I have a bit of php code that I’m not understanding why it is acting as it is. I have a variable called contactId that I want to test to see if it is empty. However even if it is empty it evaluates to true. Code is below. Thanks in advance.
print "*".$contactId."*<br/>";
if($contactId != '')
{
//queryContact($contactId);
print "Contact Present<br/>";
}
result returned to screen is:
**
Contact Present
If you want to see exactly what your string is, simply use var_dump(), like this, for instance:
var_dump($contactId)
instead of
print "*".$contactId."*<br/>";
Answer：
Couple of things you can try:
if (!empty($contactId)) {
// I have a contact Id
}
// Or
if (strlen($contactId) > 0) {
// I have a contact id
}
In my experience I have often used the latter of the two solutions because there have been instances where I would expect a variable to have the value of 0, which is valid in some contexts. For example, if I have a drink search site and want to indicate if an ingredient is non-alcoholic I would assign it a value of 0 (i.e. IngredientId = 7, Alcoholic = 0).
Answer：
Do it with if (isset($contactId)) {}.
Answer：
You likely want:
if (strlen($contactId))
You’ll want to learn the difference between '' and null, and between == and ===. See here: http://php.net/manual/en/language.operators.comparison.php
and here: http://us3.php.net/manual/en/language.types.null.php
Answer：
In future, use if(!empty($str)) { echo "string is not empty"}.
