Question

在http://trirand.com/blog/jqgrid/jqgrid.html，“版本3.7中的新功能”＆gt; “列搜索”，有一种方法被解释为搜索，但它还没有为我工作。我添加了：

jQuery("#toolbar").jqGrid('filterToolbar',{stringResult: true,searchOnEnter : false});

来自示例的不太重要的代码。我的服务器看到了稍微不同的JSON请求，但没有_search=true，也没有搜索词。

http://trirand.com/blog/jqgrid/jqgrid.html也提供了服务器端代码的不完整示例。 SQL语句在PHP示例中给出：

$SQL = "SELECT item_id, item, item_cd FROM items ".$where." ORDER BY $sidx $sord LIMIT $start , $limit";

但是，虽然$sidx，$sord，$start和$limit都有代码来定义它们，但$where未在任何地方定义（或引用）或者在页面上。

如何获取列表搜索，例如页面文档，我的服务器受到相应请求的影响？

Answer 1

应该在用于定义网格的同一元素上调用filterToolbar方法。看the working example使用它。

我无法帮助您处理问题的PHP部分，因为我自己不使用PHP。然而，来自demo files的the jqGrid download page似乎包含一些可能对您有帮助的PHP代码示例。

Answer 2

感谢前一位作者提出问题解决方案的起点。这里准备使用server-side PHP代码实现搜索请求（来自jqGrid）处理：

$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());

Answer 3

您可能希望在最简单的情况下尝试此代码：

        $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 = "";
            }
        }

在生产中使用之前，请确保在$ _GET ['filters']包含垃圾而不是json时处理案例，并且正确转义字段名称/值。否则SLQ注射有足够的空间。

Answer 4

感谢发布您的代码！

唯一的变化是我必须在'filters'参数中使用双引号才能使其正常工作：

$ filters = str_replace（'\“'，'”'，$ _ POST ['filters']）;

如何让jqGrid工具栏搜索工作？

4 个答案: