MySQL从表WHERE值> 0中选择行

Question

嘿，即时通讯在构建此mysql查询时遇到了麻烦，令我难堪了大约一个星期，即时通讯希望有人能提供答案并可以帮助我学习！

我的桌子看起来像这样：

我想要什么，我希望从表中获取所有值用户名或email ='abc123'并且攻击强度防御敏捷性等的值> 0

因此，基本上，我如何才能只获得列中的值大于0并且值与正确的用户名和电子邮件相关联的行

任何帮助将非常感激！ 我不确定我是否以最好的方式解释了这个问题，但我很爱帮助澄清！

到目前为止，这是我一直在工作的

SELECT Attack, Strength, Defense, Magic, Ranged, Prayer, Runecrafting, Dungeoneering, Construction, Constitution, Agility, Herblore, Thieving, Crafting, Slayer, Hunter, Mining, Smithing, Fishing, Cooking, Firemaking, Woodcutting, Farming, Summoning FROM experience WHERE email =  abc123 OR username = abc123

Answer 1

SELECT Attack, Strength, Defense, Magic, Ranged, Prayer, Runecrafting, Dungeoneering, Construction, Constitution, Agility, Herblore, Thieving, Crafting, Slayer, Hunter, Mining, Smithing, Fishing, Cooking, Firemaking, Woodcutting, Farming, Summoning FROM experience WHERE (email =  "abc123" OR username = "abc123") AND Attack > 0 AND Strength > 0 AND Defense > 0 AND *etc*

Answer 2

为何不起作用？ where (email = 'abc123' OR username = 'abc123') and attack > 0 and strength > 0 and 。 。 。

if( (email == 'abc123' || username == 'abc123') && stength > 0)如果），则它的工作方式如下： if( (email == 'abc123' || username == 'abc123') && stength > 0)

“（）”与“ ||”之间的第一部分 表示最后一个条件必须为真。 && outside表示最后一个条件也必须为真。

我还建议您使用用户ID，因为它确实比用户名/电子邮件好

Answer 3

您的解决方案

SELECT Attack, Strength, Defense, Magic, Ranged, Prayer, Runecrafting, Dungeoneering, Construction, Constitution, Agility, Herblore, Thieving, Crafting, Slayer, Hunter, Mining, Smithing, Fishing, Cooking, Firemaking, Woodcutting, Farming, Summoning FROM experience WHERE (email = abc123 OR username = abc123) AND (Attack + Strength + Defense + Magic + Ranged + Prayer + Runecrafting + Dungeoneering + Construction + Constitution + Agility + Herblore + Thieving + Crafting + Slayer + Hunter + Mining + Smithing + Fishing + Cooking + Firemaking + Woodcutting + Farming + Summoning > 0);

不幸的是，除非您有一列来汇总所有统计信息，否则要实现的唯一方法是手动比较所有统计信息，在这种情况下，我将使用+对其进行汇总。

Answer 4

无法找到一种方法将数据从数据库中移出我想要的方式，所以我决定最好从表中提取所有数据并使用php对其进行过滤会更好

$stmt2 = $DB_con->prepare("SELECT * FROM experience WHERE username = :uname OR email = :email");
$stmt2->execute(array(':uname' => $_SESSION['username'], ':email' => $_SESSION['email']));
$rowExperience = $stmt2->fetch(PDO::FETCH_ASSOC);

$skills = array("Attack", "Strength", "Defense", "Magic", "Ranged", "Prayer", "Runecrafting", "Dungeoneering", "Construction",
                 "Constitution", "Agility", "Herblore", "Thieving", "Crafting", "Fletching", "Slayer", "Hunter", "Mining", "Smithing","Fishing", "Cooking", "Firemaking", "Woodcutting", "Farming", "Summoning");


           for($i = 0; $i < count($skills); $i++) {
                if(intval($rowExperience[$skills[$i]]) > 0) {
                    echo '<td style="color:grey;">' . $skills[$i] . ' - ' . $rowExperience[$skills[$i]] . '</td>';   
                }           

 }