从 java 中的字符串中提取日期
Extract date from a string in java
我有一个字符串如下:
String str = "skip=anthing&id='234234432234' and (timestamp le 2017-01-17T05:05:55:358Z and timestamp ge 2017-01-17T08:05:55:358Z)&format=json";
我想从字符串中提取日期部分并将其放在单引号内并将其替换回原始字符串。
结果字符串应该是
String str = "skip=anything&id='234234432234' and (timestamp le '2017-01-17T05:05:55:358Z' and timestamp ge '2017-01-17T08:05:55:358Z')&format=json"
您可以使用以下正则表达式
String input = "skip=anthing&id='234234432234' and (timestamp le 2017-01-17T05:05:55:358Z and timestamp ge 2017-01-17T08:05:55:358Z)&format=json";
String output = input.replaceAll("(\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}:\d{3}Z)", "''");
System.out.println(output); // skip=anthing&id='234234432234' and (timestamp le '2017-01-17T05:05:55:358Z' and timestamp ge '2017-01-17T08:05:55:358Z')&format=json
我有一个字符串如下:
String str = "skip=anthing&id='234234432234' and (timestamp le 2017-01-17T05:05:55:358Z and timestamp ge 2017-01-17T08:05:55:358Z)&format=json";
我想从字符串中提取日期部分并将其放在单引号内并将其替换回原始字符串。
结果字符串应该是
String str = "skip=anything&id='234234432234' and (timestamp le '2017-01-17T05:05:55:358Z' and timestamp ge '2017-01-17T08:05:55:358Z')&format=json"
您可以使用以下正则表达式
String input = "skip=anthing&id='234234432234' and (timestamp le 2017-01-17T05:05:55:358Z and timestamp ge 2017-01-17T08:05:55:358Z)&format=json";
String output = input.replaceAll("(\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}:\d{3}Z)", "''");
System.out.println(output); // skip=anthing&id='234234432234' and (timestamp le '2017-01-17T05:05:55:358Z' and timestamp ge '2017-01-17T08:05:55:358Z')&format=json