Jelson Bonilla asked 9 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);
?>
Abu Ghufran answered 9 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 9 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">
<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?
