打印字段的名稱在表

Question

Possible Duplicate:
mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

你將如何打印表，其中一個字段值的變量的值相匹配的名稱。我有一些代碼：

$query = "select * from event WHERE eventname = $eventname"; 
$result = mysql_query($query); 
$numcolumn = mysql_num_fields($result); 
for ($i = 8; $i < $numcolumn; $i++) { 
$columnnames = mysql_field_name($result, $i); 
echo $columnnames . "<br />"; 
} 

這給了我以下錯誤：

Warning: mysql_num_fields() expects parameter 1 to be resource, boolean given in /var/www/sportevent/eventform.php on line 196 

Answer 1

您查詢失敗;你應該抓住並輸出錯誤信息。

if (!$result) { 
    die('Invalid query: ' . mysql_error()); 
} 

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements /* */ or FALSE on error.

- >http://www.php.net/manual/en/function.mysql-query.php

Answer 2

使用

$result = mysql_query($query); 

if (!$result) { 
    echo 'Could not run query: ' . mysql_error(); 
    exit; 
} 

$numcolumn = mysql_num_fields($result); 
...