Question

final EditText ed7 = (EditText)findViewById(R.id.Recycleweight);
final EditText ed8 = (EditText)findViewById(R.id.Nonweight);
final EditText ed11 = (EditText)findViewById(R.id.totalweight);

recy_wt=ed7.getText().toString();
nonrec_wt=ed8.getText().toString();
total_wt=ed11.getText().toString();

int rec = Integer.parseInt(recy_wt);
int nrec = Integer.parseInt(nonrec_wt);
ed11.setText(String.valueOf(rec + nrec));

我给的输入是，ed7 = 10kg，ed8 = 20kg，它将显示在ed11中为30公斤，但不幸的是停止了应用程序。如果我不写'公斤＆＃39;，它正确显示总重量。但我想展示＆＃39; kg＆＃39;在答案中。

Answer 1

此处已有解决方案（无法标记为重复） Find and extract a number from a string

从字符串中提取数字部分，然后将其解析为整数

Answer 2

试试这个

String regexp = "/^[0-9]+kg$/g"; //It accepts only '<number><kg>' format
int weight = 0;
if (recy_wt.matches(regexp) && nonrec_wt.matches(regex)) {
    //It's valid input
    Scanner scan = new Scanner(recy_wt);
    weight = Integer.parseInt(scan.findInLine("\\d+(\\.\\d+)?"));
    Scanner scan = new Scanner(nonrec_wt);
    weight += Integer.parseInt(scan.findInLine("\\d+(\\.\\d+)?"));
}

ed11.setText(String.valueOf(weight + "Kg"));

Answer 3

如果唯一可能的单位是kg，那么请尝试这样

final EditText ed7 = (EditText)findViewById(R.id.Recycleweight);
final EditText ed8 = (EditText)findViewById(R.id.Nonweight);
final EditText ed11 = (EditText)findViewById(R.id.totalweight);

int weight = 0;

recy_wt=ed7.getText().toString();
nonrec_wt=ed8.getText().toString();

recy_wt = recy_wt.replace("kg","");
nonrec_wt = nonrec_wt.replace("kg","");

weight += Integer.parseInt(recy_wt) + Integer.parseInt(nonrec_wt);

ed11.setText(weight + "kg");

Answer 4

这将有效： -

recy_wt=ed7.getText().toString();
nonrec_wt=ed8.getText().toString();

String n=recy_wt.replace("kg","");
String n1=nonrec_wt.replace("kg","");

weight += Integer.parseInt(n) + Integer.parseInt(n1);
ed11.setText(weight + "kg");

将替换的一个存储在字符串中，然后转换为int将有所帮助。

string to int，导致app停止

4 个答案: