Question

修复我的MySQL主要错误后，我现在认为网站上的PHP存在问题，包括我创建的MySQL表。我收到了文字：

警告：mysql_fetch_array（）要求参数1为资源，在第258行的/home/content/26/10406626/html/highscore/index.php中给出布尔值

这是我在文件中 $ result 的地方：

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");
    }

这是第254至278行：显然，第258行是 *而（$ row = mysql_fetch_array（$ result））*

<?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);
?>

Answer 1

您的查询失败，因此不会生成查询资源，而是生成FALSE。

要显示动态生成的查询的内容并显示错误，请尝试以下操作：

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

Answer 2

你的问题在这里：

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

beacuse mysql_query函数返回布尔值i suposse为false。来自php.net手册 http://pl1.php.net/mysql_query 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.

涉及MySQL的Web（PHP）错误

2 个答案: