Chris W. Parker wrote: > Dan Trainor <mailto:info@xxxxxxxxxxxxxxxx> > on Thursday, July 21, 2005 2:03 PM said: > > >>I never see "hi" even if an array is set as such: >> >>$vars = array("one","two","three","four","five"); > > > That's because your function always returns true. If it finds a missing > value it returns true. If it doesn't find a missing value it returns > true. Because of this the true code block is always executed. > > You should do this instead: > > function hasMissingVals($input) > { > // make sure we've been passed an array with values > if(is_array($input) && (count($input) > 0)) > { > foreach($input as $v) > { > if(empty($v)) > { > return true; > } > } > } > > return false; > } > > This way your function will always return false unless it finds a > missing value in which case it will return true. > > As you can see I changed the name slightly from 'findMissingVals()' to > 'hasMissingVals()' since your function is not actually finding any > missing values. Finding suggests that your function will return data > based on the location of the missing value which it does not do. > > >>if (findMissingVals($vars)) { >> if (!$var1) { ?> hi1 <? }; >> if (!$var2) { ?> hi2 <? }; >> if (!$var3) { ?> hi3 <? }; >> if (!$var4) { ?> hi4 <? }; >> if (!$var5) { ?> hi5 <? }; >> if (!$var6) { ?> hi6 <? }; >>} else { >> echo "hi"; >>} > > > 1. Where are $var1, $var2, $var3, etc. coming from? I think you mean to > write $var[1], $var[2], $var[3], etc.? > > 2. It will be a good idea to get out of the habit of breaking in and out > of PHP like that. Instead just do: echo 'hi1'; > > You'll have problems down the road with modifying the headers (cookies, > redirects, content-type, etc.) if you break in and out. > > > Hope this helps, > Chris. > Hello, Chris - I see what you're saying about always returning true. I really understand now. I'd just like to take a minute to tell you thanks for the help. Thanks! -dant -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
