从多个复选框形成查询字符串

Question

我正在尝试从多个复选框中形成一个查询字符串，用于查询我的数据库。

我有以下表格：

            <fieldset data-role="controlgroup">

            <input type="checkbox" name="wheat" id="checkbox-1a" class="custom" />
            <label for="checkbox-1a">Wheat Allergy</label>

            <input type="checkbox" name="yeast" id="checkbox-2a" class="custom" />
            <label for="checkbox-2a">Yeast Allergy</label>

            <input type="checkbox" name="sugar" id="checkbox-3a" class="custom" />
            <label for="checkbox-3a">Sugar Allergy</label>

            <input type="checkbox" name="dairy" id="checkbox-4a" class="custom" />
            <label for="checkbox-4a">Dairy Allergy</label>

我的PHP代码如下：

        if(isset($_POST['wheat']))
        {
            $str1 = 'wheatfree = 1';
        }

        if(isset($_POST['yeast']))
        {
            $str2 = 'yeastfree = 1';
        }

        if(isset($_POST['sugar']))
        {
            $str3 = 'sugarfree = 1';
        }

        if(isset($_POST['dairy']))
        {
            $str4 = 'dairyfree = 1';
        }

        $fullsearch = $str1.$str2.$str3.$str4;

        $str_SQL = "SELECT * FROM recipes WHERE ".$fullsearch;

        echo $str_SQL;

这有点像我需要的，但它不是很优雅。

首先，sql 查询如下所示：

SELECT * FROM recipes WHERE sugarfree = 1dairyfree = 1

如果用户选择不选择一个，我当然会得到一个未选择的 str 的未定义变量错误。

不确定如何解决此问题或下一步该去哪里。我想要这里的一些逻辑，它只是根据表单上检查的内容修改了字符串，然后形成一个很好的干净的 SQL 查询，我可以对我的数据库运行。但可惜我迷路了:(

帮助？

Answer 1

Dave 的进一步回答：

$options = Array();
$ingredients = Array('wheat', 'yeast', 'sugar', 'dairy');

foreach ($ingredients as $i)
   if (isset($_POST[$i]))
      $options[] = $i . 'free = 1';

$sql = "SELECT * FROM recipes";
if (count($options))
   $sql .= " WHERE " . implode(' AND ', $options);

echo $sql;

---

但是你为什么不使用复选框的value 属性呢？

<input type="checkbox" name="ingredients[]" value="wheat" />
<input type="checkbox" name="ingredients[]" value="sugar" />

等等

然后：

$options = Array();
foreach ($_POST['ingredients'] as $i)
   $options[] = $i . 'free = 1'; // don't forget to escape $i somehow!

$sql = "SELECT * FROM recipes";
if (count($options))
   $sql .= " WHERE " . implode(' AND ', $options);

echo $sql;

Answer 2

这个怎么样：

$options = array();
if(isset($_POST['wheat']))
{
    $options[] = 'wheatfree = 1';
}

if(isset($_POST['yeast']))
{
    $options[] = 'yeastfree = 1';
}

if(isset($_POST['sugar']))
{
    $options[] = 'sugarfree = 1';
}

if(isset($_POST['dairy']))
{
    $options[] = 'dairyfree = 1';
}

$fullsearch = implode(' AND ', $options);

$str_SQL = "SELECT * FROM recipes";
if ($fullsearch <> '') {
    $str_SQL .= " WHERE " . $fullsearch;
}

echo $str_SQL;