Question

我从TextView

获取文本

 String textString = textView.getText().toString();

问题是如果文本视图中的文本有这样的换行符，例如

 My 
 Sentence has
 line breaks

当我将文本记录到控制台时

 Log.v("TAG",textString);

或者如果我将文本设置为TextView

 textView.setText(textString);

结果始终显示实际换行符

 My 
 Sentence has
 line breaks

而不是我想要的转义字符

 My\nSentence has\nline breaks

我知道这是一个非常糟糕的解释，但是如何从文本视图中获取带有代表换行符的实际字符的文本。

Answer 1

试试这个

#include <iostream>
#include <fstream>
#include <string>

using namespace std;

string TelephoneNames[100];
int TelephoneNumbers[100];

void ModifyRecords(); //Function to Modify Records
void SearchRecords(); //Function to Search Records
void DeleteRecords(); //Function to Delete Records

int main()
{
    fstream inputFile;
    fstream outputFile;
    char choice;

    inputFile.open("Telephone Names.txt");  //To store
    for (int count=0;count<100;count++)     //file names
    {                                       //into a
        inputFile >> TelephoneNames[count]; //string
    }
    inputFile.close();

    inputFile.open("Telephone Numbers.txt");//To store
    for (int count=0;count<100;count++)     //file #'s
    {                                       //into a
        inputFile >> TelephoneNumbers[count];//string
    }
    inputFile.close();
    //Display options available
    cout << " Hello, do you want to:\n";
    cout << " ======================\n";
    cout << "-Modify Records|Enter M\n";
    cout << "-Search Records|Enter S\n";
    cout << "-Delete Records|Enter D\n";
    //Store choice
    cin >> choice;
    //Send to different function
    if (choice=='M'||choice=='m')
    {
        ModifyRecords();
    }
    if (choice=='S'||choice=='s')
    {
        SearchRecords();
    }
    return 0;
}

void ModifyRecords()
{
    string name;
    string newname;
    int newnumber;
    int count=0;
    cout << "Enter the name of the person: ";
    cin >> name;
    for (count=0;TelephoneNames[count]!=name;count++)//To determine where in                 the strings the new numbers need to be
    {

    }
    cout << "Enter the new name of the person: ";
    cin >> newname;
    cout << "Enter the new number of the person: ";
    cin >> newnumber;
    TelephoneNames[count]={newname};
    TelephoneNumbers[count]={newnumber};
    count=0;
    while (count<6)
    {
        cout << TelephoneNames[count] << endl;
        cout << TelephoneNumbers[count] << endl;
        cout << endl;
        count++;
    }
}

void SearchRecords()
{
    string name;
    int count=0;
    cout << "Enter the name of the person you would like to find: ";
    cin >> name;
    for (count=0;TelephoneNames[count]!=name;count++)//To determine where in         the strings the new numbers need to be
    {

    }
    cout << "Name: " << TelephoneNames[count] << endl;
    cout << "Number: " << TelephoneNumbers[count] << endl;
}

Answer 2

问题是将我们在Java中表示的String格式化为

String test = "My\nSentence has\nline breaks";

如下：

My\nSentence has\nline breaks

即。用反斜杠替换每个换行符后跟一个'n'字符。

一种解决方案是使用String.replaceAll。例如，如果您知道换行符是严格的换行符，那么。

String unbroken = broken.replaceAll("\n", "\\\\n");

如果换行符可能是特定于平台的换行符，那么：

String unbroken = broken.replaceAll("\r\n|\n|\r", "\\\\n");

请注意，Java Pattern会将正则表达式中的裸CR或NL视为文字字符。但是在替换字符串中，我们需要一个字面反斜杠字符...因此它需要双重转义。另一种写作方式是：

String unbroken = broken.replaceAll("\n", Matcher.quoteReplacement("\\n"));

有关详细信息，请参阅javadoc ...以及在替换字符串中处理特殊反斜杠需要的明确解释。

如何将换行符返回为\ n字符

2 个答案: