java 如何使用默认值将 String 解析为 int？
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How can I parse String to int with the default value?
提问by Krunal
I need to parse a string user id into integer for that I used Integer.parseInt(String s)but it returns null/nil, if there string has/holds non-decimal/integer value and in that case, I need to assign default integer value as 0.
我需要将字符串用户 ID 解析为我使用的整数，Integer.parseInt(String s)但它返回空/零，如果字符串具有/持有非十进制/整数值，在这种情况下，我需要将默认整数值分配为0.
I tried this but it (? 0)seems not working and I don't know whether it is correct syntax or not.
我试过这个，但它(? 0)似乎不起作用，我不知道它的语法是否正确。
String userid = "user001";
int int_userid = Integer.parseInt(userid) ? 0;
How can assign default value to integer if there is null assignment?
如果有空分配，如何将默认值分配给整数？
String userid is a parameter argument as part of web service function call, so I cannot update it's data type to integer.
字符串 userid 是作为 Web 服务函数调用的一部分的参数参数，因此我无法将其数据类型更新为整数。
采纳答案by dev_kd
You can try this method with a regular expression.
你可以用正则表达式试试这个方法。
public static int parseWithDefault(String s, int defaultVal) {
return s.matches("-?\d+") ? Integer.parseInt(s) : defaultVal;
}
回答by msfoster
You're most likely using apache.commons.lang3 already:
您很可能已经在使用 apache.commons.lang3：
NumberUtils.toInt(str, 0);
回答by Yannick
That syntax won't work for Integer.parseInt(), because it will result in a NumberFormatException
该语法不适用于Integer.parseInt()，因为它会导致NumberFormatException
You could handle it like this:
你可以这样处理：
String userid = "user001";
int int_userid;
try {
int_userid = Integer.parseInt(userid);
} catch(NumberFormatException ex) {
int_userid = 0;
}
Please note that your variable names do not conform with the Java Code Convention
请注意您的变量名不符合 Java 代码约定
A better solution would be to create an own methodfor this, because I'm sure that you will need it more than once:
更好的解决方案是为此创建一个自己的方法，因为我相信您将不止一次需要它：
public static int parseToInt(String stringToParse, int defaultValue) {
try {
return Integer.parseInt(stringToParse);
} catch(NumberFormatException ex) {
return defaultValue; //Use default value if parsing failed
}
}
Then you simply use this method like for e.g.:
然后您只需使用此方法，例如：
int myParsedInt = parseToInt("user001", 0);
This call returns the default value 0, because "user001" can't be parsed.
此调用返回默认值0，因为无法解析“user001”。
If you remove "user" from the string and call the method...
如果您从字符串中删除“用户”并调用该方法...
int myParsedInt = parseToInt("001", 0);
…then the parse will be successful and return 1since an int can't have leading zeros!
...然后解析将成功并返回，1因为 int 不能有前导零！
回答by mgyongyosi
It might be a little over-engineering, but you can use Guava's Ints.tryParse(String)with Java 8's Optionals like this:
这可能有点过度设计，但是您可以Ints.tryParse(String)像这样将 Guava与 Java 8 的 Optionals 一起使用：
int userId = Optional.ofNullable(Ints.tryParse("userid001")).orElse(0)
回答by YCF_L
You can use this way with String::matcheslike this :
您可以String::matches像这样使用这种方式：
String userid = "user001";
int int_userid = userid.matches("\d+") ? Integer.parseInt(userid) : 0;
You ca also use -?\d+for both positive and negative values :
您还可以使用-?\d+正值和负值：
int int_userid = userid.matches("-?\d+") ? Integer.parseInt(userid) : 0;
回答by xenteros
I believe that you can achieve what you want by the following method:
我相信您可以通过以下方法实现您想要的：
public static int parseIntWithDefault(String s, int default) {
try {
return Integer.parseInt(s);
} catch(NumberFormatException e) {
return default;
}
}
and now just assign:
现在只需分配：
int int_userid = parseIntWithDefault(userId, 0);
Please have in mind, that using Javaone should use Java good practices about formatting the code. int_useridis definitely something to improve.
请记住，使用Java一个应该使用关于格式化代码的 Java 良好实践。int_userid绝对是需要改进的地方。
回答by xoX Zeus Xox
String userid = "user001";
int int_userid = Integer.parseInt(userid) != null ? Integer.parseInt(userid) : 0);
Did you mean this syntax?
But since an intcan never be nullyou have to instead do:
你是说这个语法吗？但是因为int永远不可能是null你必须做的：
String userid = "user001";
int int_userid;
try {
int_userid= Integer.parseInt(userid);
} catch (NumberFormatexception e) {
int_userid = 0;
}
回答by deepakl
int int_userid;
try {
int_userid = Integer.parseInt(userid); // try to parse the given user id
} catch (Exception e) { // catch if any exception
int_userid = 0; // if exception assign a default value
}
