Java 忽略无效条目
我的头衔并不是最好的,但我不知道如何命名我想做的事情。不管怎样,我有一个箱子开关Java 忽略无效条目,java,switch-statement,Java,Switch Statement,我的头衔并不是最好的,但我不知道如何命名我想做的事情。不管怎样,我有一个箱子开关 switch (input) { case "A": Item item = new Item(); System.out.print("Enter a barcode: "); barCode = scan.nextLine(); item.setBarCode(barCode);
switch (input) {
case "A":
Item item = new Item();
System.out.print("Enter a barcode: ");
barCode = scan.nextLine();
item.setBarCode(barCode);
if (store.addItem(barCode)) {
System.out.println(store.stockedItems.get(barCode).getProductName()
+ " has been added to the store's inventory");
}
else {
item.setQuantity(1);
System.out.print("Enter the item's name: ");
productName = scan.nextLine();
productName = productName.toLowerCase();
item.setProductName(productName);
store.stockedItems.put(barCode, item);
System.out.println(store.stockedItems.get(barCode).getProductName()
+ " has been added to the store's inventory");
}
break;
}
这只是一个例子。这样做的目的是,当用户选择将一个对象添加到我的数据结构时,它会发现所提到的条形码是否已经在使用中
如果是,它只是增加数据结构中对象的数量
如果条形码未使用,则在检查其有效性后。它将提示用户输入对象的名称,然后继续将其添加到我的数据结构中
现在的问题是在我输入条形码字符串并在其各自的对象类中调用setter函数之后:
public void setBarCode(String code) {
if (!code.matches("[0-9]+") || code.length() != 12) {
System.out.println("The barcode entered is not in valid format. Entry ignored.");
} else {
barcode = code;
}
}
此函数仅确保它是数字且长度为12个字符。如果不是,我想忽略该条目并从菜单重新开始。我遇到的问题是,即使条形码无效且未设置,程序也会继续询问项目名称
如何跳过所有这些,然后再次打印菜单?有两种策略可以实现这一点:
setBarCode
方法之外,然后首先执行该测试(或修改setBarCode
以返回指示条形码是否有效的boolean
)addItem
以返回比boolean
更具信息性的内容,以便可以区分三种情况:坏条形码;成功;失败,因为它需要更多信息有两种策略可以做到这一点:
setBarCode
方法之外,然后首先执行该测试(或修改setBarCode
以返回指示条形码是否有效的boolean
)addItem
以返回比boolean
更具信息性的内容,以便可以区分三种情况:坏条形码;成功;失败,因为它需要更多信息setter
setBarCode()
应该(a)成功,或者(b)指示失败(可能使用IllegalArgumentException
,因为我们在Java中),而不是无声地失败。如果要使用IllegalArgumentException
,此代码将很好地工作:
boolean acceptable;
try {
item.setBarCode(barCode);
acceptable = true;
}
catch(IllegalArgumentException e) {
acceptable = false;
}
if(acceptable) {
if(store.addItem(barCode)){
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
else {
item.setQuantity(1);
System.out.print("Enter the item's name: ");
productName = scan.nextLine();
productName = productName.toLowerCase();
item.setProductName(productName);
store.stockedItems.put(barCode, item);
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
}
break;
但是,我建议您不要依赖setter的失败来判断正确性。从风格上讲,它“闻起来很有趣”。相反,我会将测试放在另一个(可能是静态
)方法中,在调用setter并做出相应反应之前进行测试,然后在setter中放入断言
。所以,更像这样:
// Somewhere up in your code -- Sorry, fixed up your regex
private static final Pattern BARCODE=Pattern.compile("^\\d{12}$");
public static boolean isValidBarcode(String candidate) {
return BARCODE.matcher(candidate).matches();
}
// Now your "real" code
case "A":
Item item = new Item();
System.out.print("Enter a barcode: ");
barCode = scan.nextLine();
if(isValidBarCode(barCode)) {
item.setBarCode(barCode);
if(store.addItem(barCode)) {
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
else {
item.setQuantity(1);
System.out.print("Enter the item's name: ");
productName = scan.nextLine();
productName = productName.toLowerCase();
item.setProductName(productName);
store.stockedItems.put(barCode, item);
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
}
else {
System.out.println("That's not a valid bar code.");
}
break;
// And, finally, your setBarCode() method
public void setBarCode(String code) {
assert isValidBarCode(code);
barcode = code;
}
setter
setBarCode()
应该(a)成功,或者(b)指示失败(可能使用IllegalArgumentException
,因为我们在Java中),而不是无声地失败。如果要使用IllegalArgumentException
,此代码将很好地工作:
boolean acceptable;
try {
item.setBarCode(barCode);
acceptable = true;
}
catch(IllegalArgumentException e) {
acceptable = false;
}
if(acceptable) {
if(store.addItem(barCode)){
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
else {
item.setQuantity(1);
System.out.print("Enter the item's name: ");
productName = scan.nextLine();
productName = productName.toLowerCase();
item.setProductName(productName);
store.stockedItems.put(barCode, item);
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
}
break;
但是,我建议您不要依赖setter的失败来判断正确性。从风格上讲,它“闻起来很有趣”。相反,我会将测试放在另一个(可能是静态
)方法中,在调用setter并做出相应反应之前进行测试,然后在setter中放入断言
。所以,更像这样:
// Somewhere up in your code -- Sorry, fixed up your regex
private static final Pattern BARCODE=Pattern.compile("^\\d{12}$");
public static boolean isValidBarcode(String candidate) {
return BARCODE.matcher(candidate).matches();
}
// Now your "real" code
case "A":
Item item = new Item();
System.out.print("Enter a barcode: ");
barCode = scan.nextLine();
if(isValidBarCode(barCode)) {
item.setBarCode(barCode);
if(store.addItem(barCode)) {
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
else {
item.setQuantity(1);
System.out.print("Enter the item's name: ");
productName = scan.nextLine();
productName = productName.toLowerCase();
item.setProductName(productName);
store.stockedItems.put(barCode, item);
System.out.println(store.stockedItems.get(barCode).getProductName() + " has been added to the store's inventory");
}
}
else {
System.out.println("That's not a valid bar code.");
}
break;
// And, finally, your setBarCode() method
public void setBarCode(String code) {
assert isValidBarCode(code);
barcode = code;
}