使用多個值搜索表單查詢 - PHP / MYSQL

Question

我在使用搜索表單時遇到了一些麻煩，我一直在創建功能。 我基本上想要一個表單（在任何頁面上）轉到此頁面，然后列出我的數據庫中的相關行。 我的問題是表單有文本字段和選擇字段（用於名稱和類別），我無法創建使這兩個值一起搜索數據庫的功能。

這就是我想要發生的事情：當你只輸入名稱而不是類別時，它將僅顯示名稱，反之亦然，類別和名稱; 然后當它們在一起時它只顯示兩個都在的行。

以下是我到目前為止的情況：

// 2. Create variables to store values
if(!$_GET['search-category'] == "") {
    $searchName = $_GET['search-name'];
}

if(!$_GET['search-category'] == "select-your-category") {
    $searchCat = $_GET['search-category'];
}



// 2. Create the query for the stored value. Matching it against the name, summary and sub type of my item.
$mainSearch = "SELECT attraction.*, type.type_name, sub_type.sub_type_name ";
$mainSearch .= "FROM attraction ";
$mainSearch .= "INNER JOIN sub_type ON attraction.sub_type = sub_type.sub_type_id ";
$mainSearch .= "INNER JOIN type ON attraction.type = type.type_id ";
$mainSearch .= "WHERE attraction.name LIKE '%" . $searchName . "%' AND (sub_type.sub_type_name LIKE '%" . $searchCat . "%' )";
$mainSearch .= "ORDER BY sub_type_name ASC";

// 2. run query
$result2 = $con->query($mainSearch);
if (!$result2) {
    die('Query error: ' . mysqli_error($result2));
}

Answer 1

我將代碼重構為類似的東西 -

foreach( $_GET['filters'] as $fname => $fval ) {

    if( !$fval ) continue;

    $where[] = "$fname LIKE '%{$fval}%'";
}

您只需要包含查詢中非空的輸入。 此外，您還需要解決安全問題，例如轉移輸入等。

Answer 2

您可以檢查相關值是否為空：

// 2. Create the query for the stored value. 
// Matching it against the name, summary and sub type of my item.
$mainSearch = "SELECT attraction.*, type.type_name, sub_type.sub_type_name ";
$mainSearch .= "FROM attraction ";
$mainSearch .= "INNER JOIN sub_type ON attraction.sub_type = sub_type.sub_type_id ";
$mainSearch .= "INNER JOIN type ON attraction.type = type.type_id ";
$mainSearch .= "WHERE ";
if ($searchName) {
    $mainSearch .= "attraction.name LIKE '%" . $searchName . "%'";
    if ($searchCat) {
        $mainSearch .= " AND ";
    }
}
if ($searchCat) {
    $mainSearch .= "sub_type.sub_type_name LIKE '%" . $searchCat . "%'"
}

$mainSearch .= "ORDER BY sub_type_name ASC";

// Double check that at least one of the search criteria is filled:
if (!$searchName && !$searchCat) {
    die("Must supply either name search or category search");
}

Answer 3

您可以做的是聲明一個名為$ search_condition的變量，並根據$ searchName或$ searchCat是否為null或不為$ search_condition賦值

例如

if (isset($searchName ) || !is_empty($searchName ))
{
   $search_condition = "WHERE attraction.name LIKE '%" . $searchName;
}
if (isset($searchCat ) || !is_empty($searchCat ))
{
   $search_condition = "sub_type.sub_type_name LIKE '%" . $searchCat . "%'";
}
if ((isset($searchName ) || !is_empty($searchName )) && (isset($searchCat ) || !is_empty($searchCat )))
{
   $search_condition = "WHERE attraction.name LIKE '%" . $searchName . "%' AND (sub_type.sub_type_name LIKE '%" . $searchCat . "%' )";
} 

希望這可能對你有所幫助

謝謝

Answer 4

這是一個評論，但我想利用格式化選項......

你知道，你可以這樣改寫......

// 2. Create the query for the stored value. Matching it against the name, summary and sub type of my item.

$mainSearch = "
SELECT a.*
     , t.type_name
     , s.sub_type_name 
  FROM attraction a 
  JOIN sub_type s
    ON a.sub_type = s.sub_type_id 
  JOIN type t
    ON a.type = t.type_id 
 WHERE a.name LIKE '%$searchName%' 
   AND s.sub_type_name LIKE '%$searchCat%' 
 ORDER 
    BY s.sub_type_name ASC;
    ";