Strange IF condition Behavior

PHP
1

Hi folks,

echo "Days Left : " . $days_left_user . "<br>";
echo "Position ID : " . $pid . "<br>";
echo "Country ID : " . $cid . "<br>";

if(($days_left_user==0) and ($pid==0) and ($cid==0)){
   echo "You must select atlest one option above to filter results.";
   exit;
}else{
   //show search results

}

Now, consider following output

Days Left : 0
Position ID : 0
Country ID : JOR

You must select atlest one option above to filter results.

So why it is showing “You must select atlest one option above to filter results” insted of search results?

2

Because you are comparing a string to an integer, which is evaluated to true.

var_dump(0 == "a"); // 0 == 0 -> true

http://php.net/manual/en/language.operators.comparison.php

You also should use “===”, as much as possible to check the type comparison is also correct. Like

if(($days_left_user === 0) and ($pid === 0) and ($cid === "")){

Only use “==”, when you need type coercion.

Scott

1 Like
3

Hi s_molinari

Worked Charm!! Thank you very much for the useful information.

if(($days_left_user==0) and ($pid==="0") and ($cid==="0")){
   echo "You must select atlest one option above to filter results.";
   exit;
}
  • days left is usually an integer and other two variables are string variables comes from html list controls.
4

You can also check it that way:

if(($days_left_user==0) && !$pid && !$cid)){
    echo "You must select atlest one option above to filter results.";
    exit;
}

Condition !$var (“not var”) equals to true when $var contains empty string, empty array or zero (integer or “0”).

2 Likes
5

Thanks buddy got it.

6

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