通过PHP API从数据库中获取图像URL时得到空白的JSON响应

Question

从服务器获取所有图像的代码。 我将服务器路径/目录中存在的图像的URL保存到SQL数据库（名称：photos，columns：id和image）

<?php

     $connection = mysqli_connect("localhost","webkuv1c_naman","naman","webkuv1c_android_api") or die("Error " . mysqli_error($connection));
     $sql = "select image from photos";
     $res = mysqli_query($con,$sql);
     $result = array();

     while($row = mysqli_fetch_array($res)){
         array_push($result,array('url'=>$row['image']));
     }

     echo json_encode(array("result"=>$result));

     mysqli_close($con);
?>

我在点击API时收到以下响应：

{"result":[]}

这是图像路径/ URL： http://webkunj.com/android_api/productUpload/0.png : http://webkunj.com/android_api/productUpload/0.png

Answer 1

您已正确完成所有操作，只需更换

$connection = mysqli_connect("localhost","webkuv1c_naman","naman","webkuv1c_android_api") or die("Error " . mysqli_error($connection));

同

 $con = mysqli_connect("localhost","webkuv1c_naman","naman","webkuv1c_android_api") or die("Error " . mysqli_error($con));