Question

我正在搜索框，我遇到了一个小问题。我已经阅读了一些与我相同的问题，并尝试了那里提供的解决方案，但它们并没有奏效。我需要回显我的SELECT查询的结果。

<?Php
$cat_id_in = (int) $_GET['cat_id'];
$cat_dropdown = Category::getCat_DropDown($cat_id_in);
if(isset($_POST['save'])) {
$search = $_POST['search'];
//connect to database
$mysqli = mysqli_connect("localhost", "php24sql", "hJQV8RTe5t", "smithside") or die('Could not    connect: ' . mysql_error());
$query2 = "SELECT * FROM lots WHERE lot_name OR lot_description LIKE '%$search%' AND cat_id =  '$cat_id_in'";
$results = mysql_query($query2);

while ($row = mysql_fetch_assoc($results)) {
  echo $row['lot_name'] . "<br />";
  }

}   
?>

错误是：

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\System_CompA2\content\search.php on line 12.

我的查询肯定有效，因为我用if语句试了一下，并且我的数据库连接没有错误。那有什么可能是错的？非常感谢！

Answer 1

你正在混合mysql和mysqli函数，你在sql查询中也有错误

试试这个：

$mysqli = mysqli_connect("localhost", "php24sql", "hJQV8RTe5t", "smithside") or die('Could not    connect: ' . mysql_error());
$query2 = "SELECT * FROM lots WHERE (lot_name LIKE '%$search%' OR lot_description LIKE '%$search%') AND cat_id =  '$cat_id_in'";
$results = mysqli_query($mysqli,$query2);

while ($row = mysqli_fetch_assoc($results)) {
     echo $row['lot_name'] . "<br />";
}

有关mysqli_*功能的更多信息，请阅读 this

Answer 2

不要混用mysql和mysqli ....

在mysqli中查询时使用连接

 //connect to database
    $mysqli = mysqli_connect("localhost", "php24sql", "hJQV8RTe5t", "smithside");
    if (!$mysqli) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $query2 = "SELECT * FROM lots WHERE lot_name LIKE '%$search%' OR lot_description LIKE '%$search%' AND cat_id =  '$cat_id_in'";
    $results = mysqli_query($mysqli ,$query2);

    while ($row = mysqli_fetch_assoc($results)) {
      echo $row['lot_name'] . "<br />";
      }

Answer 3

WHERE中有错误。

WHERE lot_name OR lot_description LIKE '%$search%'

应该是

WHERE lot_name LIKE '%$search%' OR lot_description LIKE '%$search%'"

Answer 4

您正在使用＆＃34; mysql_fetch_assoc＆＃34;和＆＃34; mysql_query＆＃34;当连接是通过＆＃34; mysqli_connect＆＃34;功能。试试＆＃34; mysqli _＆＃34;功能

如何从数据库中回显查询结果？

4 个答案: