字符串格式化中的命名占位符

Question

在Python中,当格式化字符串时,我可以按名称而不是按位置填充占位符,如下所示:

print "There's an incorrect value '%(value)s' in column # %(column)d" % \
  { 'value': x, 'column': y }

我想知道这是否可以用Java(希望没有外部库)？

Answer 1

如果您的值已经正确格式化,那么jakarta commons lang的StrSubstitutor是一种轻量级的方法.

http://commons.apache.org/proper/commons-lang/javadocs/api-3.1/org/apache/commons/lang3/text/StrSubstitutor.html

Map<String, String> values = new HashMap<String, String>();
values.put("value", x);
values.put("column", y);
StrSubstitutor sub = new StrSubstitutor(values, "%(", ")");
String result = sub.replace("There's an incorrect value '%(value)' in column # %(column)");

以上结果如下:

"第2列中的值'1'不正确"

使用Maven时,您可以将此依赖项添加到pom.xml:

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.4</version>
</dependency>

Answer 2

不完全,但您可以使用MessageFormat多次引用一个值:

MessageFormat.format("There's an incorrect value \"{0}\" in column # {1}", x, y);

以上也可以使用String.format()完成,但是如果你需要构建复杂的表达式,我发现messageFormat语法更清晰,而且你不需要关心你放入字符串的对象的类型

Answer 3

简单命名占位符的Apache Common StringSubstitutor的另一个示例。

String template = "Welcome to {theWorld}. My name is {myName}.";

Map<String, String> values = new HashMap<>();
values.put("theWorld", "Stackoverflow");
values.put("myName", "Thanos");

String message = StringSubstitutor.replace(template, values, "{", "}");

System.out.println(message);

// Welcome to Stackoverflow. My name is Thanos.

Answer 4

您可以使用StringTemplate库,它提供您想要的东西等等.

import org.antlr.stringtemplate.*;

final StringTemplate hello = new StringTemplate("Hello, $name$");
hello.setAttribute("name", "World");
System.out.println(hello.toString());

Answer 5

对于简单的情况,您可以简单地使用硬编码的String替换,不需要在那里使用库:

    String url = "There's an incorrect value '%(value)' in column # %(column)";
    url = url.replace("%(value)", x); // 1
    url = url.replace("%(column)", y); // 2

警告:我只想展示最简单的代码.当然,请不要将此用于严重的安全问题的生产代码,如评论中所述:转义,错误处理和安全性是一个问题.但在最坏的情况下,你现在知道为什么需要使用'好'的lib :-)

Answer 6

感谢你的帮助!使用你所有的线索,我写了例行程序来做我想要的 - 使用字典的类似python的字符串格式.由于我是Java新手,任何提示都受到赞赏.

public static String dictFormat(String format, Hashtable<String, Object> values) {
    StringBuilder convFormat = new StringBuilder(format);
    Enumeration<String> keys = values.keys();
    ArrayList valueList = new ArrayList();
    int currentPos = 1;
    while (keys.hasMoreElements()) {
        String key = keys.nextElement(),
        formatKey = "%(" + key + ")",
        formatPos = "%" + Integer.toString(currentPos) + "$";
        int index = -1;
        while ((index = convFormat.indexOf(formatKey, index)) != -1) {
            convFormat.replace(index, index + formatKey.length(), formatPos);
            index += formatPos.length();
        }
        valueList.add(values.get(key));
        ++currentPos;
    }
    return String.format(convFormat.toString(), valueList.toArray());
}

Answer 7

public static String format(String format, Map<String, Object> values) {
    StringBuilder formatter = new StringBuilder(format);
    List<Object> valueList = new ArrayList<Object>();

    Matcher matcher = Pattern.compile("\\$\\{(\\w+)}").matcher(format);

    while (matcher.find()) {
        String key = matcher.group(1);

        String formatKey = String.format("${%s}", key);
        int index = formatter.indexOf(formatKey);

        if (index != -1) {
            formatter.replace(index, index + formatKey.length(), "%s");
            valueList.add(values.get(key));
        }
    }

    return String.format(formatter.toString(), valueList.toArray());
}

例：

String format = "My name is ${1}. ${0} ${1}.";

Map<String, Object> values = new HashMap<String, Object>();
values.put("0", "James");
values.put("1", "Bond");

System.out.println(format(format, values)); // My name is Bond. James Bond.

Answer 8

这是一个旧线程，但是为了记录起见，您还可以使用Java 8样式，如下所示：

public static String replaceParams(Map<String, String> hashMap, String template) {
    return hashMap.entrySet().stream().reduce(template, (s, e) -> s.replace("%(" + e.getKey() + ")", e.getValue()),
            (s, s2) -> s);
}

用法：

public static void main(String[] args) {
    final HashMap<String, String> hashMap = new HashMap<String, String>() {
        {
            put("foo", "foo1");
            put("bar", "bar1");
            put("car", "BMW");
            put("truck", "MAN");
        }
    };
    String res = replaceParams(hashMap, "This is '%(foo)' and '%(foo)', but also '%(bar)' '%(bar)' indeed.");
    System.out.println(res);
    System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(foo)', but also '%(bar)' '%(bar)' indeed."));
    System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(truck)', but also '%(foo)' '%(bar)' + '%(truck)' indeed."));
}

输出将是：

This is 'foo1' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'MAN', but also 'foo1' 'bar1' + 'MAN' indeed.

Answer 9

StringSubstitutor可以使用Apache Commons Text。（有关如何将其包含在项目中的信息，请参阅依赖项信息StrSubstitutor。）请注意，已弃用。

import org.apache.commons.text.StringSubstitutor;
// ...
Map<String, String> values = new HashMap<>();
values.put("animal", "quick brown fox");
values.put("target", "lazy dog");
StringSubstitutor sub = new StringSubstitutor(values);
String result = sub.replace("The ${animal} jumped over the ${target}.");
// "The quick brown fox jumped over the lazy dog."

此类支持为变量提供默认值。

String result = sub.replace("The number is ${undefined.property:-42}.");
// "The number is 42."

要使用递归变量替换，请调用setEnableSubstitutionInVariables(true);.

Map<String, String> values = new HashMap<>();
values.put("b", "c");
values.put("ac", "Test");
StringSubstitutor sub = new StringSubstitutor(values);
sub.setEnableSubstitutionInVariables(true);
String result = sub.replace("${a${b}}");
// "Test"

Answer 10

我是一个小型库的作者，它完全可以满足您的需求：

Student student = new Student("Andrei", 30, "Male");

String studStr = template("#{id}\tName: #{st.getName}, Age: #{st.getAge}, Gender: #{st.getGender}")
                    .arg("id", 10)
                    .arg("st", student)
                    .format();
System.out.println(studStr);

或者你可以链接参数：

String result = template("#{x} + #{y} = #{z}")
                    .args("x", 5, "y", 10, "z", 15)
                    .format();
System.out.println(result);

// Output: "5 + 10 = 15"