Question

问题在于我试图获取特定的搜索结果，但它提供了所有数据库行，而不是过滤结果。我怎么能避免这个？

require_once 'db/connect.php';

if(isset($_GET['submit'])){
    $doctor = $db->escape_string($_GET['doctor']);
    $specialization = $db->escape_string($_GET['specialization']);
    $hospital = $db->escape_string($_GET['hospital']);

    $query = $db->query("
         SELECT docname, specialization,hospital
         From doctor
         WHere docname Like '%{$doctor}%' or
         specialization like '%{$specialization}%' or
         hospital like '%{$hospital}%'
        ");
?>




<div class="result-count">
    Found <?php echo $query->num_rows; ?> results.
</div>

<?php

if($query -> num_rows){
    while($r = $query->fetch_object()){
        ?>
          <div class="result-count">
             <a href="#"><?php  echo $r->docname; ?></a>
          </div>
        <?php
    }
}
}

Answer 1

您必须从WHERE子句中删除未使用的字段。否则，hospital like '%%'条件将有效地匹配表中的每一行。

因此，在您的情况下，您应该动态创建查询，仅为非空字段添加条件。

当然它极大地帮助打印出结果查询。它不仅会让您知道，您正在运行什么SQL，还会让你的问题变得合理。

Answer 2

试试这个：

require_once 'db/connect.php';

if(isset($_GET['submit'])){
   $doctor = $db->escape_string($_GET['doctor']);
   $specialization = $db->escape_string($_GET['specialization']);
   $hospital = $db->escape_string($_GET['hospital']);
   $where = array();
   if (!empty($doctor)) {
        array_push($where, "docname like '%{$doctor}%'");
   }
   if (!empty($specialization)) {
        array_push($where, "specialization like '%{$specialization}%'");
   }
   if (!empty($hospital)) {
        array_push($where, "hospital like '%{$hospital}%'");
   }
$sql = "SELECT docname, specialization, hospital "
       . "FROM doctor";
if (!empty($where)) {
    $sql .= "WHERE " . implode(' OR ', $where);
}
$query = $db->query($sql);

?>

PHP多个搜索查询返回所有行

2 个答案: