我有以下代码,但它似乎打破$result = $mysqli->query($query);
语句的页面。我究竟做错了什么?MySQL和PHP语句中的错误
$ids = join(',',$newArray);
$query = "SELECT name, image FROM logos WHERE id IN ($ids)";
$result = $mysqli->query($query);
while ($row = $result->fetch_assoc()) {
$imageLink = $row['logoImageLink'];
echo "<li><img src=\"$imageLink\"/></li>";
}
您是否尝试过'var_dump'ing你'$ query'? – Havelock 2013-04-05 19:08:45