how to json_encode 'php array'

Question

i have the following array like this :

{"label":"label1","data":[[10,55],[15,32],[16,49]]}
{"label":"label","data":[[10,55],[15,32],[16,49]]}

how to get string character (,) beetween {"label":"label1","data":[[10,55],[15,32],[16,49]]} and {"label":"label2","data":[[10,55],[15,32],[16,49]]} result like this..

{"label":"label1","data":[[10,55],[15,32],[16,49]]},
{"label":"label","data":[[10,55],[15,32],[16,49]]}

Code

while($row = mysql_fetch_assoc($result))
{   
    $int = $row['SC'];
    $join = intval($int);
    $int2 = $row['jam'];
    $join2 = intval($int2);
    $dataset1[] = array($join2,$join);

}
for ($i=0; $i <2 ; $i++) { 
$dataset = array(label => label1, data => $dataset1);
$final = json_encode($dataset);

echo $final;

but really, using mysqli_..., because msql_... has been removed from PHP. — Mike 'Pomax' Kamermans
– Mike 'Pomax' Kamermans, Commented Nov 21, 2014 at 2:09
Sure you don't mean $dataset1[$i] in that for loop? Or are you seriously dumping the same thing over and over? — Brad Christie
– Brad Christie, Commented Nov 21, 2014 at 2:11
This code only produces one {label....} object. How you would join two or more of them depends on how you generate the others. — James
– James, Commented Nov 21, 2014 at 2:14
What is $label = label doing? Should that be $label[] = $row['label']? — Barmar
– Barmar, Commented Nov 21, 2014 at 2:16

Barmar · Accepted Answer · 2014-11-21 02:18:19Z

3

You need to put everything into another array:

$dataset = array();
foreach ($dataset1 as $d) {
    $dataset[] = array('label' => $label, 'data' => $d);
}
$final = json_encode($dataset);
echo $final;

This should output:

[{"label":"label1","data":[[10,55],[15,32],[16,49]]},
 {"label":"label","data":[[10,55],[15,32],[16,49]]}]

answered Nov 21, 2014 at 2:18

Barmar

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