php - 无法跳转到 MySQL 结果索引上的第 0 行..？

Question

我收到以下错误 block (值得注意的是，它会立即重复，为了简单起见，我只复制了一次迭代):

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 89

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 91

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 93

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 95

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 97

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 99

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 101

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 9 in /home/a-slsa/www/classes/ASC/ASCStaffData.php on line 103

引用的行如下(是的，我知道使用已弃用的函数会很糟糕，但这不是我的代码 - 从其他人那里继承了一个相当大的项目):

    $ascStaff['staffId'] = (mysql_result($dbStaffData->fetch,0,"STAFF_ID") ?
                                mysql_result($dbStaffData->fetch,0,"STAFF_ID") : NULL);

    $ascStaff['ownership'] = (mysql_result($dbStaffData->fetch,0,"OWNERSHIP") ?
                                mysql_result($dbStaffData->fetch,0,"OWNERSHIP") : NULL);    

    $ascStaff['firstName'] = (mysql_result($dbStaffData->fetch,0,"FIRST_NAME") ?
                                mysql_result($dbStaffData->fetch,0,"FIRST_NAME") : NULL);

    $ascStaff['lastName'] = (mysql_result($dbStaffData->fetch,0,"LAST_NAME") ?
                                mysql_result($dbStaffData->fetch,0,"LAST_NAME") : NULL);

    $ascStaff['ritEmail'] = (mysql_result($dbStaffData->fetch,0,"RIT_EMAIL") ?
                                mysql_result($dbStaffData->fetch,0,"RIT_EMAIL") : NULL);

    $ascStaff['otherEmail'] = (mysql_result($dbStaffData->fetch,0,"OTHER_EMAIL") ?
                                mysql_result($dbStaffData->fetch,0,"OTHER_EMAIL") : NULL);

    $ascStaff['address'] = (mysql_result($dbStaffData->fetch,0,"ADDRESS") ?
                                mysql_result($dbStaffData->fetch,0,"ADDRESS") : NULL);

    $ascStaff['phoneNumber'] = (mysql_result($dbStaffData->fetch,0,"PHONE_NUMBER") ?
                                    mysql_result($dbStaffData->fetch,0,"PHONE_NUMBER") : NULL); 

令我困惑的是，当手动输入 mysql 时，查询运行得很好。

知道为什么我会收到此错误或如何修复它吗？

Answer 1

根据我的评论:mysql 扩展已被弃用，因此即使您正在使用旧的代码库，您也应该考虑升级到 mysqli。

鉴于此，通常此错误表明查询失败。

以类似于此的方式检查查询:

$result = mysql_query("SELECT foo FROM bar WHERE foo = 1");
if(!$result || !mysql_num_rows($result))
{
    die("Empty dataset.");
}