警告：mysql_result() [function.mysql-result]：无法跳转到第 11 行 profile.php 中 MySQL 结果索引 5 的第 0 行
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Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 5 in profile.php on line 11
提问by Caiapfas
When I try to access profile.php?u=destiny
当我尝试访问时 profile.php?u=destiny
//$result = mysql_query('SELECT name FROM
$imageresult = mysql_query("SELECT name FROM imagetable WHERE id = '$id'") or die(mysql_error());
$u = mysql_result($imageresult, 0 ,"name") or die(mysql_error());
//error_reporting(E_ALL);
if (isset($id) && (!isset($u))) {
}
Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 5 in profile.php on line 11
警告：mysql_result() [function.mysql-result]：无法跳转到第 11 行 profile.php 中 MySQL 结果索引 5 的第 0 行
回答by thewebguy
This warning means that there is no row in the $imageresultvar. Check it out, this should work:
此警告意味着$imageresultvar 中没有行。检查一下，这应该有效：
$imageresult = mysql_query("SELECT name FROM imagetable WHERE id = '$id'") or die(mysql_error());
if (mysql_num_rows($imageresult) > 0) {
$u = mysql_result($imageresult, 0 ,"name") or die(mysql_error());
if (isset($id) && (!isset($u))) {
}
}
