我试图将MySQL选择的输出写入变量。问题是我得到错误“数组到字符串转换”。“将数组写入字符串”同时将MySQL输出写入变量
$user = mysql_query("select username from user");
echo "<table border='1'>";
if (isset($_POST['winneron'])) {
echo "<tr>";
while ($printuser = mysql_fetch_array($user)) {
echo "<th align='center'>". $printuser['username'] . "</th>";
}
echo "</tr>";
$user = mysql_query("select username from user");
while ($printuser = mysql_fetch_array($user)) {
$games = mysql_query("SELECT s.spielid, date, team1, team2, sieger, wettid, u.userid, w.spielid, team
FROM user u, spiele s, wette w
WHERE u.userid = w.userid
AND u.username = '$printuser'
AND w.spielid = s.spielid"); <-- Error line
while ($printgames = mysql_fetch_array($games)) {
if ($printgames['sieger'] == $printgames['team1']) {
echo "<tr><td align='center'><b>". strtoupper($printgames['sieger']) . "</b></td></tr>";
}
else {
echo "<tr><td align='center'>". strtoupper($printgames['sieger']) . "</td></tr>";
}
}
}
}
什么问题更换呢? – Dev
在哪一行?请注意,如果您尝试从相同的$用户获取两次,下次可能会得到一个空数组 – Fallen
编辑代码使其更易于阅读。 已经认为$ user在第一次读取后可能是空的,所以我在while循环之后再次添加了它。 – ProfGhost