How can I get jqGrid toolbar search working?
Solution 1
The filterToolbar method should be called on the same element which you use to define the grid. Look at the working example used it.
I can't help you with the PHP part of your question, because I don't use PHP myself. Nevertheless the demo files from the jqGrid download page seems to contain some PHP code examples which could be helpful for you.
Solution 2
Thanks to the previous author for the starting point for the problem solution. Here is ready to use piece of the server-side PHP
code implementing search request (from jqGrid
) processing:
$filters = $_POST['filters'];
$search = $_POST['_search'];
$where = "";
if(($search==true) &&($filters != "")) {
$filters = json_decode($filters);
$where = " where ";
$whereArray = array();
$rules = $filters->rules;
$groupOperation = $filters->groupOp;
foreach($rules as $rule) {
$fieldName = $rule->field;
$fieldData = mysql_real_escape_string($rule->data);
switch ($rule->op) {
case "eq":
$fieldOperation = " = '".$fieldData."'";
break;
case "ne":
$fieldOperation = " != '".$fieldData."'";
break;
case "lt":
$fieldOperation = " < '".$fieldData."'";
break;
case "gt":
$fieldOperation = " > '".$fieldData."'";
break;
case "le":
$fieldOperation = " <= '".$fieldData."'";
break;
case "ge":
$fieldOperation = " >= '".$fieldData."'";
break;
case "nu":
$fieldOperation = " = ''";
break;
case "nn":
$fieldOperation = " != ''";
break;
case "in":
$fieldOperation = " IN (".$fieldData.")";
break;
case "ni":
$fieldOperation = " NOT IN '".$fieldData."'";
break;
case "bw":
$fieldOperation = " LIKE '".$fieldData."%'";
break;
case "bn":
$fieldOperation = " NOT LIKE '".$fieldData."%'";
break;
case "ew":
$fieldOperation = " LIKE '%".$fieldData."'";
break;
case "en":
$fieldOperation = " NOT LIKE '%".$fieldData."'";
break;
case "cn":
$fieldOperation = " LIKE '%".$fieldData."%'";
break;
case "nc":
$fieldOperation = " NOT LIKE '%".$fieldData."%'";
break;
default:
$fieldOperation = "";
break;
}
if($fieldOperation != "") $whereArray[] = $fieldName.$fieldOperation;
}
if (count($whereArray)>0) {
$where .= join(" ".$groupOperation." ", $whereArray);
} else {
$where = "";
}
}
// evaluating $sidx, $sord, $start, $limit
$SQL = "SELECT id, brandName, name, description FROM products".$where." ORDER BY $sidx $sord LIMIT $start , $limit";
$result = mysql_query( $SQL ) or die("Couldn't execute query.".mysql_error());
Solution 3
You may want to try this code for simplest case:
$filters = $_GET['filters'];
$where = "";
if (isset($filters)) {
$filters = json_decode($filters);
$where = " where ";
$whereArray = array();
$rules = $filters->rules;
foreach($rules as $rule) {
$whereArray[] = $rule->field." like '%".$rule->data."%'";
}
if (count($whereArray)>0) {
$where .= join(" and ", $whereArray);
} else {
$where = "";
}
}
Before using in production make sure you are handling cases when $_GET['filters'] contains garbage instead of json, and field names/values are properly escaped. Otherwise there are plenty of space for SLQ injections.
Christos Hayward
Jonathan Hayward is a recovering geek. He holds master's degrees bridging math and computer science (UIUC) and philosophy and theology (Cambridge), and is considered to be in the profoundly gifted range. He is presently learning Node and Russian. Read full biography—it's interesting.
Updated on July 09, 2022Comments
-
Christos Hayward almost 2 years
At http://trirand.com/blog/jqgrid/jqgrid.html, under "New in Version 3.7" > "Column Search", there is a method explained to search, and it hasn't worked yet for me. I've added:
jQuery("#toolbar").jqGrid('filterToolbar',{stringResult: true,searchOnEnter : false});
and less essential code from the example. My server saw slightly different JSON requests, but no
_search=true
and no search term, ever.http://trirand.com/blog/jqgrid/jqgrid.html also gives an incomplete example of server-side code. The SQL statement is given in the example PHP:
$SQL = "SELECT item_id, item, item_cd FROM items ".$where." ORDER BY $sidx $sord LIMIT $start , $limit";
but, while
$sidx
,$sord
,$start
, and$limit
all have code to define them,$where
is not defined (or referenced) anywhere else on the page.How can I get a column search like the page documents, where my server is being hit by the appropriate requests?
-
jeffery_the_wind over 11 yearsYes the PHP code on the demo page for jqgrid is not correct, because it doesn't incorporate any of the searching functionality. References an undefined
$where
clause variable for the sql... :-/