擅长:python、mysql、java
<pre><code>s = 'gfdhbobobyui'
bob = 0
for x in range(len(s)):
if s[x:x+3] == 'bob': # From x to three 3 characters ahead.
bob += 1
print('Number of times bob occurs is: ' + str(bob))
</code></pre>
<p>正在工作<a href="http://ideone.com/sQVAcc" rel="nofollow">example</a>。</p>
<p>但是,最好的方法是这样,但是它不适用于重叠字符串:</p>
<pre><code>s.ount('bob')
</code></pre>