<div>呵呵，还有如下的方法：）</div>
<div> </div>
<div>my $a = '/sts3c_au4-j=[1..64]/vt2_tu12-k=[1..3]-l=[1..7]-m=[1..3]';</div>
<div>
<p>my $i;<br>$i = split /=/,$a;</p>
<p>print $i-1;</p>
<p> </p><br><br> </div>
<div><span class="gmail_quote">On 8/4/06, <b class="gmail_sendername">潘建锋</b> <<a href="mailto:panjianfeng1979@gmail.com">panjianfeng1979@gmail.com</a>> wrote:</span>
<blockquote class="gmail_quote" style="PADDING-LEFT: 1ex; MARGIN: 0px 0px 0px 0.8ex; BORDER-LEFT: #ccc 1px solid">
<div>谢谢你了，没有专门的函数吗？<br><br>
<div><span class="gmail_quote">2006/8/4, fayland <<a onclick="return top.js.OpenExtLink(window,event,this)" href="mailto:fayland@gmail.com" target="_blank">fayland@gmail.com</a>>:</span></div>
<div><span class="e" id="q_10cd82ffb1408de6_1">
<blockquote class="gmail_quote" style="PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: rgb(204,204,204) 1px solid">潘建锋 wrote:<br>> 比如：/sts3c_au4-j=[1..64]/vt2_tu12-k=[1..3]-l=[1..7]-m=[1..3]<br>> 中"="出现的次数。
<br>> 谢谢！<br><br>my $a = '/sts3c_au4-j=[1..64]/vt2_tu12-k=[1..3]-l=[1..7]-m=[1..3]';<br><br>my $i;<br>$i++ while ($a =~ /=/g); <br><br>print $i;<br><br>> ------------------------------------------------------------------------
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