我不知道为什么,但我在这一行收到错误:$row = mysql_fetch_array($run_games)){
这是代码:
<div class='topnav'> <!--Start of the Top Navigation-->
<?php
include("includes/connect.php");
$select_games = "SELECT * FROM games";
$run_games = mysql_query($select_games);
$row = mysql_fetch_array($run_games)){
$game_category = $row['game_category'];
?>
<!--some links here with the variable of "game_category"-->
<?php } ?>
</div> <!--End of the Top Navigation-->
请帮帮我:(
答案 0 :(得分:0)
应该是......
while($row = mysql_fetch_array($run_games))
{
echo $game_category = $row['game_category'];
}
您需要遍历结果集!
答案 1 :(得分:0)
使用 while 循环遍历结果集。
$row = mysql_fetch_array($run_games)){
应该是
while($row = mysql_fetch_array($run_games)) {
// do stuff ...
}
答案 2 :(得分:0)
这可能是剪切和粘贴错误。缺少while
循环的第一部分:
$row = mysql_fetch_array($run_games)){
应该是
while ($row = mysql_fetch_array($run_games)){
答案 3 :(得分:0)
不推荐使用这些功能,因此请使用最新功能
$row = mysql_fetch_array($run_games));//if only one row
//other wise should use while
while($row = mysql_fetch_array($run_games))
{
}