How to add a array variable parameter in mysql query?(PHP)
I need to do a query like:
UPDATE screening
SET maileddate = NOW(),
reference = '{$reference[$user_id]}'
W开发者_JS百科HERE user_id IN(....)
And I want to do the judgement, when reference[$user_id] is empty, reference[$user_id] = NULL. My code is :
if(empty($reference[$user_id]) || $reference[$user_id] == ''){
$reference[$user_id] = NULL;
}
But when I execute the query, if $reference[$user_id] is empty, the value of reference in database is empty but not null. What should I do to make it null? thanks
You may need to pass NULL
as a string to MySQL if the variable is empty. Use a different variable to hold the possibly NULL contents and quote the non-null contents. Don't forget to escape it otherwise:
$refid = empty($reference['user_id']) ? "NULL" : "'" . mysql_real_escape_string($reference['user_id']) . "'";
UPDATE screening SET maileddate = NOW(), reference = '{$refid}'
WHERE user_id IN(....)
Just make it a string saying null. If it's null $reference[$user_id] = 'null';
Oh, also with the query you should be using reference IS null {$reference[$user_id]}
instead of the equals sign
$reference = 'reference = ';
$reference .= ' empty($reference[$user_id]) ? 'NULL' : "'". mysql_real_escape_string($reference[$user_id]) . "'";
$query = "UPDATE screening
SET maileddate = NOW(),
$reference
WHERE user_id IN(....)";
Without seeing the rest of your code, my guess is it has something to do with reference = '{$reference[$user_id]}'
having {$reference[$user_id]}
in single quotes. MySQL is going to see whatever is in there as what should be in the database. So if $reference[$user_id]
prints out as nothing (because it's NULL), that bit of your query will be reference = ''
rather than reference = NULL
. You need to use the keyword NULL rather than the actual value NULL for the variable you use in the query.
As an example:
$query = "UPDATE screening " .
"SET maileddate = NOW(), " .
"reference = " . ($reference[$user_id] === NULL ? 'NULL ' : "'{$reference[$user_id]}' ") .
"WHERE user_id IN(....)";
Try this:
$user_id = empty($reference[$user_id]) ? 'NULL' : $reference[$user_id];
$sql = "UPDATE screening
SET maileddate = NOW(),
reference = " . $user_id . "
WHERE user_id IN(....)";
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