How do you remove last comma in a PHP for loop to build a JSON string [duplicate]

Question

When building a JSON string from a php for loop from an SQL query there is an unnessary comma at the end of the build string. How do I control the final comma.

$count =  label::grabAll()->count();

echo '{"data": { "graph": {';

for ($x = 0; $x < $count; $x++){

if($x <= 3){
    $cm = ',';
}else{
    $cm .= '';
}

    echo '"'.$x.'": "'.label::grabAll()->results()[$x]->count.'"'.$cm;

}

I get this result from the code above.

How do I remove the final comma?

I recommend not building JSON strings by hand. Instead, populate an array and use json_encode() — Phil
– Phil, Commented Jun 5, 2017 at 23:12
what @Phil says. If you really want to concat data like that to s string rather use implode() with an array. (for any other usecase. just as a hint for the future) — Jeff
– Jeff, Commented Jun 5, 2017 at 23:15

Phil · Accepted Answer · 2017-06-05 23:26:17Z

1

Use plain old objects / arrays and use json_encode. For example

$graph = array_map(function($result) {
    return $result->count;
}, label::graball()->results());

echo json_encode(['data' => ['graph' => $graph]]);

Demo ~ https://eval.in/812302

edited Jun 5, 2017 at 23:26

answered Jun 5, 2017 at 23:17

Phil

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Blackspade · Accepted Answer · 2017-06-05 23:24:47Z

0

$count =  label::graball()->count();

$arr = array();

for ($x = 0; $x < $count; $x++){

 array_push($arr, label::graball()->results()[$x]->count);

}


echo json_encode($arr);

Thanks everyone for the quick responses. This works.

answered Jun 5, 2017 at 23:24

Blackspade

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1 Comment

Phil Over a year ago

Are you sure that executing label::graball() and label::graball()->results() multiple times is the correct thing to do? Is each execution querying your data source?

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