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[已解決]否則如果總是從文件讀取時執行
通過添加是否已達到語句的標記來修復它。
if(contains) // condition reached. marker "found" is 1.
found = 1;
else if(found != 1){ // If found is not 1, not found. break.
System.out.println("Not found"); break;
編寫一個程序,包括讀取從一個文件的示例SSN。我試圖說明無效輸入(即字符串不在文件內)。但是,這些陳述並未執行我想要的方式。 (忽略一切,但我的if語句是如何構造的)。
public static void getNameNum(String SocSec_input){
try{
reader1 = new BufferedReader(new FileReader("src\\DEPARTMENT.txt"));
reader2 = new BufferedReader(new FileReader("src\\EMPLOYEE.txt"));
/**
* reads from employee text file
*/
while((curr = reader2.readLine()) != null){
/**
* checks which lines contain user input and split the name from the code into storage
*/
if(curr.contains(SocSec_input)){
String[] parts = curr.split(",");
String FNAME = parts[0];
String LNAME = parts[1];
String SSN = parts[2];
String DNO = parts[9];
if(SSN.equals(SocSec_input)){
while((curr = reader1.readLine()) != null){
/**
* searches the file for the user input
*/
if(curr.contains(DNO)){
/**
* splits the line containing user input at each comma and store the values
*/
String[] parts2 = curr.split(",");
String DNAME = parts2[0];
String DNUMBER = parts2[1];
if(DNUMBER.equals(DNO))
System.out.println(FNAME + " " + LNAME + " works in department " + DNO + ", " + DNAME);
}
}
}
}
// ALWAYS EXECUTING STATEMENT HERE*****************
else if(!(curr.contains(SocSec_input))){
System.out.println("Invalid SSN entered. We could not find
that SSN in our database.");
break;
}
}
}
catch (IOException e){
e.printStackTrace();
}
}
輸出:
Please enter the employee's SSN: 123
Invalid SSN entered. We could not find that SSN in our database.
Please enter the employee's SSN: 888665555
Invalid SSN entered. We could not find that SSN in our database.
--------------------
但888665555是在文件中!到底是怎麼回事?
檢查你嘗試使用字符串上CURR格式,以確保您的字符串是你認爲他們是誰? – Tosh
@Tosh感謝您的快速回復,您是什麼意思的字符串格式? – pmcg521
基本上做了一些快速調試,可以像幾個打印語句一樣簡單,以確保您認爲您使用的字符串與計算機傳遞的字符串相同,如果不是,則使用類似字符串的字符串.format可以讓你的字符串看起來像你期望的那樣 – Tosh