php - 在 php 中获取 json 对象而不是 json 数组

Question

我希望将 json_encode() 的结果作为一个数组，例如:

[
   {
      "url":"http://localhost/.....",
      "name":"abc"
   },
   {
      "url":"http://localhost/.....",
      "name":"xyz"
   },
]

但我得到的结果是这样的对象:

{"images":[{"url":"http:\/\/192.168.0.100\/1.JPG","name":"abc"},{"url":"http:\/\/192.168.0.100\/2.JPG","name":"xyz"}]}

php代码:

<?php 

//Importing dbdetails file 
 require_once 'dbDetails.php';

 //connection to database 
 $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect...');

 //sql query to fetch all images 
 $sql = "SELECT * FROM images";

 //getting images 
 $result = mysqli_query($con,$sql);

 //response array 
 $response = array();  
 $response['images'] = array(); 

 //traversing through all the rows 
 while($row = mysqli_fetch_array($result)){
 $temp = array(); 
 $temp['url']=$row['url'];
 $temp['name']=$row['name'];
 array_push($response['images'],$temp);

 }

 //displaying the response 
 echo json_encode($response);

我试过这样使用 array_values:

 echo json_encode(array_values($response));

但它会导致在 json 字符串之前附加一个 html 代码...

Answer 1

你需要这样做:-

 $response = array();  
 //$response['images'] = array();  not needed

 //traversing through all the rows 
 while($row = mysqli_fetch_assoc($result)){ //since you are using name indexes so use _assoc()
   $temp = array(); 
   $temp['url']=$row['url'];
   $temp['name']=$row['name'];
   $response[] =$temp;
 }

 //displaying the response 
 echo json_encode($response);