如何从数据库中的数据得到的网址,然后将其放置到毕加索负载
问题描述:
enter image description here如何从数据库中的数据得到的网址,然后将其放置到毕加索负载
这是我的PHP代码..
<?php
if(!isset($_GET['s'])){
$s = "Berita";
}else{
$s = $_GET['s'];
}
if(!isset($_GET['id'])){
header("Location: content_berita_list");
}else{
$parent = $_GET['id'];
}
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'manteb_user');
define('DB_PASSWORD', 'manteb_pass123');
define('DB_DATABASE', 'manteb_core');
$db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
$sql2 = "SELECT * FROM manteb_posts WHERE post_type = 'attachment' && post_parent = $parent";
$result2 = $db->query($sql2);
if ($result2->num_rows > 0) {
// output data of each row
while($row2 = $result2->fetch_assoc()) {
$id = $row2['ID'];
$type = $row2['post_type'];
$judul = $row2['post_title'];
$teks = $row2['post_content'];
$date = $row2['post_date'];
$gambar = $row2['guid'];
$parent = $row2['post_parent'];
}
} else {
echo "0 results";
}
?>
<html><body><?php echo $gambar; ?></body></html>
这个班里
public class Event implements Serializable {
public Event(int ID, String post_title, String post_content, String guid, String post_date, String post_parent) {
this.ID = ID;
this.post_title = post_title;
this.post_content = post_content;
this.guid = guid;
this.post_date = post_date;
this.post_parent = post_parent;
}
public int ID;
public String post_title, post_content, guid, post_date, post_parent;
}
这是我的毕加索
public static final String BASE_URL = "http://manteb.com/";
public static final String GAMBAR = BASE_URL + "content_gambar_get.php?id=";
public void onBindViewHolder(EventAdapter.EventViewHolder holder, int position) {
final Event event = eventlist.get(position);
String fullUrl = PHPLink.GAMBAR + event.ID;
Picasso.with(context)
.load(fullUrl)
.placeholder(R.drawable.mantebbw)
.error(android.R.drawable.stat_notify_error)
.into(holder.image_url);
holder.txtTitle.setText(event.post_title);
holder.txtDate.setText(event.post_date);
holder.image_url.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
Intent in = new Intent(context, EventDetail.class);
in.putExtra("event", event);
in.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
context.startActivities(new Intent[]{in});
}
});
}
,我想要$ gambar放在毕加索加载(url) 我真的很迷惑cz这个数据库是从wordpress默认.. tia
答
它没有显示任何图片,因为你只是打印的URL。毕加索将尝试请求manteb.com/android/content_gambar.get.php?id=2039
,但它得到
<html>
<body>
http://www.manteb.com/wp-content/uploads/2016/05/14edonor4_slo.jpg
</body>
</html>
毕加索不知道如何将其转换成图像。你可以做的是直接显示图像,只要你打开content_gambar.get.php
。尝试使用this method
PS:这是更好,当你正在使用的生产机器:)
+0
ty很多家伙..它的工作很好..需要一两天才能使它工作.. cz我不擅长php .. hehe .. –
发生了什么错误删除凭据您的数据库? –
有没有错误..他没有显示immage ..我只是beginer和不知道如何解析从我的PHP放在毕加索加载(网址)的url的数据..所以这个网址在毕加索成为数据从我的网址就像在我的张贴图片..或这.. manteb.com/android/content_gambar_get.php?id=2039 –
在我的代码它没有显示图像cz它只是网址,显示我想要的图像url show .. tia .. –