Php 使用json解析设置textview的样式
我有一段代码,将我的db值解析为我的android活动,但我希望它的样式与现在不同 这是我的密码Php 使用json解析设置textview的样式,php,android,json,parsing,textview,Php,Android,Json,Parsing,Textview,我有一段代码,将我的db值解析为我的android活动,但我希望它的样式与现在不同 这是我的密码 public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.leesverslag); /* Werkt altijd */ // Create a crude view - this should reall
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.leesverslag);
/* Werkt altijd */
// Create a crude view - this should really be set via the layout resources
// but since its an example saves declaring them in the XML.
LinearLayout rootLayout = new LinearLayout(getApplicationContext());
txt = new TextView(getApplicationContext());
rootLayout.addView(txt);
setContentView(rootLayout);
/* Werkt altijd */
// Set the text and call the connect function.
//txt.setText("Connecting...");
//call the method to run the data retrieval
txt.setText(getServerData(KEY_121));
}
我认为它即将到来,因为我说,LinearLayout rootLayout=newlinearlayout(getApplicationContect())
但我不知道如何解决这个问题,因此它将与我自己的布局一起工作
新的布局向我展示
{“introtext”:“android应用程序的自定义值文本”}
我想看看
android应用程序的自定义值文本
这是我的php代码
<?php
header('Content-type: application/json');
$user = '***';
$pswd = '***';
$db = '***';
$conn = mysql_connect('localhost', $user, $pswd);
mysql_select_db($db, $conn);
$sql = mysql_query("SELECT jos_content.introtext FROM jos_content, jos_categories WHERE jos_categories.id = '80' AND jos_content.state = '1' AND jos_categories.id = jos_content.catid AND jos_content.title = '".$_REQUEST['titel']."'");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print str_replace('\/','/',json_encode($output));
mysql_close;
?>
有人能帮我做些什么,让它与我自己的布局工作吗
亲切问候,,
帕特里克
<?php
header('Content-type: application/json');
$user = '***';
$pswd = '***';
$db = '***';
$conn = mysql_connect('localhost', $user, $pswd);
mysql_select_db($db, $conn);
$sql = mysql_query("SELECT jos_content.introtext FROM jos_content, jos_categories WHERE jos_categories.id = '80' AND jos_content.state = '1' AND jos_categories.id = jos_content.catid AND jos_content.title = '".$_REQUEST['titel']."'");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print str_replace('\/','/',json_encode($output));
mysql_close;
?>