CODEWITHSUNDEEP
javascriptphpchart.jschartjs-2.6.0
Kwiznix
Kwiznix

Reputation: 23

ChartJS sees date in string as numbers

I am trying to get some strings from a csv file, to put under a graph from chartjs.

I wrote this php function:

function getTitle($id) {
 $stations = fopen("stations.csv", 'r');
 //$csv_data = fopen(date("G") . '.csv', 'r');
 $csv_data = fopen('16.csv', 'r'); //fopen("/mnt/weather-data/" . 
 date("Y/m/d/G") . ".csv", 'r');
 $data = array();
 $DAY_LENGTH = 60 * 60;
 $currentDate = date('U');

while (($line = fgetcsv($csv_data)) !== FALSE) {
    $date = date('U', strtotime($line[1]));
    if ($date >= ($currentDate-$DAY_LENGTH) && $line[2] == $id) {
        array_push($data, $line[0]);
    }
}

for ($i = 0; $i < count($data); $i++) {
    echo $data[$i] . ",";
}
fclose($stations);
fclose($csv_data);
}

which is working fine, everywhere I call this function, it prints the necesarry strings, which is a date, so 2018-02-08 for example.

When I try to call this function in the label part where i need them in ChartJS. it returns this date as a sum, so like 2018-02-08 will return as 2008.

labels: [<?php getTitle(($_GET['id']))?>],

What is it what I'm doing wrong?

Another problem when I try to change $line[0] to $line[1] which is a time with the following format: 20:17:36, my whole application stops working..

Hope anyone can help me!!

Upvotes: 0

Views: 380

Answers (1)

Joe Pi
Joe Pi

Reputation: 382

The Problem with your code is that the echo will just output the content of the variable but as you are parsing it's value into a javascript array you need to surround your value with quotation marks. So if you change

for ($i = 0; $i < count($data); $i++) {
    echo $data[$i] . ",";
}

into

for ($i = 0; $i < count($data); $i++) {
    echo "'" . $data[$i] . "',";
}

it should work. This will also solve your other problem. So with this change you should also be able to use the time.

Upvotes: 1

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