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java × 926620php × 807008 echo 'Android = '.mysqli_query($conn, "SELECT COUNT(*) as tot FROM data where category = 'ajax
我尝试将它插入到代码中,以便通过sql表生成下拉菜单,但它在下拉列表中没有响应。当我执行sql查询时,它也不会显示任何错误。请你纠正我的错误。谢谢 $db = mysqli_connect('localhost', 'root', '', 'registration');
$sql = "Select(unitid) from unit where