我想從字串中替換除數字字符以外的所有內容，但如果 出現在字串的開頭，則應忽略它。例如
1234 !@#$%^&*()_ =-;',.><:" becomes 1234
1234 becomes 1234
1234 becomes 1234
1234 !@#$%^&*()_ =-;',.><:" becomes 1234
1234 !@#$%^&*()_ =-;',.><:" becomes 1234
1234ABCabc !@#$%^&*()_ =-;',.><:" becomes 1234
1234ABCabc !@#$%^&*()_ =-;',.><:" becomes 1234
Aa1234ABCabc !@#$%^&*()_ =-;',.><:" becomes 1234
a 1234ABCabc !@#$%^&*()_ =-;',.><:" becomes 1234
1 1234ABCabc !@#$%^&*()_ =-;',.><:" becomes 11234
你能建議嗎？
uj5u.com熱心網友回復：
你可以試試：
str.replace(/((?<!^)\D)|(^[^ 0-9])/g, '');
這取代（什么都沒有）：
- 任何不在字串開頭的非數字。
uj5u.com熱心網友回復：
這會從字串中提取數字，包括可選的前導 ：
var tests = [
'1234 !@#$%^&*()_ =-;\',.><:"', // becomes 1234
' 1234', // becomes 1234
' 1234', // becomes 1234
' 1234 !@#$%^&*()_ =-;\',.><:"', // becomes 1234
' 1234 !@#$%^&*()_ =-;\',.><:"', // becomes 1234
' 1234ABCabc !@#$%^&*()_ =-;\',.><:"', // becomes 1234
'1234ABCabc !@#$%^&*()_ =-;\',.><:"', // becomes 1234
'Aa1234ABCabc !@#$%^&*()_ =-;\',.><:"', // becomes 1234
'a 1234ABCabc !@#$%^&*()_ =-;\',.><:"', // becomes 1234
'1 1234ABCabc !@#$%^&*()_ =-;\',.><:"' // becomes 11234
];
tests.forEach(str => {
console.log(str ' => ' str.replace(/^.*?(\ ?[0-9] ).*$/, '$1'));
});
輸出：
1234 !@#$%^&*()_ =-;',.><:" => 1234
VM1036:2 1234 => 1234
VM1036:2 1234 => 1234
2VM1036:2 1234 !@#$%^&*()_ =-;',.><:" => 1234
VM1036:2 1234ABCabc !@#$%^&*()_ =-;',.><:" => 1234
VM1036:2 1234ABCabc !@#$%^&*()_ =-;',.><:" => 1234
VM1036:2 Aa1234ABCabc !@#$%^&*()_ =-;',.><:" => 1234
VM1036:2 a 1234ABCabc !@#$%^&*()_ =-;',.><:" => 1234
VM1036:2 1 1234ABCabc !@#$%^&*()_ =-;',.><:" => 1
您描述的目標和最后一個示例相互矛盾。請更具體地說明您需要什么。
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