涉及MySQL的Web（PHP）错误

修复了我的主MySQL错误之后，我现在发现包含我创建的MySQL表的网站上的PHP存在问题。我收到的文字：涉及MySQL的Web（PHP）错误

警告：mysql_fetch_array（）预计参数1是资源，布尔在在线/home/content/26/10406626/html/highscore/index.php给出258

这是我$结果文件中：

if ($skill == "Attack" || $skill == "Defence" || $skill == "Strength" || $skill == "Hitpoints" || $skill == "Range" || $skill == "Prayer" || $skill == "Magic" || $skill == "Cooking" || $skill == "Woodcutting" || $skill == "Fletching" || $skill == "Fishing" || $skill == "Firemaking" || $skill == "Crafting" || $skill == "Smithing" || $skill == "Mining" || $skill == "Herblore" || $skill == "Agility" || $skill == "Thieving" || $skill == "Slayer" || $skill == "Farming" || $skill == "Runecraft" || $skill == "Hunter" || $skill == "Construction" || $skill == "Summoning" || $skill == "Dungeoneering") { 
    mysql_select_db("scores", $con); 
    $result = mysql_query("SELECT * FROM skills WHERE ". $PLAYERS_TO_NOT_SHOW . " ORDER BY " . $skill . "lvl DESC, " . $skill . "xp DESC"); 
    } 

else { 
    $skill = ""; 
    mysql_select_db("scores", $con); 
    $result = mysql_query("SELECT * FROM skillsoverall WHERE ". $PLAYERS_TO_NOT_SHOW . " ORDER BY lvl DESC, xp DESC"); 
    }

钍是为线254至278：显然，线258是*而（$行= mysql_fetch_array（$结果））*

<?php 

$rank = 1; 

while($row = mysql_fetch_array($result)) 
    { 
    echo "<a name=\"" . $rank . "\"></a>"; 
    echo "<a class=\"row\">"; 
    echo "<span class=\"columnRank\">"; 
    echo "<span>" . $rank . "</span>"; 
    echo "</span>"; 
    echo "<span class=\"columnName\">"; 
    echo "<span>" . $row['playerName'] . "</span>"; 
    echo "</span>"; 
    echo "<span class=\"columnLevel\">"; 
    echo "<span>" . $row[$skill . 'lvl'] . "</span>"; 
    echo "</span>"; 
    echo "<span class=\"columnXp\">"; 
    echo "<span>" . $row[$skill . 'xp'] . "</span>"; 
    echo "</span>"; 
    echo "</a>"; 
    $rank++; 
    } 
mysql_close($con); 
?>

来源

2014-02-20 user2619426

你能还供应其中'$ result'设置的线路？ – EWit

@EWit是的，我只是做到了。抱歉! – user2619426

什么是$ PLAYERS_TO_NOT_SHOW？我认为这是你的问题来自 – CodeBird

您的查询失败，因此不会产生查询的资源，而是生产假。

透露自己的动态生成的查询是什么样子，揭示错误，试试这个：

$result2 = mysql_query($result) or die($result."<br/><br/>".mysql_error());

来源

2014-02-20 23:20:56 user3329793

你的问题躺在这儿：

$result = mysql_query("SELECT * FROM skills WHERE ". $PLAYERS_TO_NOT_SHOW . " ORDER BY " . $skill . "lvl DESC, " . $skill . "xp DESC");

怎么一回事，因为的mysql_query函数返回布尔值，我suposse是假的。从php.net手册 http://pl1.php.net/mysql_query的mysql_query函数返回：

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning 
resultset, mysql_query() returns a **resource** on success, or FALSE on error. 

For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, 
mysql_query() returns TRUE on success or FALSE on error.

来源

2014-02-20 23:22:44 ZiupeX

涉及MySQL的Web（PHP）错误

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