Question

我想搜索一些产品并获得结果。这是我的PHP代码：

function Item_Search($AppID, $Keyword = '', $Wears)
{
    // expected $Wears = [1,0,1,0,1] , numbers can be diffrent from 0 to 1

    $WearNames = ['red', 'green', 'blue', 'yellow', 'black'];
    $FinalWear = [];

    foreach ($Wears as $i => $Wear) {
        if ($Wear == 1) {
            $FinalWear[] = $WearNames[$i];
        }
    }
    $FinalWear = json_encode($FinalWear);
    $FinalWear = str_replace(str_split('[]'), '', $FinalWear);

$ItemList = Query("SELECT * FROM items WHERE appid=$AppID 
AND name LIKE '%$Keyword%'
AND wear IN ($FinalWear)
");
}

此代码可以正常工作，但是有些产品没有任何颜色。在这种情况下，我想说一下是否所有颜色均为1（真），然后显示也没有颜色的产品。

所以我的问题是，我可以在$FinalWear中放入wear IN ($FinalWear)这样的东西，以便获得所有结果吗？

Answer 1

您可以取$Wear中元素的总和，如果与计数相同，则将设置所有元素。然后，您可以使用它向OR测试中添加wear条件，在设置wear中的所有元素后检查$Wear中的空字符串：

function Item_Search($AppID, $Keyword = '', $Wears)
{
    // expected $Wears = [1,0,1,0,1] , numbers can be different from 0 to 1

    $WearNames = ['red', 'green', 'blue', 'yellow', 'black'];
    $FinalWear = [];

    foreach ($Wears as $i => $Wear) {
        if ($Wear == 1) {
            $FinalWear[] = $WearNames[$i];
        }
    }
    $FinalWear = "'" . implode("','", $FinalWear) . "'";
    $EmptyWear = array_sum($Wears) == count($Wears) ? "OR wear = ''" : '';

    $ItemList = Query(<<<EOD
SELECT * 
FROM items 
WHERE appid=$AppID 
  AND name LIKE '%$Keyword%'
  AND (wear IN ($FinalWear) $EmptyWear)
EOD
);
}

3v4l.org上的（查询输出的）演示

请注意，如果$AppID和$Keyword来自外部资源，则您的查询容易受到SQL注入的攻击，因此应使用准备好的语句来避免这种可能性。参见此question。

如果wear为空意味着实际上是NULL，请将生成$EmptyWear的行更改为

 $EmptyWear = array_sum($Wears) == count($Wears) ? "OR wear IS NULL" : '';

按多个字段搜索SELECT IN（所有内容）

1 个答案: