Get a url like "http://TEST/Data.php?Name=ABC&Number=12"
Want to post this url ,but not open this link.
Then Name and Number columns will change into ABC & 12.
setOnClickListener:
btn_send.setOnClickListener{
val i = Intent(Intent.ACTION_VIEW, Uri.parse("http://TEST/Data.php?Name=ABC&Number=12"))
startActivity(i)
}
This will open the link in a browser. How can I just post this url but not open it?
with this code my app crushed after click button:
btn_send.setOnClickListener{
URL(url).readText()
}
9/18 update----------------------------------
tring OKHTTP but nothing happened
fun get() {
val client = OkHttpClient()
val url = URL("http://TEST/Data.php?Name=ABC&Number=12")
val request = Request.Builder()
.url(url)
.build()
OkHttpClient().newCall(request)
root.btn_submit.setOnClickListener {
get()
}
if I change OkHttpClient().newCall(request)
into OkHttpClient().newCall(request).execute()
app will crush
I think is there a problem when i send request to HTTP which is unsafe cause the error
Answer
You just cannot directly open a POST URL in browser.
You have to use Library like one of the following:
- Retrofit
- Volley
- OkHttp
Or you can open it in WebView like below
WebView webview = new WebView(this);
setContentView(webview);
byte[] post = EncodingUtils.getBytes("Name=ABC&Number=12", "BASE64");
webview.postUrl("http://TEST/Data.php", post);
convert the above code into Kotlin
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