2

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;
asked Nov 21, 2014 at 2:06
6
  • Using mysql_fetch_assoc! Then read this. Commented Nov 21, 2014 at 2:08
  • but really, using mysqli_..., because msql_... has been removed from PHP. 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? 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. Commented Nov 21, 2014 at 2:14
  • What is $label = label doing? Should that be $label[] = $row['label']? Commented Nov 21, 2014 at 2:16

1 Answer 1

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
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