Php test empty string

Question

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

• Kamil Sindi: I think it's because the variable is NULL if it's not defined

Answer 1

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/>";

• FloydThreepwood: The snippet works fine for me via php cli, would be interesting whats exactly inside the variable.
• jcmitch: Thank you very much!!! var_dump showed me what I actually had in the variable. Turns out that I had an xml tag in there that wouldn't print because it was in <> but var_dump showed it. The above code works now!
• greg0ire: Cool! Now I guess you won't use the "*" . $var . "*" method again ;)
• Kingsolmn: How does this response answer the OP's question?
• greg0ire: @Kingsolmn: Let's sum up the question : "I want to test to see if it is empty". How does my response not answer this question?

Answer 2

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 3

Do it with if (isset($contactId)) {}.

• Fatih Donmez: then it means $contacId isnt set yet. just set it before using
• pomaxa: if you got '' <- empty string, then isset() return true!
• greg0ire: @jcmitch : if this evaluates to false, $contactId is not set so you should get a notice telling you so. Make sure your php setting display_errors is set to 1, and that error_reporting is set to -1

Answer 4

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: https://www.php.net/manual/en/language.types.null.php

• pomaxa: not really good, cuz if you haven't yet got any value into $contactId, - you'll get notice

Answer 5

In future, use if(!empty($str)) { echo "string is not empty"}.

• Jonathan M: OP, FYI, This will exclude the possibly legitimate $contactId="0";.
• pomaxa: Yes. empty will return true for: "" (an empty string), 0 (0 as an integer), 0.0 (0 as a float), "0" (0 as a string), NULL, FALSE, array() (an empty array), var $var; (a variable declared, but without a value in a class), for your case - $contactId, if it is numeric, then just check it like is_numeric