如何从不同的搜索字段输出信息？

Question

我有以下三个搜索字段的表单。两个文本输入(关键字和距离)和一个下拉列表(运输)。

我的搜索函数适用于关键字文本输入，但不适用于其他两个字段。当我"echo“时，我可以看到数据正在输出，但它实际上不会搜索这些术语。

    <form action="searchresults.php" method="POST">
  <h3> Keyword </h3>
    <input type="text" name="keyword-search">

  <h3> Primary Function </h3>
  <?php
  $sql = "SELECT pf_id, primary_function FROM primary_function ORDER BY pf_id;";
  $result = mysqli_query($conn, $sql);

  echo "<select name='function-search' id = 'function-search'>";
  echo '<option value=""></option>';

  while ($row = mysqli_fetch_assoc($result)) {
                unset($id, $name);
                $id = $row['pf_id'];
                $name = $row['primary_function'];
                echo '<option value="'.$name.'">'.$name.'</option>';

              }
  echo "</select>";
  ?>

  <h3> Distance </h3>
    within <input type="text" name="distance-search"> miles of PCC
    <br>
    <button type="submit" name="submit-search">Search</button>
</form>

下面是数据被推送到的位置。

    <?php
if (isset($_POST['submit-search'])) {

$functionSearch = mysqli_real_escape_string($conn, $_POST['function-search']);
$keySearch = mysqli_real_escape_string($conn, $_POST['keyword-search']);
$distanceSearch = mysqli_real_escape_string($conn, $_POST['distance-search']);
$sql = "SELECT * FROM resource WHERE resource_name LIKE '%$keySearch%'
OR  description LIKE '%$keySearch%' OR username LIKE '%$keySearch%'
OR primary_function LIKE '%$keySearch%'
OR distance_in_miles LIKE '%$keySearch%'
ORDER BY distance_in_miles";
$result = mysqli_query($conn, $sql);
$queryResult = mysqli_num_rows($result);
// Checking for search result page errors
//echo $result;
echo $queryResult;
// echo $functionSearch;

if ($queryResult > 0){
    while ($row = mysqli_fetch_assoc($result)) {
      echo "<tr><td>". $row['resource_id'] ."</td><td>". $row['resource_name'] ."</td><td>". $row['username'] ."</td><td>". "$" .$row['cost_in_usd'] ."/". $row['cost_per'] ."</td><td>". $row['distance_in_miles'] ."</td></tr>";
    } echo "</table>";
  } else {
    echo "<br>No results matching your search";
  }
}
?>

Answer 1

附注:这篇评论太长了。

我将删除foreach循环，并使用单独的变量。

现在，您的foreach将在每个like查询中显示类似search1search2search3的内容，这本身不会抛出错误，只是(可能)不会匹配任何内容。

下面是一个示例，将$search_x替换为您选择的条件/帖子：

$sql = "SELECT * FROM resource WHERE resource_name LIKE '%$search_1%'
        OR description LIKE '%$search_2%' OR username LIKE '%$search_3%'
        OR primary_function LIKE '%$search_4%'
        OR distance_in_miles LIKE '%$search_5%'
        ORDER BY distance_in_miles";

$result = mysqli_query($conn, $sql);
if(!$result){
   // handle no results/failure here
}

您可能还必须通过在两端、开头和/或结尾使用%来玩弄它。

附注:正如注释中所述，对于任何类型的循环函数，始终使用适当的大括号。

顺便说一句，我建议您为此使用准备好的语句。可以绕过real_escape_string()，这里有一个关于这个主题的问答：

• SQL injection that gets around mysql_real_escape_string()