Am trying to build a dynamic table. Where the backend holds the colModel and the table data.
For example:
colModel:[
{
name:'Forename',
index:'forename',
align: 'left'
},
{
name:'Surname',
index:'surname'
},
{
name:'Date Of Birth',
index:'dob',
width: 80
}
]
is on the front end but wanna be able to construct it via PHP.
My Answer:
Yes, it is possible to pass colModel to jQgrid from a PHP backend. You can generate the colModel array in PHP and then pass it to the frontend as a JSON object. Here is an example of how you can do this:
In your PHP backend, you can generate the colModel array like this:
php
$colModel = array(
array(
'name' => 'Forename',
'index' => 'forename',
'align' => 'left'
),
array(
'name' => 'Surname',
'index' => 'surname'
),
array(
'name' => 'Date Of Birth',
'index' => 'dob',
'width' => 80
)
);
echo json_encode($colModel);
Then, in your frontend, you can make an AJAX request to fetch the colModel array from the PHP backend and use it to initialize the jQgrid:
javascript
$.ajax({
url: 'your_php_backend_url.php',
dataType: 'json',
success: function(data) {
$('#your_grid_id').jqGrid({
colModel: data,
// other jQgrid options
});
}
});
This way, you can dynamically generate the colModel array in PHP and pass it to the frontend for use in jQgrid.
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