Use imploded result from MySql Query in PHP in_array() statement

Question

Trying to use a string of coma separated values compiled from a mysql query within an in_array() statement. The problem is that the statement fails because the output of $array does not work in the in_array() function even though the pasted output of array does work...

$quiz = mysql_query("SELECT id FROM quiz WHERE quiz_page = '2'");
$array1=array();
while($quiz2 = mysql_fetch_assoc($quiz))
{
    $array1[]=$quiz2["id"];
}
$array = implode(",", $array1);
echo $array;
echo '<br>';

if (in_array(184, array($array))) {
echo '184 in $array';
} else {
echo '184 not in $array';
}
echo '<br>';

if (in_array(184, array(77,82,85,90,180,181,182,183,184,185))) {
echo '184 in array';
} else {
echo '184 not in array';
}

The result of the code above:

77,82,85,90,180,181,182,183,184,185
184 not in $array
184 in array

Don't Panic · Accepted Answer · 2016-03-17 22:59:54Z

2

You need to be checking if (in_array(184, $array1)) {... instead. $array1 already contains each of numbers in a separate elements, like you have in your second if condition, e.g.

array(77,82,85,90,180,181,182,183,184,185)

When you do array($array) You get an array with one element, the entire imploded string, e.g

array("77,82,85,90,180,181,182,183,184,185")

I'm not sure exactly what you're doing with this, but it looks like you might be able to eliminate the whole in_array problem by just including the 184 in your query

SELECT id FROM quiz WHERE quiz_page = '2' AND id = 184

And checking whether or not the query returns anything.

edited Mar 17, 2016 at 22:59

answered Mar 17, 2016 at 22:51

Don't Panic

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2 Comments

Joshua Over a year ago

This worked, thank you. The '184' was just for demonstration purposes. Is there a way to use the mysql querry in the in_array() function with a while loop?

Don't Panic Over a year ago

I'm sorry, I am not really sure what you mean. From what I've seen, fetching in a while loop is the normal way to get your query results.

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