Reputation: 73
Could anyone assist me with how I can pass one JSON object as a field to another without having quotes added? Basically I have a function that needs to be able to append a 'header' to a set of data pre parsed into JSON in some cases or just parse the data in others.
Problem is everything works fine until I try to pass a JSON object to be stored as a "payload" for a header, at which point the JSON becomes invalid because of the extra set of quotations attached.
The object that I am trying to use is:
{
"header": {
"table": "user",
"action": "insert",
"opType": "string",
"message": "Insert sucessful for user: 6",
"start of records": false,
"end of records": true
},
"data": "[
{
"id": "6",
"Surname": "Peter",
"Forename": "Kane",
"Email": "[email protected]",
"Personal_Email": "[email protected]",
"Home_Phone_No": "01216045429",
"Work_Phone_No": "087852489",
"Mobile_Phone_No": "77245455598",
"Address_Line_1": "1aplace",
"Address_Line_2": "thistown",
"Address_Line_3": "Someplace",
"Address_Line_4": "whereever",
"Post_Code": "W549AJ",
"Mode_ID": "44",
"Route_ID": "g12",
"image": ""
}
]"
}
The problem is the quotes after the "data" key and before the last curley brace without these everything validates fine. As I've said Im using PHP Ive tried regex expressions substring etc but nothing seems to work.
my PHP is as follows:
public function dataToJSON($operationType, $table, $action, $data, $message, $header = true, $firstRecord = null) {
if ((!($operationType) === 'recordSet') and (!($operationType === 'error')) and (!($operationType === 'string') )) {
throw new Exception("Operation type:" . ' ' . $operationType . ' ' . 'passed to the dataToJSON function not recogonised');
}
if (!(is_null($firstRecord))) {
$isFirstRecord = $firstRecord;
$isLastRecord = !$firstRecord;
} else {
$isFirstRecord = false;
$isLastRecord = false;
}
if ($header) {
$jsonData = array('header' => array(
'table' => "$table",
'action' => "$action",
'opType' => "$operationType",
'message' => "$message",
'start of records' => $isFirstRecord,
'end of records' => $isLastRecord),
);
} else {
$jsonData = array();
}
$recordSet = array();
if ($operationType === 'recordSet') {
while ($row = mysql_fetch_assoc($data)) {
array_push($recordSet, $row);
}
if ($header) {
$jsonData ['data'] = $recordSet;
return json_encode($jsonData);
} else {
return json_encode($recordSet);
}
} else if (($operationType === 'error') || ($operationType === 'string')) {
if ($header) {
$jsonData ['data'] = $data;
return stripslashes(json_encode($jsonData));
} else {
return $data;
}
}
}
Upvotes: 0
Views: 323
Reputation: 265585
A JSON object is nothing more than a mere string. To achieve what you are trying to achieve, you would probably need to decode and then re-encode your JSON string:
$jsonData['data'] = json_decode($data, TRUE);
Upvotes: 0
Reputation: 21830
in order to use / parse json, it needs to be valid json... and those **"**
chars make it invalid.
paste and process here to see what i mean: http://jsonformat.com/
Upvotes: 1