Question

我在使用搜索表单时遇到了一些麻烦，我一直在为其创建功能。我基本上想要一个表单（在任何页面上）转到此页面，然后列出我的数据库中的相关行。我的问题是表单有文本字段和选择字段（用于名称和类别），我无法创建使这两个值一起搜索数据库的功能。

所以我想要发生的事情：当你只输入名称而不是类别时，它只显示名称，反之亦然，类别和名称;然后，当它们在一起时，它只显示包含它们的行。

到目前为止我所拥有的：

// 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

我将代码重构为 -

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

    if( !$fval ) continue;

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

您只需要在查询中包含那些非空的输入。此外，您还需要解决安全问题，例如转义输入等。

Answer 2

您可以检查相关值是否为空：

// 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

您可以做的是声明一个名为$ search_condition的变量，并根据$ searchName或$ searchCat是否为null或者为$ search_condition分配值

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 . "%' )";
}

希望这可以帮到你

由于

Answer 4

这是评论，但我想利用格式化选项......

你知道，可以以这种方式改写......

// 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;
    ";

使用多个值搜索表单查询 - PHP / MYSQL

4 个答案: