Question

我有这个PHP代码

if(isset($_POST['submit'])){

$likeString = '%' . $_POST['search'] . '%';
$query = $conn->prepare("SELECT * FROM images WHERE image_caption LIKE ?");
$query->bind_param('s', $likeString);
$query->execute();

    var_dump($likeString);
    if (!$query) {
      printf("Query failed: %s\n", $mysqli->error);
      exit;
    }   
    if($res->num_rows > 0) {
       while ($row = $res->fetch_assoc()) {          
             echo "<br>Title: " . $row['image_caption'];          
       }
    } else { 
          echo " <br> 0 results"; 
    }

}

var_dump($likeString)显示我通过搜索表单正确发布的单词。我也试过直接运行phpmyadmin来运行这个查询

SELECT * 从图像在哪里image_caption喜欢“％Volvo％”

我收到了1个正确的结果。在页面上，我看到0 results。试图玩fetch：

$res->fetch_assoc()
$res->fetchAll()
$res->fetch()

他们都没有显示任何结果。我确信这是一个非常愚蠢和简单的错误，但看不到它。请帮忙。

我没有Call to a member function bind_param() on a non-object当我从其中一个答案中提出改变时，这是我的错误。问题仍然存在 - 0结果

更新：当前代码

$likeString = "%{$_POST['search']}%";
$query = $conn->prepare("SELECT * FROM images WHERE image_caption LIKE ? ");
$query->bind_param('s', $likeString);
$query->execute();

    if($query->num_rows > 0) {
       while ($row = $query->fetch()) {          
             echo "<br>Title: " . $row['image_caption'];          
       }
    } else { 
          echo " <br> 0 results"; 
    }

}

更新2：检查数据库连接 - ＆gt;结果是Connected successfully

$servername = "localhost";
$username = "mydbUsername"; // it's changed for the question
$password = "myPass"; // it's changed for the question
$dbname = "myDbName"; // it's changed for the question

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";

Answer 1

您使用的是$res，但未定义...您必须使用$query。下次打开错误报告以查看这些愚蠢的错误

Answer 2

试试这个:(更新你的代码）

$likeString= "%{$_POST['search']}%";
$stmt = $db->prepare("SELECT * FROM images WHERE image_caption LIKE ?");
$stmt->bind_param('s', $likeString);
$stmt->execute();

$result = $stmt->get_result();
    while ($row = $result->fetch_array(MYSQLI_NUM))
    {
        foreach ($row as $r)
        {
           echo "<br>Title: " . $r['image_caption'];                
        }
        print "\n";
    }

OR

<?php
  $conn = new mysqli("localhost","mydbUsername","myPass","myDbName");
  /* check connection */
  if ($conn->connect_errno) { 
    printf("Connect failed: %s\n", $conn->connect_error); 
    exit(); 
  }

  $query = "SELECT * FROM images WHERE image_caption LIKE %".$_POST['search']."%";

  if ($result = $conn->query($query)) {

    /* fetch associative array */
    while ($row = $result->fetch_assoc()) {
        echo "<br> Title: ". $row["image_caption"]);
    }print "\n";

    /* free result set */
    $result->free();
  }

  /* close connection */
  $conn->close();
?>

简单搜索功能总是返回0结果

2 个答案: