通过php api从数据库中获取图像URL时获取空白的json响应

时间:2017-06-06 08:18:10

标签: php mysql json api

从服务器获取所有图像的代码。我正在保存的网址 我的服务器路径/目录中存在的图像到SQL数据库(名称:照片,列:id和图像)

<?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时得到以下回复:

{"result":[]}

这是图片路径/网址:http://webkunj.com/android_api/productUpload/0.png

1 个答案:

答案 0 :(得分:0)

你已经把所有事情都做好了 只需替换

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

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