Please help with conditional defining grid

QuestionsPlease help with conditional defining grid
Michael W Olson asked 2 years ago

Why does this work:

$whichGrid=”fuel_usage.php”;
$myTable=”fuel_usage”;
include(“grids/”.$whichGrid);

But this doesn’t?

$whichGrid=”fuel_usage.php”;
if(isset($_POST[‘section_loc’])){
$myTable=”fuel_usage”;
}
include(“grids/”.$whichGrid);

1 Answers
Abu Ghufran Staff answered 2 years ago

Hello,

Perhaps, You need to move the include inside the ‘if’ statement.
e.g.

$whichGrid=”fuel_usage.php”;
if(isset($_POST[‘section_loc’])){
    $myTable=”fuel_usage”;
    include(“grids/”.$whichGrid);
}

Basically, setting table property is necessary for data to show. Currently you are just unsetting the table and all grid code is still in execution.
I don’t know how you are implementing file fuel_usage.php, If above does not solve, you can share that file as well and then I can suggest better.

_________________________
Abu Ghufran - Dev Team
Grid 4 PHP Framework
 
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