how can display any variable contant integer value is_int()?

Question

I have this code:

$showCountSql    = "select cad_count from counteraccountdtl WHERE cad_userid =".$_SESSION['UID']." LIMIT 1";
$showCountresult = mysql_query($showCountSql);
$showCountrow    = mysql_fetch_array($showCountresult);
$newCount        = $showCountrow[cad_count];

if(is_int($newCount))
 echo "Value  is Integer";
else
 echo "Value not Integer";

I am fetching the value from MySql as "cad_count integer(5)", then I check whether this value is an integer or not and show the "Value not Integer" accordingly. What's wrong in it?

You can always use `var_dump($newCount)` to see exactly what the variable's type is, along with contents. — Marc B, Sep 12 '11 at 15:14
For completness to the answers, `mysql_fetch_array` returns an array of strings. See docs: http://www.php.net/manual/en/function.mysql-fetch-array.php — enricog, Sep 12 '11 at 15:22

score 2 · Answer 1 · answered Sep 12 '11 at 15:12

2

Use is_numeric() or ctype_digit(). These functions test if the given variable is a valid representation of a number or contains only digits.

is_int tests if the variable's type is int, and mysql_fetch* functions return integers as strings.

answered Sep 12 '11 at 15:12

Arnaud Le Blanc

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score 0 · Answer 2 · answered Sep 12 '11 at 15:14

0

is_numeric() is better for you. To make is_int() work, you have to make int value from old one

$var = intval($var);

answered Sep 12 '11 at 15:14

genesis

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how can display any variable contant integer value is_int()?

2 Answers2