base64 json_decode issue - php - php

Question

Fetching BLOB from mysql
$sql = "SELECT data as datajson FROM tablename";
$res=$obj->_executeQuery($sql);
$res=$obj->getAll($res);
$blobdata = $res[0]['datajson'];
My json data from BLOB looks like this
[{"name":"propose","value":"Propose~sample"},
{"name":"expenseamount","value":"Expense Amount~15246"},
{"name":"paymenttype","value":"Payment Type~ Cash"},
{"name":"img0","value":"/9j/4AAQSkZHIjks........61A/3qBo/9k="},
{"name":"img1","value":"NO"}]
Now
{"name":"img0","value":"/9j/4AAQSkZHIjks........61A/3qBo/9k="}
This pair is a base 64 string which is included in my BLOB. And when i did json_decode($blobdata); i am getting null
The exact base64 string can be found in JSFIDDLE
I have added json_last_error(); in my code
switch (json_last_error()) {
case JSON_ERROR_NONE:
echo ' - No errors';
break;
case JSON_ERROR_DEPTH:
echo ' - Maximum stack depth exceeded';
break;
case JSON_ERROR_STATE_MISMATCH:
echo ' - Underflow or the modes mismatch';
break;
case JSON_ERROR_CTRL_CHAR:
echo ' - Unexpected control character found';
break;
case JSON_ERROR_SYNTAX:
echo ' - Syntax error, malformed JSON';
break;
case JSON_ERROR_UTF8:
echo ' - Malformed UTF-8 characters, possibly incorrectly encoded';
break;
default:
echo ' - Unknown error';
break;
}
Which returns - Syntax error, malformed JSON and
json_decode does not work, since this is not a proper json
I want the details other than img0 and img1
1 . Now how do i remove the last two pairs (img0, img1) BEFORE json_decode, so that this becomes valid json.
2 . Or how to convert the base64 string to become proper json
UPDATE
This is how BLOB data is storing in mysql
foreach($_REQUEST as $key=>$value)
{
$raw_data[] = array('name' => $key, 'value' => $value);
}
$json = json_encode($raw_data);
$sql="INSERT INTO tablename (data) VALUES ('$json')";