python regex: match a string with only one instance of a character

Question

Suppose there are two strings:

$1 off delicious ham.
$1 off delicious $5 ham.

In Python, can I have a regex that matches when there is only one $ in the string? I.e., I want the RE to match on the first phrase, but not on the second. I tried something like:

re.search(r"\$[0-9]+.*!(\$)","$1 off delicious $5 ham.")

..saying "Match where you see a $ followed by anything EXCEPT for another $." There was no match on the $$ example, but there was also no match on the $ example.

Thanks in advance!

Simple test method for checking:

def test(r):
  s = ("$1 off $5 delicious ham","$1 off any delicious ham")    
  for x in s:
    print x
    print re.search(r,x,re.I)
    print ""

Answer 1

>>> import re
>>> onedollar = re.compile(r'^[^\$]*\$[^\$]*$')
>>> onedollar.match('$1 off delicious ham.')
<_sre.SRE_Match object at 0x7fe253c9c4a8>
>>> onedollar.match('$1 off delicious $5 ham.')
>>>

Breakdown of regexp:
^ Anchor at start of string
[^\$]* Zero or more characters that are not $
\$ Match a dollar sign
[^\$]* Zero or more characters that are not $
$ Anchor at end of string

Answer 2

re.search("^[^$]*\$[^$]*$",test_string)

Answer 3

^.*?\$[^$]*$

this should make the trick

Answer 4

You want to use the complement of a character class [^] to match any character other than $:

re.match(r"\$[0-9]+[^\$]*$","$1 off delicious $5 ham.")

The changes from your original are as follows:

1. .* replaced with [^\$]*. The new term [^\$] means any character other than $
2. $ appended to string. Forces the match to extend to the end of the string.
3. re.search replaced with re.match. Matches the whole string, rather than any subset of it.

Answer 5

>>> '$1 off delicious $5 ham.'.count('$')
2
>>> '$1 off delicious ham.'.count('$')
1