如何使用 MYSQL 在 PHP 中獲取嵌套的 json 值？

Question

我正在使用 MySQL 數據庫和 PHP。

我正在使用 PHP 從數據庫中獲取這些

PHP 代碼：這是我用來運行查詢的 PHP 代碼，然后將結果集編碼為 JSON

<?php
   $con=mysqli_connect("localhost","root","root","BASIC");

   if (mysqli_connect_errno($con)) {
      echo json_encode("Failed to connect to MySQL: ") . mysqli_connect_error();
   }

   $categoryID=$_GET['categoryID'];
   $areaID=$_GET['areaID'];
//SELECT a.tutorID FROM TutorAreas a, TutorForCategories c WHERE c.categoryID='$categoryID' AND a.areaID='$areaID' and c.tutorID=a.tutorID

   $result = mysqli_query($con,"SELECT * FROM Tutor INNER JOIN TutorForCategories ON Tutor.tutorID=TutorForCategories.tutorID INNER JOIN TutorAreas ON Tutor.tutorID=TutorAreas.tutorID WHERE TutorAreas.areaID='$areaID' AND TutorForCategories.categoryID='$categoryID'");


$rows=array();
    while($row=$result->fetch_assoc())
    {
       $rows[]=(object)$row;

    }
   //$row = mysqli_fetch_array($result);

    echo json_encode($rows); 
   mysqli_close($con);
?>

[{"tutorID":"2",
"tutorFirstName":"Nasreen",
"tutorLastName":"Shah",
"tutorDOB":"1983-05 13",
"tutorMobileNumber":"03452530200",
"tutorCity":"Karachi",
"tutorCountry":"Pakistan",
"tutorGender":"Female",
"tutorRating":"0",
"tutorAvailabilityTime":"13.45",
"tutorQualification":"BSc. Microbiology",
"tutorEmail":"nasreen2145@gmail.com",
"tutorAddress":"AX 25 Shama Gardens Saddar Karachi",
"tutorLat":null,
"tutorLong":null,
"scvurl":"",
"tutorProfilePic":null,
"tutorVideo":null,
"categoryID":"9",
"areaID":"3"}]

我想要嵌套的值。

["Tutor":
{"tutorID":"2",
"tutorFirstName":"Nasreen",
"tutorLastName":"Shah",
"tutorDOB":"1983-05 13",
"tutorMobileNumber":"03452530200",
"tutorCity":"Karachi",
"tutorCountry":"Pakistan",
"tutorGender":"Female",
"tutorRating":"0",
"tutorAvailabilityTime":"13.45",
"tutorQualification":"BSc. Microbiology",
"tutorEmail":"nasreen2145@gmail.com",
"tutorAddress":"AX 25 Shama Gardens Saddar Karachi",
"tutorLat":null,
"tutorLong":null,
"scvurl":"",
"tutorProfilePic":null,
"tutorVideo":null},
"categpry":{
"categoryID":"9"},
"area":{
"areaID":"3"}}]

如何使用 PHP 實現這一目標？ 我正在嘗試嵌套它們，以便稍后在 Android 工作室中使用它們。

Answer 1

您可以將json_decode與array_shify和foreach一起使用

$f = json_decode($json, true);
foreach(array_shift($f) as $v){
 echo $v;
}

演示： https://3v4l.org/ZktqI