String format - Logic issue

Question

Possible Duplicate:
PHP : Create array for JSON

I need to output data in the following JSON format.

Output :

[
    {
        "name": "jake",
        "age": "20"

}
]

In the following code, i need to set the key and value pairs in a way where it will give the above JSON output. How could i do this ?

Code :

$result = array();
$key = array("name", "age");
$value = array("jake", "20");

while($i>2)
{
    $result [] = HERE I NEED TO SET <key : value> COMBINATION, I NEED IT TO SET IN A WAY IT WILL GIVE THE FORMAT OF THE JSON OUTPUT GIVEN ABOVE.
}


echo json_encode($result );

score 1 · Answer 1 · answered Jul 12 '12 at 07:13

1

You can use array_combine

$result = array_combine($key, $value);

Results:

var_dump($result);

array(2) {
  ["name"]=>
  string(4) "jake"
  ["age"]=>
  string(2) "20"
}

var_dump(json_encode($result));

string(26) "{"name":"jake","age":"20"}"

answered Jul 12 '12 at 07:13

hsz

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score 1 · Accepted Answer · answered Jul 12 '12 at 07:16

1

$key = array("name", "age");
$value = array("jake", "20");
$json=array();
for($i=0;$i<=count($key)-1;$i++)
{
    $json[$key[$i]]=$value[$i];
}
echo json_encode($json); // {"name":"jake","age":"20"}

answered Jul 12 '12 at 07:16

The Alpha

143,660
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1

Micro-optimization: `$count = count($key); for($i=0;$i<$count;++$i){}` – hsz Jul 12 '12 at 07:18
Only for 2 items, I thought it's not necessary, but it's a good practice and thanks. :-) – The Alpha Jul 12 '12 at 07:19
1

@hsz: Or just `for($i=0,$count=count($key);$i<$count;++$i){}`. – Alix Axel Jul 12 '12 at 08:22

score 0 · Answer 3 · answered Jul 12 '12 at 07:13

0

Try array_combile.

$result = array();
$key = array("name", "age");
$value = array("jake", "20");

$result[] = array_combine($key, $value);

echo json_encode($result);

answered Jul 12 '12 at 07:13

xdazz

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String format - Logic issue

3 Answers3