I'm sure this is a simple fix, I want to run a code block if an sql query comes back with a positive result. something like:
if($query = mysqli_query($cxn,"[query]"))
{
Code to be executed if the query returns a positive result
}
I have tried this format but doesn't work. I'm sure I have done this before but I'm running in to a wall here.
Hope you can help.
As stated in php.net/manual/mysqli.query.php, mysqli_query will:
Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.
You should for you next question specify what you mean by positive result. But this code will see if there was an error, if the query returned an empty set, and so on... (I have not tried to run the code)
$result = mysqli_query($cxn,"[query]");
if ($result === FALSE) {
echo "There was an error";
}
else if ($result === TRUE) {
echo "The query was successful (a query that didn't return anything)";
}
else if (mysqli_num_rows($result) == 0) {
echo "The result is empty"
}
else {
echo '$result contains data';
}
So you mean if the query doesn't fail? then:
$query = mysqli_query($cxn, "[query]");
if($query !== false) {
// Code to be executed if the query returns a positive result
}
else { print mysqli_error($cxn); }