将数组转换为 JSON 字符串 php
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Convert Array to JSON string php
提问by Blessan Kurien
I have an array like this
我有一个这样的数组
$prebook=array(
'sourceCity'=>$_POST['source'],
'destinationCity'=>$_POST['dest'],
'doj'=>$_POST['doj'],
'routeScheduleId'=>$_POST['routeid'],
'boardingPoint'=>array(
'id'=>$id,
'location'=>$location,
'time'=>$time
),
'customerName'=>$_POST['fname'],
'customerLastName'=>$_POST['lname'],
'customerEmail'=>$_POST['email'],
'customerPhone'=>$_POST['mobileno'],
'emergencyPhNumber'=>$_POST['emc-number'],
'customerAddress'=>$_POST['address'],
'blockSeatPaxDetails'=>array(array(
'age'=>$_POST['age'][$key],
'name'=>$value,
'seatNbr'=>$_POST['seat-no'][$key],
'Sex'=>$_POST['gender'.$no],
'fare'=>$_POST['base-fare'][$key],
'totalFareWithTaxes'=>$_POST['amount'][$key],
'ladiesSeat'=>$ladies,
'lastName'=>$_POST['plname'][$key],
'mobile'=>$_POST['mobileno'],
'title'=>'Mr',
'email'=>$_POST['email'],
'idType'=>$_POST['idtype'],
'idNumber'=>$_POST['id-number'],
'nameOnId'=>$value,
'primary'=>true,
'ac'=>$ac,
'sleeper'=>$sleeper
)),
'inventoryType'=>$_POST['invtype']
)
From this i want to get Json string look like this
从这里我想让 Json 字符串看起来像这样
apiBlockTicketRequest:{"sourceCity":"Hyderabad","destinationCity":"Bangalore","doj":"2016-01-22","routeScheduleId":"6717","boardingPoint":{"id":"2889","location":"Mettuguda,Opp. Mettuguda Church","time":"04:50PM"},"customerName":"jj","customerLastName":"jjj","customerEmail":"[email protected]","customerPhone":"7779","emergencyPhNumber":"7878","customerAddress":"gjgj","blockSeatPaxDetails":[{"age":"22","name":"hjhj","seatNbr":"G4","Sex":"F","fare":"900","totalFareWithTaxes":"945","ladiesSeat":false,"lastName":"hjhj","mobile":"7779","title":"Mr","email":"[email protected]","idType":"Aadhar Card","idNumber":"jkjk","nameOnId":"hjhj","primary":true,"ac":false,"sleeper":false}],"inventoryType":"0"}
Here is my code
这是我的代码
$data =json_encode($prebook);
$json='apiBlockTicketRequest:'.$data;
echo $json;
But when i Validate the JSON string using thisI will get the following error
但是当我使用它验证 JSON 字符串时，我会收到以下错误
Expecting object or array, not string.[Code 1, Structure 1]
Error:Strings should be wrapped in double quotes.
需要对象或数组，而不是字符串。[代码 1，结构 1]
错误：字符串应该用双引号括起来。
回答by Robert
You creating invalid json by adding apiBlockTicketRequestto output
您通过添加apiBlockTicketRequest到输出来创建无效的 json
$json='apiBlockTicketRequest'.$data;
instead you can do
相反，你可以做
$json = json_encode(['apiBlockTicketRequest' => $prebook]);
回答by zilth
Your code is correct to make that stringbut the entire output string ($json)will never parse as json, only the $datayou encoded.
您的代码是正确的，string但整个输出字符串($json)永远不会解析为json，只有$data您编码的。
