Question

我有两个查询计算数据库中“艺术家”和“组”的数据数量。我想显示一条消息，如果有艺术家或团体（或两者）都有数据显示，如果两个数据都返回0，那么就不显示任何内容。

我有以下代码似乎不起作用：

<?php if (($numrowsartists==0)OR($numrowsgroups==0)) {

 } else {
        echo "There is information to display.";
        }
?>

以下是我的查询：

$query = "SELECT COUNT(*) FROM `Credit_To_Artist` AS c2a 
INNER JOIN `Credits` AS cr ON cr.credit_id = c2a.credit_id
INNER JOIN `Artist` AS a ON a.artist_id = c2a.artist_id
WHERE c2a.song_id = $id";

$res = mysql_query($query);

$numrowsartists = mysql_fetch_assoc($res);

$query = "SELECT COUNT(*) FROM `Credit_To_Artist` AS c2a 
INNER JOIN `Credits` AS cr ON cr.credit_id = c2a.credit_id
INNER JOIN `Artist_Group` AS ag ON ag.group_id = c2a.group_id
WHERE c2a.song_id = $id
ORDER BY ag.group_name ASC";

$res = mysql_query($query);

$numrowsgroups = mysql_fetch_assoc($res);

提前致谢。我敢肯定它可能是一个超级基本的修复，但我仍然是非常新的PHP，并希望得到一些帮助。

Answer 1

你应该得到行的值，例如使用列名

的别名

$query = "SELECT COUNT(*) as num_artists  FROM `Credit_To_Artist` AS c2a 
INNER JOIN `Credits` AS cr ON cr.credit_id = c2a.credit_id
INNER JOIN `Artist` AS a ON a.artist_id = c2a.artist_id
WHERE c2a.song_id = $id";

$res = mysql_query($query);

$row = mysql_fetch_assoc($res);
$numrowsartists = row['num_artists'];

$query = "SELECT COUNT(*) as  num_groups FROM `Credit_To_Artist` AS c2a 
INNER JOIN `Credits` AS cr ON cr.credit_id = c2a.credit_id
INNER JOIN `Artist_Group` AS ag ON ag.group_id = c2a.group_id
WHERE c2a.song_id = $id
ORDER BY ag.group_name ASC";

$res = mysql_query($query);
$row = mysql_fetch_assoc($res);
$numrowsgroups = row['num_groups'];

Answer 2

有几种解决方案，最简单的方法如下：

if($numrowsartists[0]+$numrowsgroups[0] > 0)

然而，正如人们所说，你不应该再使用mysql_ *函数了。假设ID是用户输入，您应该使用预准备语句。此外，您可以在单个查询中处理这两个测试：

$stmt = $con->mysqli_prepare("SELECT COUNT(1) as `count` FROM `Credit_To_Artist` AS c2a 
      INNER JOIN `Credits` AS cr ON cr.credit_id = c2a.credit_id
      INNER JOIN `Artist` AS a ON a.artist_id = c2a.artist_id
      INNER JOIN `Artist_Group` AS ag ON ag.group_id = c2a.group_id
      WHERE c2a.song_id = ?");
$stmt->bind_param('i',$id);
$stmt->execute();
if($stmt->get_result()->fetch_array()[0] > 0){
    ...
}else{
    //message that nothing was found
}

使用php if ... else语句和两个查询

2 个答案: