java.lang.IllegalArgumentException异常非法字符的网址 [英] java.lang.illegalargumentexception illegal character in url

问题描述

我想请求到Web服务
像这样我的参数

i want to request into web service
my parameter like this

urlString = http://ip/autodownload/andro.php?key=apps.apk|2|bla.bla.bla

public void getRequest(String Url) {
    Toast.makeText(this, "Tambah Data " + Url + " ", Toast.LENGTH_SHORT).show();
    HttpClient client = new DefaultHttpClient();
    HttpGet request = new HttpGet(urlString);
    try {
        System.out.println("tes");
        HttpResponse response = client.execute(request);
        Toast.makeText(this,request(response),Toast.LENGTH_SHORT).show();
        String res = EntityUtils.toString(response.getEntity());
        System.out.println(res);
        Update(res);
    } catch (Exception ex) {
        Toast.makeText(this, "Gagal Konek Server !", Toast.LENGTH_SHORT).show();
    }       
}

如果我运行的浏览器，它是好的，但是当我运行在Android中是这样的错误

if i run in browser it's ok, but when i run in android it's error like this

10-25 10:24:49.862: ERROR/AndroidRuntime(14602): FATAL EXCEPTION: main
10-25 10:24:49.862: ERROR/AndroidRuntime(14602): java.lang.IllegalArgumentException: Illegal character in query at index 67: http://10.234.152.120/autodownload/andro.php?key=DeliverReceipt.apk|2|com.sat.deliver
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at java.net.URI.create(URI.java:970)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at org.apache.http.client.methods.HttpGet.<init>(HttpGet.java:75)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at com.sat.deliver.MenuUtama.getRequest(MenuUtama.java:140)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at com.sat.deliver.MenuUtama.requestParam(MenuUtama.java:118)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at com.sat.deliver.MenuUtama.onClick(MenuUtama.java:355)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at android.view.View.performClick(View.java:2408)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at android.view.View$PerformClick.run(View.java:8816)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at android.os.Handler.handleCallback(Handler.java:587)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at android.os.Handler.dispatchMessage(Handler.java:92)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at android.os.Looper.loop(Looper.java:123)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at android.app.ActivityThread.main(ActivityThread.java:4627)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at java.lang.reflect.Method.invokeNative(Native Method)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at java.lang.reflect.Method.invoke(Method.java:521)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:868)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:626)
10-25 10:24:49.862: ERROR/AndroidRuntime(14602):     at dalvik.system.NativeStart.main(Native Method)

我已经尝试用URL编码，并替换字符，但它不工作
我应该怎么办？谢谢
*解决
我使urlString替换字符，如：

i have try with Urlencoding, and replacement character, but it's not working
what should i do?thank you
*SOLVED
i make replace character in urlString like :

urlString+="?key="+appName.trim().replace(".", "%2E")+"|2|".trim().replace("|", "%7C")+packageName.trim().replace(".", "%2E");

和它的做工精细：）

推荐答案

尝试编码您的网址

String link="http://example.php?string1="+URLEncoder.encode(string1)+"&string2="+URLEncoder
.encode(string2)+"&string3="+URLEncoder.encode(string3)+"&string4="+URLEncoder.encode(string4)+"";

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