HttpURLConnection 无法打开连接
HttpURLConnection Cannot openConnection
@Override
protected Void doInBackground(String... params) {
String type = params[0];
String url= "http://10.0.2.2/login.php" ;
if(type.equals("Singin")) {
try {
URL Singin_url = new URL(url);
//Cannot not resolve method 'openConnection()'
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
} catch (MalformedURLException e) {
e.printStackTrace();
}
}
return null;
}
为什么它不让我打开连接??
我无法继续前进,因为它说 'connection' 未初始化
您需要在 URL
对象 Singin_url
上调用 openConnection()
方法。您在 String
对象上调用它。
@Override
protected Void doInBackground(String... params)
{
String type = params[0];
String url= "http://10.0.2.2/login.php" ;
if(type.equals("Singin"))
{
try {
URL Singin_url = new URL(url);
HttpURLConnection connection = (HttpURLConnection)Singin_url.openConnection();
} catch (MalformedURLException e) {
e.printStackTrace();
}
}
return null;
}
另外你的变量名第一个字符应该是小写的,比如signInUrl
.
@Override
protected Void doInBackground(String... params) {
String type = params[0];
String url= "http://10.0.2.2/login.php" ;
if(type.equals("Singin")) {
try {
URL Singin_url = new URL(url);
//Cannot not resolve method 'openConnection()'
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
} catch (MalformedURLException e) {
e.printStackTrace();
}
}
return null;
}
为什么它不让我打开连接?? 我无法继续前进,因为它说 'connection' 未初始化
您需要在 URL
对象 Singin_url
上调用 openConnection()
方法。您在 String
对象上调用它。
@Override
protected Void doInBackground(String... params)
{
String type = params[0];
String url= "http://10.0.2.2/login.php" ;
if(type.equals("Singin"))
{
try {
URL Singin_url = new URL(url);
HttpURLConnection connection = (HttpURLConnection)Singin_url.openConnection();
} catch (MalformedURLException e) {
e.printStackTrace();
}
}
return null;
}
另外你的变量名第一个字符应该是小写的,比如signInUrl
.