我收到错误登录无法解析或不是字段错误,密码和 ip 也一样。 我在执行 android 程序时在 eclipse 中遇到此错误。
public void onClick(View v) {
if (v == this.viewcam)
{
Log.i("login", this.login.getText().toString());
Log.i("Passwd", this.passwd.getText().toString());
Intent localIntent1 = new Intent(v.getContext(), MjpegSample.class);
Log.i("My ip", this.editTextIp.getText().toString());
localIntent1.putExtra("ip", "http://" + this.editTextIp.getText().toString() + "/");
localIntent1.putExtra("user", this.login.getText().toString());
localIntent1.putExtra("passwd", this.passwd.getText().toString());
startActivityForResult(localIntent1, 1000);
}
}
});
下面是上面对应的main.xml文件
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical" >
<LinearLayout
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:orientation="vertical" >
<TextView
android:id="@+id/textView1"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="IP CAM VIEWER"
android:textAppearance="?android:attr/textAppearanceLarge"
/>
</LinearLayout>
<TableLayout
android:layout_width="match_parent"
android:layout_height="wrap_content" >
<TableRow
android:id="@+id/tableRow1"
android:layout_width="wrap_content"
android:layout_height="wrap_content" >
<TextView
android:id="@+id/textView2"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Username : "
android:textAppearance="?android:attr/textAppearanceLarge" />
<EditText
android:id="@+id/username"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:ems="10" >
<requestFocus />
</EditText>
</TableRow>
<TableRow
android:id="@+id/tableRow2"
android:layout_width="wrap_content"
android:layout_height="wrap_content" >
<TextView
android:id="@+id/textView3"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Password : "
android:textAppearance="?android:attr/textAppearanceLarge" />
<EditText
android:id="@+id/password"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:ems="10"
android:inputType="textPassword" />
</TableRow>
<TableRow
android:id="@+id/tableRow3"
android:layout_width="wrap_content"
android:layout_height="wrap_content" >
<TextView
android:id="@+id/textView4"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="IP : "
android:textAppearance="?android:attr/textAppearanceLarge" />
<EditText
android:id="@+id/ip"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:ems="10" />
</TableLayout>
<RelativeLayout
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_weight="0.62" >
<Button
android:id="@+id/viewcam"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_alignParentRight="true"
android:layout_alignParentTop="true"
android:layout_marginRight="96dp"
android:layout_marginTop="16dp"
android:text="View cam" />
</RelativeLayout>
</LinearLayout>
最佳答案
我假设这段代码在 OnClickListener 中,在这种情况下,删除 this
关键字,因为它指的是 OnClickListener 实例,而不是 Activity 实例:
public void onClick(View v) {
if (v == viewcam)
{
Log.i("login", login.getText().toString());
Log.i("Passwd", passwd.getText().toString());
Intent localIntent1 = new Intent(v.getContext(), MjpegSample.class);
Log.i("My ip", editTextIp.getText().toString());
localIntent1.putExtra("ip", "http://" + editTextIp.getText().toString() + "/");
localIntent1.putExtra("user", login.getText().toString());
localIntent1.putExtra("passwd", passwd.getText().toString());
startActivityForResult(localIntent1, 1000);
}
}
另一种选择是“完全限定”this
,例如:
Log.i("login", TheActivity.this.login.getText().toString());
关于java - 无法解决或者不是eclipse中android中的字段错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10456668/