0

我有一个以下字符串,我必须在某些条件下拆分并替换该值

http://localhost:8080/api/upload/form/{uploadType}/{uploadName}

我必须只替换 {uploadType}/{uploadName} 的值。我真的不知道如何用价值替换它们。

其中的字符串{uploadType}/{uploadName}可以是任何要替换的类型。

我尝试过类似以下的方法:

包 com.test.poc;

public class TestString {
public static void main(String[] args) throws ClassNotFoundException {
    String toBeFixed = "http://localhost:8080/api/upload/form/{uploadType}/{uploadName}"; 
    String[] toReplaceWith =toBeFixed.split("{");
for (String string : toReplaceWith) {
    System.out.println("string : "+string);
}   
}

}

但我得到以下异常:

Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition
{
    at java.util.regex.Pattern.error(Pattern.java:1713)
    at java.util.regex.Pattern.closure(Pattern.java:2775)
    at java.util.regex.Pattern.sequence(Pattern.java:1889)
    at java.util.regex.Pattern.expr(Pattern.java:1752)
    at java.util.regex.Pattern.compile(Pattern.java:1460)
    at java.util.regex.Pattern.<init>(Pattern.java:1133)
    at java.util.regex.Pattern.compile(Pattern.java:823)
    at java.lang.String.split(String.java:2292)
    at java.lang.String.split(String.java:2334)
    at com.test.poc.TestString.main(TestString.java:9)

编辑 :

这是我根据Sean Patrick Floyd答案尝试的方法

public String doPOSTPathVariable(String uri,List paramsList) throws IllegalArgumentException, IllegalAccessException, InstantiationException, Exception{
        String uriString="";
        UriComponents uriComponents = UriComponentsBuilder.fromUriString(uri).build();
        for (int j = 0; j<= paramsList.size(); j++) {
            System.out.println("path variable");
            MethodParams methodParams;
            methodParams =(MethodParams) paramsList.get(j);

            if(methodParams.isPrimitive() && methodParams.getDataType()=="boolean"){
                  uriString = uriComponents.expand(true).encode().toUriString();
            }
            if(methodParams.isPrimitive() && methodParams.getDataType()=="java.math.BigDecimal"){
                uriString = uriComponents.expand(123).encode().toUriString();
            }
            if(methodParams.isPrimitive() && methodParams.getDataType()=="java.lang.String"){
                uriString = uriComponents.expand("hexgen").encode().toUriString();
            }
       } 
        return uriString;
    }

但我得到以下异常:

java.lang.IllegalArgumentException: Not enough variable values available to expand 'uploadName'
    at org.springframework.web.util.UriComponents$VarArgsTemplateVariables.getValue(UriComponents.java:1025)
    at org.springframework.web.util.UriComponents.expandUriComponent(UriComponents.java:443)
    at org.springframework.web.util.UriComponents.access$1(UriComponents.java:431)
    at org.springframework.web.util.UriComponents$FullPathComponent.expand(UriComponents.java:800)
    at org.springframework.web.util.UriComponents.expandInternal(UriComponents.java:413)
    at org.springframework.web.util.UriComponents.expand(UriComponents.java:404)
    at com.hexgen.tools.HexgenClassUtils.doPOSTPathVariable(HexgenClassUtils.java:208)
    at com.hexgen.reflection.HttpClientRequests.handleHTTPRequest(HttpClientRequests.java:77)
    at com.hexgen.reflection.HexgenWebAPITest.main(HexgenWebAPITest.java:115)

有人可以帮我解决这个问题吗?

4

5 回答 5

2

也许UriBuilder

UriBuilder.fromPath("http://localhost:8080/api/upload/form/{uploadType}/{uploadName}").build("foo", "bar");

(方便地使用您正在使用的确切格式)

于 2013-05-02T10:20:44.413 回答
2

如果您使用 Spring MVC,则可以使用新的 URI 构建器技术开箱即用地使用此功能。

UriComponents uriComponents =
    UriComponentsBuilder.fromUriString(
        "http://example.com/hotels/{hotel}/bookings/{booking}").build();

URI uri = uriComponents.expand("42", "21").encode().toUri();
// or:
String uriString = uriComponents.expand("42", "21").encode().toUriString();

(事实上​​你仍然可以使用这项技术,即使你不使用 Spring MVC,但显然你需要在 Classpath 上有 Spring MVC jar)

于 2013-05-02T10:40:04.993 回答
1

You can try this one :

public class TestString {
public static void main(String[] args) throws ClassNotFoundException {
String toBeFixed = "http://localhost:8080/api/upload/form/{uploadType}/{uploadName}"; 
String[] toReplaceWith =toBeFixed.split("\\{");
for (String string : toReplaceWith) {
System.out.println("string : "+string);
}   
}

}
于 2013-05-02T10:23:15.187 回答
1

UrlBuilder 是解决方案,但如果您只是查看分隔符,您也可以使用子字符串:

String url = "http://localhost:8080/api/upload/form/{uploadType}/{uploadName}";
int lastSlash = url.lastIndexOf('/');
int secondLastSlash = url.substring(0, lastSlash).lastIndexOf('/');
System.out.println(url.substring(secondLastSlash+1, lastSlash));
System.out.println(url.substring(lastSlash+1));

要删除花括号,您可以在字符串上使用 String.replace('{', '') 和 String.replace('}', '')

于 2013-05-02T10:40:56.077 回答
0

A{meta-character用于范围重复的正则表达式。

在这样做之前

toBeFixed = toBeFixed.replaceAll("\\{", "\n");   

另见文档:http ://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html

类似的问题:Java String ReplaceAll 方法给出非法重复错误?

于 2013-05-02T10:20:09.970 回答