我有以下json响应，我使用php创建它。
[{"id":"1","title":"Clean Code","author":"Robert Martin","bookUrl":"http:\/\/amzn.to\/1DJybxH","imageUrl":"http:\/\/adavis.github.io\/adept-android\/images\/clean_code.jpg","displayDate":"August 11, 2008","numberOfPages":"464"},{"id":"2","title":"Effective Java","author":"Joshua Bloch","bookUrl":"http:\/\/amzn.to\/1Ku8Xel","imageUrl":"http:\/\/adavis.github.io\/adept-android\/images\/effective_java.jpg","displayDate":"May 28, 2008","numberOfPages":"346"},{"id":"3","title":"Working Effectively with Legacy Code","author":"Michael Feathers","bookUrl":"http:\/\/amzn.to\/1Jqe1PA","imageUrl":"http:\/\/adavis.github.io\/adept-android\/images\/legacy_code.jpg","displayDate":"October 2, 2004","numberOfPages":"456"},{"id":"4","title":"Refactoring: Improving the Design of Existing Code","author":"Martin Fowler","bookUrl":"http:\/\/amzn.to\/1Lx4cjR","imageUrl":"http:\/\/adavis.github.io\/adept-android\/images\/refactoring.jpg","displayDate":"July 8, 1999","numberOfPages":"464"}]
在这个响应中，我有一个查询，它将一本书作为json对象返回。
因此，使用该URL，将返回id = 2的书。
{"id":"2","title":"Effective Java","author":"Joshua Bloch","bookUrl":"http:\/\/amzn.to\/1Ku8Xel","imageUrl":"http:\/\/adavis.github.io\/adept-android\/images\/effective_java.jpg","displayDate":"May 28, 2008","numberOfPages":"346"}
但是，当我不在我的网址中使用id路径时，总会返回第一本书（id = 1）。我不确定这是否正确。
这是我的PHP代码。
<?php
include("init.php");
$string="";
$newString="";
$id = mysqli_real_escape_string($con,$_GET['id']);
$get_posts = "select * from books";
if($id != '') $get_posts .= " WHERE id LIKE '%{$id}%'";
$run_book = mysqli_query($con,$get_posts);
$string =
json_encode(mysqli_fetch_object($run_book),JSON_UNESCAPED_UNICODE);
echo $string;
?>
有什么建议吗？
谢谢，
西奥。