Question

我的代码是关于如果用户输入pH值（正常pH值= 7.35-7.45）它会给他答案是酸性pH值还是碱性pH值或正常，那么我想保存一个名为phStatus的字符串作为在onclick方法之后，在resultTextView中最后调用其他字符串的答案，这里是：

Button resultButton = (Button) findViewById ( R.id.resultButton );
        final TextView resultTextView = (TextView) findViewById ( R.id.resultTextView );

        resultButton.setOnClickListener ( new OnClickListener ( ) {
            @Override
            public void onClick ( View v )
                            {
            EditText phET = (EditText) findViewById ( R.id.phEditText );
            int phEntered = Integer.parseInt ( phET.getText ( ).toString ( ) );
            String phStatus = "";
            if ( phEntered < 7.35 )
                {phStatus = "Acidosis";}
            else if ( phEntered > 7.45 )
                {phStatus = "Alkalosis";}
            else 
                {phStatus = "Normal acid-base balance";}


                String resultMSG = "This is " + phStatus;

                                resultTextView.setText ( resultMSG );
                            }

Answer 1

试试这个......
添加try catch块来处理异常。

Button resultButton = (Button) findViewById ( R.id.resultButton );
    final TextView resultTextView = (TextView) findViewById ( R.id.resultTextView );

    resultButton.setOnClickListener ( new OnClickListener ( ) {
        @Override
        public void onClick ( View v ) {
          // TODO Auto-generated method stub
          EditText phET = (EditText) findViewById ( R.id.phEditText );
          try{
            double phEntered = Double.parseDouble ( phET.getText ( ).toString ( ) );
            String phStatus = "";
            if ( phEntered < 7.35 ) {
               phStatus = "Acidosis";
            }else if ( phEntered > 7.45 ) {
               phStatus = "Alkalosis";
            }else {
               phStatus = "Normal acid-base balance";
            }
             String resultMSG = "This is " + phStatus;
             resultTextView.setText ( resultMSG );
             Toast.makeText ( getApplicationContext ( ), "Not Long Enough :(", 1000 ).show ( );
          }catch(Exception e){
            // exception handle here
          }
       }

Answer 2

将int与double进行比较是有效的 - 它会将int提升为a 在进行比较之前加倍。您只需要@deepak提到的试试捕获.. 您也可以尝试为EditText指定一个默认字符串（可能是“0”）。

Answer 3

我在你们的帮助下想出来了，非常感谢，现在这是代码：

公共类ABG扩展了Activity { @覆盖 protected void onCreate（Bundle savedInstanceState） { super.onCreate（savedInstanceState）; setContentView（R.layout.abg）;

        }
    String phStatus;

    @Override
    public String ph ( String phStatus )
        {
            EditText phET = (EditText) findViewById ( R.id.phEditText );
            try
                {
                    double phEntered = Double.parseDouble ( phET.getText ( ).toString ( ) );
                    if ( phEntered < 7.35 )
                        {
                            phStatus = "Acidosis";
                        }
                    else if ( phEntered > 7.45 )
                        {
                            phStatus = "Alkalosis";
                        }
                    else
                        {
                            phStatus = "Normal acid-base balance";
                        }

                }
            catch (Exception e)
                {
                    // exception handle here
                }
            return phStatus;
        }



    @Override
    public void result ( View v )
        {
            // TODO Auto-generated method stub

            String resultMSG = "This is " + ph ( phStatus );

            TextView resultTextView = (TextView) findViewById ( R.id.resultTextView );
            resultTextView.setText ( resultMSG );
        }    
}

int EditText＆amp;如果/ else语句得到一个字符串，有什么问题？

3 个答案: