這是我的代碼。我想讀取一個名為“write.txt”的文件，然后一旦它讀取。將其與一個單詞進(jìn)行比較，在這里我使用“目標(biāo)變量（字符串類型）”一旦在名為findTarget的方法內(nèi)完成比較，它將在條件為真后返回1。我嘗試調(diào)用該方法，但我不斷收到錯誤。test.java：88： 錯誤： 找不到符號 字符串測試 = findTarget（target1， source1）;^ 符號： 變量 target1 位置： 類測試 1 錯誤 有人可以糾正我的錯誤。我對編程很陌生。import java.util.*;import java.io.*;public class test {public static int findTarget( String target, String source ) {int target_len = target.length();int source_len = source.length();int add = 0;for(int i = 0;i < source_len; ++i) // i is an varialbe used to count upto source_len.{int j = 0; // take another variable to count loops while(add == 0) { if( j >= target_len ) // count upto target length { break; } else if( target.charAt( j ) != source.charAt( i + j ) ) { break; } else { ++j; if( j == target_len ) { add++; // this will return 1: true } } }}return add;//System.out.println(""+add);}public static void main ( String ... args ) {//String target = "for";// function 1 try{// read the fileFile file = new File("write.txt"); //establising a file objectBufferedReader br = new BufferedReader(new FileReader(file)); //reading the files from the file object "file"String target1; while ((target1 = br.readLine()) != null) //as long the condition is not null it will keep printing.System.out.println(target1);//target.close();}catch (IOException e) { System.out.println("file error!"); }String source1 = "Searching for a string within a string the hard way.";// function 2test ob = new test();String testing = findTarget(target1, source1);// end //System.out.println(findTarget(target, source));System.out.println("the answer is: "+testing);}}
2 回答
溫溫醬
TA貢獻(xiàn)1752條經(jīng)驗 獲得超4個贊
錯誤是因為是類函數(shù)。findTarget
所以，你有這個：
test ob = new test();
String testing = findTarget(target1, source1);
...應(yīng)更改為從靜態(tài)上下文調(diào)用函數(shù)：
//test ob = new test(); not needed, the function is static
int testing = test.findTarget(target1, source1);
// also changed the testing type from String to int, as int IS findTarget's return type.
我沒有您的文件內(nèi)容進(jìn)行試運行，但這至少應(yīng)該有助于克服錯誤。
===== 更新：
你離你很近！
在main中，更改循環(huán)中的代碼，使其如下所示：
String target1;
int testing = 0; // move and initialize testing here
while ((target1 = br.readLine()) != null) //as long the condition is not null it will keep printing.
{
//System.out.println(target1);
testing += test.findTarget(target1, source1);
//target1 = br.readLine();
}
System.out.println("answer is: "+testing);
ibeautiful
TA貢獻(xiàn)1993條經(jīng)驗 獲得超6個贊
我終于能夠解決我的問題。但擴展了功能。我想將加法遞增 1。但是在我的編程中，它不斷給我輸出
答案是：1 答案是：1
相反，我希望我的程序打印的不是兩個1，而是1 + 1 = 2
有人可以解決這個遞增的問題嗎？
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.*;
public class test {
public static int findTarget(String target, String source) {
int target_len = target.length();
int source_len = source.length();
int add = 0;
// this function checks the character whether it is present.
for (int i = 0; i < source_len; ++i) // i is a varialbe used to count upto source_len.
{
int j = 0; // take another variable to count loops
while (add == 0)
{
if (j >= target_len) // count upto target length
{
break;
}
else if (target.charAt(j) != source.charAt(i + j))
{
break;
}
else
{
++j;
if (j == target_len)
{
add++; // this will return 1: true
}
}
}
}
return add;
//System.out.println(""+add);
}
public static void main(String... args) {
//String target = "for";
// function 1
try {
// read the file
Scanner sc = new Scanner(System.in);
System.out.println("Enter your review: ");
String source1 = sc.nextLine();
//String source1 = "Searching for a string within a string the hard way.";
File file = new File("write.txt"); //establising a file object
BufferedReader br = new BufferedReader(new FileReader(file)); //reading the files from the file object "file"
String target1;
while ((target1 = br.readLine()) != null) //as long the condition is not null it will keep printing.
{
//System.out.println(target1);
int testing = test.findTarget(target1, source1);
System.out.println("answer is: "+testing);
//target1 = br.readLine();
}
br.close();
}
catch (IOException e)
{
System.out.println("file error!");
}
}
}
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