PHP Json Help need? - php

Question

I need to convert the below result. I generate this JSON array using PHP but now I need the same result into a single curly brace. Below is my PHP code to generate the JSON.
$arr = array();
#If no data was returned, check for any SQL errors
if ($res == false)
{
echo 'Query not Executed : '. die(print_r(sqlsrv_errors(), TRUE));
}
else
{
while($obj = sqlsrv_fetch_array($res, SQLSRV_FETCH_ASSOC))
{
$arr[] = array(
'code' => array_search($obj['name'], $countries_iso),
'total' => $obj['Total']
);
}
}
header("Content-type: application/json");
#Output the JSON data
print (json_encode($arr));
Output of above PHP code:
[
{
code: "AF",
total: 1
},
{
code: "DZ",
total: 1
},
{
code: "AS",
total: 2
}
]
But I want to show like below result:
{
"AF":1
"DZ": 1,
"AS": 2
}

Answer 1

I guess you want $arr[array_search($obj['name'], $countries_iso)] = $obj['Total'];
Your mistake was the fact that you were declaring a new array each time you add something to your original array. This way you assign the correct way :)
Try this code:
$arr = array();
#If no data was returned, check for any SQL errors
if ($res == false)
{
echo 'Query not Executed : '. die(print_r(sqlsrv_errors(), TRUE));
}
else
{
while($obj = sqlsrv_fetch_array($res, SQLSRV_FETCH_ASSOC))
{
$arr[array_search($obj['name'], $countries_iso)] = $obj['Total'];
}
}
header("Content-type: application/json");
#Output the JSON data
//$json = json_encode($arr);
$json = json_encode($arr);

Answer 2

Just do this:
$arr[array_search($obj['name'], $countries_iso)] = $obj['Total'];
But you can get rid of the whole array_search($obj['name'], $countries_iso) if you restrict the query to only return rows WHERE name IN (list of $countries_iso). Look up WHERE and IN for SQL Server.

Answer 3

You need to transform your data structure to an associative array, where the value you are currently storing in code will be the key and the value stored in total will be value.
I'll be able to help better if you post the code that generates your current JSON object