Posted by Dan Trainor on 07/21/05 23:51

Mike Johnson wrote:
> From: Dan Trainor [mailto:info@hostinthebox.net]
>
>
>>Hello, all -
>>
>>I've been looking around for a function that would tell me if a $value
>>in a $key=>$value array was empty, and I could not find one. So I
>>decided to make my own. Even if I am re-inventing the wheel,
>>I thought that the practice might be good for me.
>>
>>However, my function doesn't *quite* work, and I'm having a difficult
>>time finding out why. The code is as follows:
>>
>>function findMissingVals($workingArray) {
>> $newcount = count($workingArray);
>> for ($i = 0; $i <= $newcount; $i++) {
>> if (empty($workingArray['$i'])) {
>> return 1;
>> }
>> }
>>}
>>
>>So it takes in $workingArray as an array, runs a loop, checks $i, yada
>>yada. The thing is, that sometimes the function does not
>>return 1, even when it should.
>>
>>I was hoping some experienced eyes could take a gander at
>>this and give me some pointers.
>
>
> PHP doesn't eval code in single-quotes, so what you want to do is
> simply:
>
> if (empty($workingArray[$i])) {
> return 1;
> }
>
> With the single-quotes, it's looking for the string $i as a key.
>
> HTH!
>


Hey there, Mike -

Your tips were very helpful, thank you. I saw my error, but I am still
having problems. Being somewhat novice to PHP, I think my error might
very well just be in my implementation of the function, as follows:

function findMissingVals($workingArray) {
$newcount = count($workingArray);
for ($i = 0; $i <= $newcount; $i++) {
if (empty($workingArray[$i])) {
return 1;
}
}
}

if (findMissingVals($vars)) {
if (!$var1) { ?> hi1 <? };
if (!$var2) { ?> hi2 <? };
if (!$var3) { ?> hi3 <? };
if (!$var4) { ?> hi4 <? };
if (!$var5) { ?> hi5 <? };
if (!$var6) { ?> hi6 <? };
} else {
echo "hi";
}


I never see "hi", even if I have an array as $vars as such:

$vars = array("one","two","","four","five");

so I'm a bit confused.

If you wouldn't mind taking another few minutes here, I would greatly
appreciate it.

Thanks!
-dant

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