Error In Javascript code with php

Question

I wanna make statistics in my website for the last 3 years

I want to show result like this

2016 : 159
2015 : 132
2014 : 200

I try my code (this's)

$date2 = date('Y');
$n = $date2;
for($i=$n-2;$i<=$n;$i++) {
$sql = "SELECT sum(count) AS value_sum FROM statistics where YEAR(st_date) = $i ";
$sql_sel = mysqli_query($conn,$sql);
echo '
    <script>
    var pieData = [
   ';

while($rows = mysqli_fetch_assoc($sql_sel)) {
if($i == $n-2) {
echo '{
    value: '.$rows['value_sum'].',
    color:"#337AB7"
  },';
}
else if($i == $n-1) {
echo '{
    value: '.$rows['value_sum'].',
    color:"#FC8213"
    },';
  }
else if($i == $n) {
  echo '{
    value: '.$rows['value_sum'].',
    color:"#8BC34A"
  },';
}
echo'];
new Chart(document.getElementById("pie").getContext("2d")).Pie(pieData);
</script>';
}}
?>

but this code give me just 1 one row like this

2016 : 159

I wanna see all result, any help ?

PROTIP: You can build an array in PHP then use json_encode() to create a JSON string, which can actually be echoed into the JavaScript and parsed as an array/object. — gen_Eric
– gen_Eric, Commented Jun 15, 2016 at 15:28
@RocketHazmat I'm beginner Can you help me ? and I need to change color for Each year — sayousaad
– sayousaad, Commented Jun 15, 2016 at 15:31

aynber · Accepted Answer · 2016-06-15 15:33:16Z

1

You're creating 3 charts within your for loop. If you allow php to encode the data for you, you can echo the script outside of the PHP like so:

$date2 = date('Y');
$n = $date2;
$data = array();
for($i=$n-2;$i<=$n;$i++) {
    $sql = "SELECT sum(count) AS value_sum FROM statistics where YEAR(st_date) = $i ";
    $sql_sel = mysqli_query($conn,$sql);
    while($rows = mysqli_fetch_assoc($sql_sel)) {
        if($i == $n-2) {
            $data[] = array('value'=> $rows['value_sum'], 'color'=>'#337AB7');
        }
        else if($i == $n-1) {
            $data[] = array('value'=> $rows['value_sum'], 'color'=>'#FC8213');
        }
        else if($i == $n) {
            $data[] = array('value'=> $rows['value_sum'], 'color'=>'#8BC34A');
        }
    }
}
?>

<script>
    var pieData = <?php json_encode($data) ?>;
    new Chart(document.getElementById("pie").getContext("2d")).Pie(pieData);
</script>

answered Jun 15, 2016 at 15:33

aynber

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2 Comments

aynber Over a year ago

Does it throw any errors? What doesn't work about it?

aynber Over a year ago

If you look at the source code, does the script look correct?

MonkeyZeus · Accepted Answer · 2016-06-15 16:04:01Z

0

This should help you out:

<?php
// Store the array for the pie data within PHP at first
$js_pie_data = [];

// Define colors
$colors = [
    date('Y') => '8BC34A',  // current year
    (date('Y') - 1) => 'FC8213', // last year
    (date('Y') - 2) => '337AB7' // 2 years ago
];

$sql = "SELECT
            YEAR(st_date) as `year`, -- get year from DB, makes life simpler :)
            sum(count) AS value_sum
        FROM
            statistics
        WHERE
            YEAR(st_date) >= (YEAR(CURDATE())-2) -- get all records from 2 years ago to today
        GROUP BY
            YEAR(st_date) -- group by year so that our sum() above will work";

$sql_sel = mysqli_query($conn,$sql);

while($rows = mysqli_fetch_assoc($sql_sel))
{
    // keep adding entries to the pie data
    $js_pie_data[] = [
        'value' => $rows['value_sum'],
        'color' => '#'.$colors[$rows['year']] // based on year from DB, pick a color
    ];
}

echo '<script>
var pieData = '.json_encode($js_pie_data).'; // this will output a properly formatted JS array which will be understood by JS with no problem
new Chart(document.getElementById("pie").getContext("2d")).Pie(pieData);
</script>';

edited Jun 15, 2016 at 16:04

answered Jun 15, 2016 at 15:39

MonkeyZeus

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5 Comments

sayousaad Over a year ago

That's perfect, can I make the same for month ? without any problem ?

MonkeyZeus Over a year ago

@sayousaad I'm not sure what you mean but I'm sure it's possible.

Jay Blanchard Over a year ago

Why should this "help out" the OP? A good answer will always have an explanation of what was done and why it was done in such a manner, not only for the OP but for future visitors to SO. ¯\_(ツ)_/¯

MonkeyZeus Over a year ago

@JayBlanchard Would you like me to comment more thoroughly?

Jay Blanchard Over a year ago

Not so much comment but explain why you do what you do.

Dest · Accepted Answer · 2016-06-15 15:39:52Z

1)You can create array in php side.

2)Then convert it as json string.

3)Define javascript variable with that json string

4)Convert it as object with JSON.parse

5)Insert it in option

 <?php

    $db = Array("1","2","3");

    $out = Array();

    $x = 0;
        foreach($db as $vals){

            if($x == 1){
            $out[] = Array("value"=>$vals,"color"=>"#bfbfbf");
            }

            if($x == 2){
            $out[] = Array("value"=>$vals,"color"=>"#00ffff");
            }

            if($x == 3){
            $out[] = Array("value"=>$vals,"color"=>"#fff00");
            }


        if($x == 3){$x = 0;}
        $x++;
        }



    echo "
    <script language='javascript'> 

    var stat_str = '".json_encode($out)."';
    var stat_obj = JSON.parse(stat_str);

    // then you can insert stat_obj if you need object into stats
    </script>";
    ?>

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