簡單搜索函數返回總是0結果

Question

我有這樣的PHP代碼

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)顯示我已經通過搜索表單提交正確的字。我也試過在phpmyadmin直接運行這個查詢

SELECT * FROM images WHERE image_caption LIKE "%Volvo%"

而且我收到1個結果是正確的。在頁面上，我看到0 results。試圖用取玩：

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

他們沒有表現出任何結果。我相信是非常愚蠢和簡單的錯誤，但看不到它。請幫助。

我沒有Call to a member function bind_param() on a non-object這是我的錯誤，而我從一個答案中提出了修改。問題仍然存在 - 0結果

UPDATE：當前代碼

$likeString = "%{$_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：DB連接checked->結果是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

試試這個：（代碼更新）

$likeString= "%{$_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(); 
?> 

Answer 2

您正在使用$res而沒有定義它...您必須改用$query。錯誤報告 下一次輪到看到這樣愚蠢的錯誤