel">
<?=page_selector($total, $page, PERPAGE)?>
</div>
<table border='1'>
<?=$results?>
</table>
<?php } ?>
Hi! I need to retrieve data from a WCF Web Service and don't seem to be connecting when the "new SoapClient" is called. I am using php version 5.2. The service requires 2 Parmameters, an account ID and a comma separated email list. Can someone please tell me what I am doing wrong? Thanks! Code: [Select] <?php header('Content-Type: text/xml'); $client = new SoapClient('http://www.zbestlistings.com/WCF/Primedia.svc?wsdl'); $response = $client->GetPropertyXML(array( "AccountID" => "12807175152", "eMailList" => "Sheryl@Rentingslc.com, Tenants@Rpmwestvalley.com, Tom@Rpmsouthernutah.com" )); echo $response; ?> Dear All, I am having 2 different DB on 2 different hosts. I am running MySQL Server on my local PC where the user is entering data in the tables. I have a website which has the identical DB on the web. I am able to connect to the database by using the codes on the server. I want to update the server DB with the local system DB by running one update command. I get the error "-SELECT command denied to user 'localusername'@'localhost' for table 'pst_data'" Given below is the code used for the process : //connecting the remote system DB $link = mysql_connect('IPAddress:3306', 'remoteusername', 'Password'); if (!$link) { die('Not connected : ' . mysql_error()); } $db_selected = mysql_select_db('remotedb', $link); if (!$db_selected) { die ('Can\'t use Remote System DB: ' . mysql_error()); } //connecting the local database on the webstite $weblink = mysql_connect("localhost","localusername","password"); if (!$weblink) { die('Not connected : ' . mysql_error()); } $webdb_selected = mysql_select_db('localwebdb', $weblink); if (!$webdb_selected) { die ('Can\'t use WebServer Database : ' . mysql_error()); } //query to fetch the records from remote ystem and insert into local website database $upd_Query=mysql_query("INSERT INTO localwebdb.`table` SELECT * FROM remotedb.`table` where field=' some condition' "); --------------------------------------------------- I have tested that I have connected the remote DB by running queries on the webserver. Could anyone bail me out so that i can copy the DB from remote to local Any help would be appreciated hi, i`m trying to connect a cms to a gateway but in my country i cant use paypal or any other known gateways so i use a local bank for payments . i want to change cms`s paypal gateway to the local one these are the sample codes they gave me : sender.php
: send.php
and this one is for cms `s paypal gate way
define("_VALID_PHP", true);i`ll be grateful if you help me on this hello,
uhm, im trying to make a log-in page that will connect to the areas wifi. i can make a log-in page using php and mysql but i was wondering how to make so that when the user "logs in" he/she will be connected to a certain areas wifi.
for example, our school has a wifi hot spot in which, when you try to connect to it your'e sent to a log-in page to determine whether your'e a student (or facilitator). if you are, once you log - in your'e sent to google page and then you can use the schools wifi. if you aren't then you can't connect to it.
so now my question is how to i make a log-in page that will allow a user to connect to wifi? do i have to contact the company (the one that manages web servers like pldt) for permission? are there any policies regarding this? can it be coded using php and others?
if you can tell me then please do. thank you in advance
So I've followed this code, corrected about 12 error, talked to my hosting and I am so done. I have tried everything but the error won't go away. The code is pasted below. As it's really late any help would be appreciated that would make my day the next day...
<?PHP Hi I am trying to connect to my database but i am having some problems where the server( WAMP) crashes and does not load this page. here is the php coding
<?php
$db = new PDO("mysql:host=localhost;dbname=test;port=80", "root", "");
var_dump($db);
when i run this no error comes the page doesnt load. anyone know why? when using PDO do i have to do some sort of configuration changes to wamp? i have even changed the directory of the file and still same issue. any suggestions would be wonderful.
<?php $username="a7564065_si"; $password="*****"; $database="a7564065_food"; mysql_connect(mysql4.000webhost.com,$username,$password); ... I am trying to connect to mysql but having trouble. The host name: mysql4.000webhost.com I get the error: Parse error: syntax error, unexpected T_DNUMBER in /home/a7564065/public_html/Index.html on line 17 Any ideas? Im trying to connect to a database from php. Heres the code: <?php $dbc = mysqli_connect('192.168.0.122', 'boyyo', 'KiaNNa11', 'aliendatabase') or die('Error connecting to MySQL server.'); $query = "INSERT INTO aliens_abduction (first_name, last_name, " . "when_it_happened, how_long, how_many, alien_description, " . "what_they_did, fang_spotted, other, email) " . "VALUES ('Sally', 'Jones', '3 days ago', '1 day', 'four', " . "green with six tentacles', 'We just talked and played with a dog', " . "'yes', 'I may have seen your dog. Contact me.', " . "'sally@gregs-list.net')"; $result = mysqli_query($dbc, $query) or die('Error querying database.'); ?> I think i have a theory that my MySQL server location is wrong but i dont know. I use HostGator to do this and im using Phpmyadmin. But everytime i type in a form that i created it says Error querying database. Can someone tell me whats wrong with this code. Oh by the way im using head first into PHP and MySQL when I try to connect to my database with CLI I get this error Code: [Select] Access denied for user 'bommobco_cwbtest'@'cpe-123.45.789.123.neo.res.rr.com' (using password: YES) 123.45.789.123 being my ip address, how can I fix this issue so I can connect? If the script is not uploaded onto the server you cant use $dbc = mysql_connect('localhost', 'username', 'pass'); so how could I connect in cli? this did not work $dbc = mysql_connect('http://123.456.789.10', 'username', 'pass'); So what can I do? Thanks |
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