Question

我有sql语句将结果添加到变量中但当我将该变量（$ team）放入我的第二个sql语句时，它不起作用WHERE id ='$ team'。我该怎么做呢？

$query = mysql_query("SELECT team_name FROM team");
$team = array();
while($row = mysql_fetch_assoc($query)){
$team[] = $row;}
echo $arra[0];
$loop=count($team);
for($x=0;$x<$loop;$x++)
foreach($team[$x] as $child) {



$result = mysql_query("SELECT * FROM members 
WHERE id =  '$team'") 
or die(mysql_error());  
while($row = mysql_fetch_array( $result )) {
echo '<td>' . $row['first_name'] . '</td>'; 
}

Answer 1

如果$ team是一个数组，则使用join php函数：

$team = join(',', $team);
$result = mysql_query("SELECT * FROM members WHERE id in ($team)") or die(mysql_error());

Answer 2

我同意jetawe使用join，但我建议使用不同类型的join。

SELECT m.first_name
FROM team AS t
LEFT OUTER JOIN members AS m ON m.id=t.team_name;

PS。请务必使用表“团队”的代理键，而不要使用名称。

Answer 3

我认为你的问题不清楚，你的代码有点乱。

$query = mysql_query("SELECT team_name FROM team");
// loop through each team
while($team = mysql_fetch_assoc($query)){
    // find members for this team
    $query = mysql_query("SELECT * FROM members WHERE id = '" . $team['team_name'] . "'") or die(mysql_error());
    while ($member = mysql_fetch_array($query)) {
        echo '<td>' . $member['first_name'] . '</td>'; 
    }
}

但是，最好的方法是使用user1032531之类的联接，而不是多次调用数据库。

sql语句中的数组变量

3 个答案: