var str = “I learned to play the Ukulele in Lebanon.” var regex = /le/gi, result, indices = []; while ( (result = regex.exec(str)) ) { indices.push(result.index); }
我未能在原始问题中发现搜索字符串需要是一个变量。我编写了另一个版本来处理使用的这种情况indexOf，因此您回到了起点。正如Wrikken在评论中所指出的那样，要对具有正则表达式的一般情况执行此操作，您需要转义特殊的正则表达式字符，这时我认为正则表达式解决方案变得比其价值更令人头疼。
function getIndicesOf(searchStr, str, caseSensitive) {
var searchStrLen = searchStr.length;
if (searchStrLen == 0) {
return [];
}
var startIndex = 0, index, indices = [];
if (!caseSensitive) {
str = str.toLowerCase();
searchStr = searchStr.toLowerCase();
}
while ((index = str.indexOf(searchStr, startIndex)) > -1) {
indices.push(index);
startIndex = index + searchStrLen;
}
return indices;
}
var indices = getIndicesOf("le", "I learned to play the Ukulele in Lebanon.");
document.getElementById("output").innerHTML = indices + "";
<div id="output"></div>
602392714
清零编程群