Oh, try the true cycle, find the presented key. Then make the mark.
This method is not high in efficiency :(
For ($ I = 0; $ i
{
For ($ J = 0, $ Numb = 0; $ i
{
IF (or 4) == ORD ($ key [j])) / / determine if each ACI is equal
{
$ NUMB ;
}
IF ($ NUMB == (count ($ key) -1)))
{
// means that I want $ key I need, then the value of $ i can be recorded in another array variable.
}
}
}
The above code is not debugged. Just an idea.
Http://search.9cbs.net/expert/topic/2576/2576471.xml?temp=.4132349
