I have the following code for exporting all the items in one of my pods to json. The thing is I don't need all the 130 columns in the json file, but only about 20. Since this will be done for about 150 items I thought I could save some loading time by not printing out all the fields, but I do not know how to do this. For example I only want to print the column value named 'title' for all items in the pod. My code is attached bellow.
<?php
$pods = pods('name', array('orderby' => 'name asc', 'limit' => -1));
$all_companies = $pods->export_data();
if ( !empty( $all_companies ) ) {
die(json_encode($all_companies);
}else{
die(json_encode(array('error' => 'No cars found.')));
}
?>
I thought about doing something like this:
if ( 0 < $all_companies->total() ) {
while ($all_companies->fetch()) {
$json .= $all_companies->field('title');
}
$json = rtrim($json, ",");
$json .= '}}';
}
echo $json;
But it doesn't work and also the code becomes very long.
I'd make an array of the names of the twenty fields you want then build an array of those fields for each item, by doing a foreach of those field names passed to Pods::field() inside the while loop. Like this:
$pods = pods('name', array('orderby' => 'name asc', 'limit' => -1));
$fields = array( 'field_1', 'field_2' );
if ( $pods->total() > 0 ) {
while ( $pods->fetch() ) {
foreach ( $fields as $field ) {
$json[ $pods->id() ] = $pods->field( $field );
}
}
$json = json_encode( $json );
}
Alternatively, you could hack the /pods/<pod> endpoint of our JSON API to accept a list of fields to return as the body of the request. Wouldn't be hard to do, make sure to submit a pull request if you make it work.