I'm using jqGrid, on edit/add function I want to have a drop down list in one of these fields.
This works if i use the setSelect function as this:
$grid->setSelect("title", "SELECT DISTINCT name,name as TestingName FROM template", true, true, false, array(""=>"All"));
How can I pass parameters to my query? I tried these:
1-"SELECT DISTINCT name,name as TestingName FROM template where tempid = ?"
2- "SELECT DISTINCT name,name as TestingName FROM template where tempid = $rowid"
3-"SELECT DISTINCT name,name as TestingName FROM template where tempid = ". $rowid
none of the above worked while having:
if(isset ($_REQUEST["tempid"]))
$rowid = jqGridUtils::Strip($_REQUEST["tempid"]);
else
$rowid = "";
If I correct understand your question you use editoptions with dataUrl
. You want to have the URL which has additional parameter tempid
which value should be the rowid of the current selected row.
From the syntax of your question I suppose that you use some commercial jqGrid for PHP product from trirand.net. In the case you should use tag [jqgrid-php]. jqGrid is pure JavaScript open source product. So I answer how you can add dataUrl
parameter in JavaScript.
jqGrid has ajaxSelectOptions option which can be used to modify the jQuery.ajax
options of the call which use dataUrl
. You can do following
var myGrid = $("#list");
myGrid.jqGrid({
// all your current parameters of jqGrid and then the following
ajaxSelectOptions: {
data: {
tempid: function () {
return myGrid.jqGrid('getGridParam', 'selrow');
}
}
}
});
If the data
parameter of jQuery.ajax contain a method instead of a property the method will be called every time of the corresponding jQuery.ajax
call. I used the same trick in the answer.
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