Question

我是php和sql的新手，所以请耐心等待。我确实尝试过搜索，但没找到相当于我正在寻找的。希望你能帮助我。所以这里去了！

我有一定数量的带有值的下拉菜单。我想包括一个select all选项，它将选择所有选项并对所有值执行sql查询。我已经使用if条件来执行查询但是if cond的数量基于下拉菜单的数量呈指数增长。 2 ^ n我猜。我想知道是否有更好的方法来做到这一点。请找到下面的代码。

$sql = mysql_query("SELECT distinct `Num. of Parts` from `$verb`");
echo "<select name='filter1'>";
echo "<option value='1'> Select all </option>";
while($row = mysql_fetch_assoc($sql))
{
$filter1 = $row['Num. of Parts'];
echo "<option value='$filter1'> $filter1 </option>";
}
echo "</select>";
$sql = mysql_query("SELECT distinct `Type of positioning` from `$verb`");
echo "<select name='filter2'>";
echo "<option value='1'> Select all </option>";
while($row = mysql_fetch_assoc($sql))
{
$filter2 = $row['Type of positioning'];
echo "<option value='$filter2'> $filter2 </option>";
}
echo "</select>";
echo "<input type=\"submit\" name=\"submitC\" value=\"SUBMIT\">";

submitC值在此处传递：

if($filter1==1 && $filter2!=1)
{$sql = mysql_query("SELECT * from `$verb` where `Num. of Parts` like '%' and `Type of positioning`='$filter2' and `Distance range`='$dist'");
}
elseif($filter1!=1 && $filter2==1)
{$sql = mysql_query("SELECT * from `$verb` where `Num. of Parts`='$filter1' and `Type of positioning` like '%' and `Distance range`='$dist'");
}
elseif($filter1==1 && $filter2==1)
{$sql = mysql_query("SELECT * from `$verb` where `Num. of Parts` like '%' and `Type of positioning` like '%' and `Distance range`='$dist'");
}
else
{
$sql = mysql_query("SELECT * from `$verb` where `Num. of Parts`='$filter1' and `Type of positioning`='$filter2' and `Distance range`='$dist'");
}

先谢谢你们！ ps EXCLUDE $ dist

Answer 1

您可以按name='filter1[]'

选择多个选项

您可以按$filter1= serialize($_POST['filter1']) ;

存储这些内容

Answer 2

试试这个

if($filter1==1) {  if(&& $filter2!=1){
$sql = "SELECT * from `$verb` where `Num. of Parts` like '%' and `Type of positioning`='$filter2' and `Distance range`='$dist'"; }
elseif($filter2==1){$sql = "SELECT * from `$verb` where `Num. of Parts` like '%' and `Type of positioning` like '%' and `Distance range`='$dist'";  }  }

elseif($filter1!=1 && $filter2==1) { 
$sql = "SELECT * from `$verb` where `Num. of Parts`='$filter1' and `Type of positioning` like '%' and `Distance range`='$dist'";  }

else{
$sql = "SELECT * from `$verb` where `Num. of Parts`='$filter1' and `Type of positioning`='$filter2' and `Distance range`='$dist'";  }

$result = mysql_query($sql);

Answer 3

如果我找对你，你正在搜索这样的东西：

<?php

$mapping = array ( 'column1' => 'Num_of_Parts' , 
      'column2' => 'Type_of_positioning'
);  
    $sql = "select * from table ";
    $connector = " AND ";
    if ( isset($_POST['filters']) ) {
        $sql = $sql . "where ";
        foreach ($_POST['filters'] as $key=>$val) {
            $sql .= $mapping[$key] ." like '%"  . $val. "%'" . $connector ; 
        }

    }//end of if
    $sql.= " Distance range = 'value'";
    var_dump($sql);


?>


<html>
<body>

<form action="" method="post">
<select name = "filters[column1]">
  <option value="">Select All</option>
  <option value="saab">Saab</option>
  <option value="mercedes">Mercedes</option>
  <option value="audi">Audi</option>
</select>

<select name = "filters[column2]">
  <option value="">Select All</option>
  <option value="saab">Car</option>
  <option value="mercedes">Bus</option>
  <option value="audi">Cycle</option>
</select>

<input type="submit" value="submit">
</form>

</body>
</html>

注意

始终使用mysqli_real_escape_string或类似的东西清理用户输入，但不要使用mysql系列函数，因为鼓励他们不要在php doc中使用。
出于安全目的，请勿使用列名 作为元素的名称。您可以在代码中看到如何使用一个简单的技巧重新映射列名。

快乐编码:)

如何为多个下拉列表选择所有选项并在php和mysql中执行查询？更方便吗？

3 个答案:

注意