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Php 使用json解析设置textview的样式_Php_Android_Json_Parsing_Textview - Fatal编程技术网

Php 使用json解析设置textview的样式

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

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 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;
?>