PHP，sql有趣的查詢選項

Question

<input type="checkbox" name="average" value="average" <? if (get_option('average') == 'average'){ echo 'checked="checked"';} ?>>Average 

    <input type="checkbox" name="petite" value="petite" <? if (get_option('petite') == 'petite'){ echo 'checked="checked"';} ?>>Petite 



    if (get_option('average') == 'average'): // choose category 
    $average = "AND build = '".get_option('average')."'"; 
    endif; 

    if (get_option('petite') == 'petite'): // choose category 
    $petite = "OR build = '".get_option('petite')."'"; 
    endif; 


    $qry = mysql_query("SELECT performerid,pic0 FROM ".$table." WHERE **$average $petite** ORDER BY RAND() LIMIT 20"); 

我怎樣才能使這段代碼工作？我需要SQL查詢工作，如果選擇$平均，是$平均和$嬌小被選中，或者只是$ petite被選中？

TY！

Answer 1

你可以使用這樣的事情：

$selectedCategories = array(); 
foreach(array('petite', 'average', 'athletic') as $category) 
{ 
    if (get_option($category) == $category) 
    { 
    $selectedCategories[] = $category; 
    } 
} 

$qry = mysql_query(
"SELECT performerid,pic0 
    FROM ".$table." 
    WHERE build IN('" . implode("', '", $selectedCategories) . "') 
ORDER BY RAND() LIMIT 20;"); 

注：如果您選擇不分類這是不行的。

Answer 2

你需要使用php函數isset（）;

if(isset($_POST)){ //checks if post is set 
    if(isset($)POST['average']) && !isset($_POST['petite'])){ 
     //do staff only average is set 
    } 
    elseif(isset($_POST['petite']) && !isset($_POST['average'])){ 
     //do staff only petite is set 
    } 
    elseif(isset($_POST['average'] && isset($_POST['average']))){ 
     //do staff both is set 
    } 

}