Question

我想在jqgrid中添加自己的列名，我也希望防止php-jqgrid根据sql查询自动添加的列名。

我正在使用php-jqgrid。

实际上jqgrid从mysql表的列名中获取列名，而我的db表有大约55列，但所有列都是进一步计算所必需的。所以我想只打印12个选定的列，而不是所有其他的休息列。

我正在禁用其他列： -

$grid->setColProperty("lat", array("hidden"=>true));
$grid->setColProperty("lng", array("hidden"=>true));
$grid->setColProperty("clinic_id", array("hidden"=>true));
$grid->setColProperty("id", array("hidden"=>true));
.....
.....
.....

但是设置每个列隐藏/禁用是很复杂的。需要为每个列设置隐藏/禁用代码。

是否有任何排序方法可以隐藏/禁用网格中的休息列。

我正在使用此代码来执行此操作，但它也获取了我未在方法$ grid-＆gt; setColModel（null，null，$ mylabels）中声明的列的名称;

任何人都可以告诉我应该写什么短代码来删除jqgrid中额外添加的列。

非常感谢。

require_once '/var/www/html/zbajtmp/public/jqgrid/jq-config.php';
// include the jqGrid Class
require_once "/var/www/html/zbajtmp/public/jqgrid/php/jqGrid.php";
// include the driver class
require_once "/var/www/html/zbajtmp/public/jqgrid/php/jqGridPdo.php";
// Connection to the server
$conn = new PDO(DB_DSN,DB_USER,DB_PASSWORD);
// Tell the db that we use utf-8
$conn->query("SET NAMES utf8");

// Create the jqGrid instance
$grid = new jqGridRender($conn);
 // Write the SQL Query
//$grid->SelectCommand = 'SELECT OrderID, OrderDate, CustomerID, Freight, ShipName  FROM orders';
$grid->SelectCommand = 'SELECT *  FROM clinic';
// set the ouput format to json
$grid->dataType = 'json';
// Let the grid create the model
//$grid->setColModel();
$mylabels = array(
"clinic_name"=>"Clinic ame",
"clinic_address"=>"Address",
"HomePhone"=>"Home Phone",
"WorkPhone"=>"Work Phone",
"Email_Id"=>"Email",
);
// Let the grid create the model with the desired labels
$grid->setColModel(null, null, $mylabels);

$grid->setColProperty("lat", array("hidden"=>true));
$grid->setColProperty("lng", array("hidden"=>true));
$grid->setColProperty("clinic_id", array("hidden"=>true));
$grid->setColProperty("id", array("hidden"=>true));
.....
.....
.....
// Set the url from where we obtain the data
//$grid->setUrl('/var/www/html/zbajtmp/application/views/scripts/clinic/grid.php');
$grid->setUrl('http://sunlinux/zbajtmp/application/views/scripts/clinic/grid.php');
// Set grid caption using the option caption
$grid->setGridOptions(array(
    "caption"=>"This is my custom Caption...",
    "rowNum"=>50,
    "sortname"=>"id",
    "hoverrows"=>true,
    "rowList"=>array(20,50,100,1000),
    "width"=>"100%",
    "height"=>350,
"footerrow"=>true,
"rownumbers"=>true,
"multiselect"=>true,
"altRows"=>true,
"altclass"=>'clsalt',
"loadtext"=>"<div class='loadingbox'>Please wait. Loading...</div>",

    ));


$grid->renderGrid('#grid','#pager',true, null, null, true,true);


$conn = null;

非常感谢。

Answer 1

jqGrid使用colModel构建selectCommand - 因此您需要隐藏手动呈现不需要的额外列 - 我建议您从{{*中删除select 1}}语句，只需选择所需的列。

从jqgrid中删除列

1 个答案: