當我運行這段代碼,它返回話題罰款...獲取SQL記錄
$query = mysql_query("SELECT topic
FROM question
WHERE id = '$id'");
if(mysql_num_rows($query) > 0) {
$row = mysql_fetch_array($query) or die(mysql_error());
$topic = $row['topic'];
}
但是當我改變這一點,它不會運行在所有。爲什麼發生這種情況?
$query = mysql_query("SELECT topic, lock
FROM question
WHERE id = '$id'");
if(mysql_num_rows($query) > 0) {
$row = mysql_fetch_array($query) or die(mysql_error());
$topic = $row['topic'];
$lockedThread = $row['lock'];
echo "here: " . $lockedThread;
}
究竟是如何做的不跑?有沒有錯誤信息? – 2012-04-13 17:26:37
此外,您正在爲查詢插入'$ id',而不是將其作爲安全參數傳遞。 – vol7ron 2012-04-13 17:28:13
我有這個:error_reporting(1); – droidus 2012-04-13 17:28:21