我正在使用Android应用程序,其中包含从服务器检索的一些文本的列表视图源是source link。这是我的PHP代码
<?php
require_once("dbConnect.php");
$sql = "SELECT image,fullname,location from uploadfinding";
$res = mysqli_query($conn,$sql);
$result = array();
while($row = mysqli_fetch_array($res)){
array_push($result, array(
"image"=>$row[0],
"fullname"=>$row[1],
"location"=>$row[2]));
echo " over";
}
echo json_encode($result);
mysqli_close($conn);
?>
这是我的json回复
Connected successfully over over over[{"image":myurl\/uploadfinding\/uploads\/2016-04-25 06:38:051461584281226.jpg","fullname":"adi","location":"fgh"},{"image":myurl\/uploadfinding\/uploads\/2016-04-25 06:38:201461584297706.jpg","fullname":"adi2","location":"fgh2"},{"image":myurl\/uploadfinding\/uploads\/2016-04-25 06:45:441461584739479.jpg","fullname":"adi23","location":"cn"}]
我尝试了很多事情但结果却没有。完整的来源就在这里:
我不是在这里发布任何java文件,因为这个例子适用于这个网址:
http://api.androidhive.info/json/movies.json
但是当我用我的网址替换它时:
http://myurl/PhotoUpload/getAllImages.php
它只返回空活动。
答案 0 :(得分:0)
只是为了澄清: 在您的应用程序中,您可以简单地将模拟器称为
&#39;本地主机&#39;或127.0.0.1。
或
如果您使用真实设备运行,请传递正确的网址,如
编辑:
尝试比较您的网址,
myurl / uploadfinding / uploads / 2016-04-25 06:38:051461584281226.jpg&#34;
与
&#34; http://api.androidhive.info/json/movies/1.jpg&#34;此
根据您的回复,图片代码前面缺少&#34; 双引号
传递图片链接,
&#34; your_link&#34;
答案 1 :(得分:0)
试试这个
<?php
header("content-type:application/json");
require_once("dbConnect.php");
$sql = "SELECT x1,x2,x3 from table_name";
$res = mysqli_query($conn,$sql);
$result = array();
while($row = mysqli_fetch_array($res)){
array_push($result, array(
"x1"=>$row["x1"],
"x2"=>$row['x2'],
"x3"=>$row["x3"]));
echo " over";
}
echo json_encode($result);
$fp = fopen('results.json', 'w');
fwrite($fp, json_encode($result));
fclose($fp);
mysqli_close($conn);
?>
替换您的网址&#34; http://myurl/PhotoUpload/getAllImages.php&#34;使用&#34; http://myurl/PhotoUpload/results.json&#34;