当我回显sql查询时，它显示
the volunteer id is 312INSERT INTO volunteer_events (event_id, volunteer_id) VALUES ('91
', '312')
它在'91之后添加了一个新行，而不是在一行中。我想知道我在做什么，导致它添加了一个新的行。
public function addUserToEvent($event,$volunteer)
{
$event_id = $event;
echo "the event id is ". $event_id;
$volunteer_id = $volunteer;
echo "the volunteer id is ". $volunteer_id;
$sql = "INSERT INTO volunteer_events (event_id, volunteer_id) VALUES ('$event_id', '$volunteer_id')";
echo $sql;
return mysql_query($sql);
}
我在这里调用addUserToEvent
if(isset($_POST["events"])){
$eventIds = $_POST["events"];
$N = count($eventIds);
for($i=0; $i < $N; $i++) {
echo "</br>";
echo "The events IDs are";
echo "</br>";
echo($eid = $eventIds[$i]);
$sql = "SELECT distinct v.id
FROM volunteers v
WHERE v.first_name = '$fname'
AND v.last_name = '$lname'
AND v.email = '$email'";
$results = $db->q($sql);
if($row = $results->getAssoc())
{
$vID = $row['id'];
echo "the vid is ". $vID;
}
echo "</br>";
echo " the eid is blah ".$eid;
echo $db->addUserToEvent($eid, $vID);
}
}
很容易修复，只需将其转换为int。
public function addUserToEvent($event,$volunteer)
{
$event = (int) $event;
$volunteer = (int) $volunteer;
$event_id = $event;
echo "the event id is ". $event_id;
[...]
我想的是实地志愿者的数据类型。id是一个varchar类型，它应该是INT类型。