如何逃脱MySQL'无法跳转0行'错误？

My question is how can I be sure of a row that I'd like to return is exists? I don't want to suppress it with PHP's @ option or count rows before every query to find out the row is exists or not.

So there's a simple query like this:

"SELECT `title`, `id` FROM `event` WHERE `id` = '234'";

and the table cannot contain the row with id 234.

You're probably using mysql_result() to fetch the fields. Consider mysql_fetch_array instead. It returns FALSE if there are no more rows to fetch.

<?php
$mysql = mysql_connect('..', '..', '..') or die(mysql_error());
mysql_select_db('..', $mysql) or die(mysql_error());

$query = "SELECT `title`, `id` FROM `event` WHERE `id` = '234'";
$result = mysql_query($query, $mysql) or die(mysql_error());
$row = mysql_fetch_array($result, MYSQL_ASSOC);

if ( !$row ) {
  echo 'no such record';
}
else {
  echo $row['title'], ' ', $row['id'];
}

You don't have to count rows before every query - generally you do it after.

What about something like this

$query = "SELECT `title`, `id` FROM `event` WHERE `id` = '234'";

$results = mysql_query($query);

if (mysql_num_rows($results)) {
    // do something because it was found
}