我有一個變數 requestID,我必須用空值替換它。
任何人都可以幫助我使用適合以下字串模式的正則運算式。例如- requestID="DEABCD-1196745-000214557" requestID="0195789_ledabj_6156096"
uj5u.com熱心網友回復:
你可以使用[A-Z-a-z0-9] (?:[_-][A-Z-a-z0-9] ) :
List<String> inputs = Arrays.asList(new String[] { "DEABCD-1196745-000214557",
"0195789_ledabj_6156096",
"BANANAS" });
for (String input : inputs) {
if (input.matches("[A-Z-a-z0-9] (?:[_-][A-Z-a-z0-9] ) ")) {
System.out.println("MATCH: " input);
}
else {
System.out.println("NO MATCH: " input);
}
}
這列印:
MATCH: DEABCD-1196745-000214557
MATCH: 0195789_ledabj_6156096
NO MATCH: BANANAS
uj5u.com熱心網友回復:
/[\w|-] /能滿足你的需要嗎?
uj5u.com熱心網友回復:
您可以通過Regex101、Regexr等在線工具嘗試和測驗您的正則運算式
uj5u.com熱心網友回復:
這可能會有所幫助,請嘗試以下正則運算式:
"([A-Z] -\\d -\\d |\\d _[a-z] _\\d )"
背景關系和測驗平臺中的正則運算式:
public static void main(String[] args) {
String requestID_1 = "DEABCD-1196745-000214557";
String requestID_2 = "0195789_ledabj_6156096";
Matcher matcher1 = Pattern.compile("([A-Z] -\\d -\\d |\\d _[a-z] _\\d )").matcher(requestID_1);
if (matcher1.find()) {
requestID_1 = null;
System.out.println("Variable requestID_1 is set to: " requestID_1);
}
Matcher matcher2 = Pattern.compile("([A-Z] -\\d -\\d |\\d _[a-z] _\\d )").matcher(requestID_2);
if (matcher2.find()) {
requestID_2 = null;
System.out.println("Variable requestID_2 is set to: " requestID_2);
}
}
輸出:
Variable requestID_1 is set to: null
Variable requestID_2 is set to: null
轉載請註明出處,本文鏈接:https://www.uj5u.com/caozuo/470742.html
標籤:javascript 爪哇 正则表达式 细绳
上一篇:如何從C 中的字串中減去字符?
下一篇:C#基礎和字串問題
