Unable to connect to database

QuestionsUnable to connect to database
Jelson Bonilla asked 5 years ago

I am getting the error below.

Couldn't execute query. No database selected – SELECT * FROM clients WHERE 1=1 LIMIT 1 OFFSET 0

——————————————————————————–
Config:

<?php
$db_conf = array();
$db_conf["type"] = "mysqli";
$db_conf["server"] = "localhost"; // or your mysql ip
$db_conf["user"] = "root"; // username
$db_conf["password"] = "test"; // password
$db_conf["database"] = "griddemo"; // database

// include and create object
$base_path = strstr(realpath("."),"demos",true)."lib/";
include($base_path."inc/jqgrid_dist.php");
$g = new jqgrid($db_conf);

?>

3 Answers
Abu Ghufran answered 5 years ago

This code looks fine. If you share complete code file it would help in debugging.

$g = new jqgrid($db);

$g->con->debug = 1; // changed from 0 to 1

Try after enabling debug to see exact error. Make sure credentials are correct.

Jelson Bonilla answered 5 years ago

This is what I have.

index.php

<?php
/**
* PHP Grid Component
*
* @author Abu Ghufran <[email protected]> – http://www.phpgrid.org
* @version 1.5.2
* @license: see license.txt included in package
*/

// include db config
include_once("config.php");

// set up DB
mysql_connect('localhost','user','pass');
mysql_select_db(PHPGRID_DBNAME);

// include and create object
include(PHPGRID_LIBPATH."inc/jqgrid_dist.php");
$g = new jqgrid();

// set few params
$grid["caption"] = "Sample Grid";
$g->set_options($grid);

// set database table for CRUD operations
$g->table = "clients";

// render grid
$out = $g->render("list1");

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"&gt;
<html>
<head>
<link rel="stylesheet" type="text/css" media="screen" href="lib/js/themes/redmond/jquery-ui.custom.css"></link>
<link rel="stylesheet" type="text/css" media="screen" href="lib/js/jqgrid/css/ui.jqgrid.css"></link>

<script src="lib/js/jquery.min.js" type="text/javascript"></script>
<script src="lib/js/jqgrid/js/i18n/grid.locale-en.js" type="text/javascript"></script>
<script src="lib/js/jqgrid/js/jquery.jqGrid.min.js" type="text/javascript"></script>
<script src="lib/js/themes/jquery-ui.custom.min.js" type="text/javascript"></script>
</head>
<body>
<div style="margin:10px">
<?php echo $out?>
</div>
</body>
</html>

config.php

<?php
$db_conf = array();
$db_conf["type"] = "mysqli";
$db_conf["server"] = "localhost"; // or your mysql ip
$db_conf["user"] = "root"; // username
$db_conf["password"] = ""; // password
$db_conf["database"] = "griddemo"; // database

// include and create object
$base_path = strstr(realpath("."),"demos",true)."lib/";
include($base_path."inc/jqgrid_dist.php");
$g = new jqgrid($db_conf);
$g->con->debug = 1; // changed from 0 to 1

?>

Still getting same error. How am I supposed to use the debug feature?

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