準備聲明插入在PHP失敗在我的情況

Question

我使用ajax，我不能告訴錯誤是什麼，但我確信數據已插入我的數據庫。這裏是我的嘗試：

if(isset($_POST['user_id']) && isset($_POST['content']) && isset($_POST['date']) && isset($_POST['category_id'])){ 

    $content = $_POST['content']; 
    $date = $_POST['date']; 
    $user_id = $_POST['user_id']; 
    $category_id = $_POST['category_id']; 

    $stmt = $db->prepare("INSERT INTO post_items(`post_id`,`content`,`date`,`user_id`,`category_id`) 
     VALUES (?,?,?,?)"); 
    $stmt = bind_param('ssii',$content,$date,$user_id,$category_id); 

    if($stmt->execute()) { 
     echo mysqli_insert_id($db); 

    }else{ 
     echo "Something is wrong. Insert failed.."; 
    } 
} 

我的舊的工作版本（與缺乏安全的）如下所示：

if(isset($_POST['user_id']) && isset($_POST['content']) && isset($_POST['date']) && isset($_POST['category_id'])){ 

    $content = $_POST['content']; 
    $date = $_POST['date']; 
    $user_id = $_POST['user_id']; 
    $category_id = $_POST['category_id']; 

/* $result = $db->query("INSERT INTO post_items(`content`,`date`,`user_id`,`category_id`) 
     VALUES (".$content."', '".$date."', '".$user_id."', '".$category_id."')");*/ 

    $stmt = $db->prepare("INSERT INTO post_items(`post_id`,`content`,`date`,`user_id`,`category_id`) 
     VALUES (?,?,?,?)"); 
    $stmt = bind_param('ssii',$content,$date,$user_id,$category_id); 

    if($stmt->execute()) { 
     echo mysqli_insert_id($db); 

    }else{ 
     echo "Something is wrong. Insert failed.."; 
    } 
} 

我不知道我做錯了什麼在這裏，因爲它是第一次我已經使用這個材料。

Answer 1

首先，你不綁定相同數量的值作爲分配參數...

此外，您使用的bind_param的方式是錯誤的。它應該是

if(isset($_POST['user_id']) && isset($_POST['content']) && isset($_POST['date']) && isset($_POST['category_id'])){ 

    $content = $_POST['content']; 
    $date = $_POST['date']; 
    $user_id = $_POST['user_id']; 
    $category_id = $_POST['category_id']; 

    $stmt = $db->prepare("INSERT INTO post_items(`content`,`date`,`user_id`,`category_id`) 
     VALUES (?,?,?,?)"); 
    $stmt->bindParam(1, $content); 
    ... 
    $stmt->bindParam(4, $category_id); 

    if($stmt->execute()) { 
     echo mysqli_insert_id($db); 

    }else{ 
     echo "Something is wrong. Insert failed.."; 
    } 
} 

，或者你也可以做到以下幾點：

if(isset($_POST['user_id']) && isset($_POST['content']) && isset($_POST['date']) && isset($_POST['category_id'])){ 

    $content = $_POST['content']; 
    $date = $_POST['date']; 
    $user_id = $_POST['user_id']; 
    $category_id = $_POST['category_id']; 

    $stmt = $db->prepare("INSERT INTO post_items(`content`,`date`,`user_id`,`category_id`) 
     VALUES (?,?,?,?)"); 

    if($stmt->execute(array($content,$date,$user_id,$category_id))) { 
     echo mysqli_insert_id($db); 

    }else{ 
     echo "Something is wrong. Insert failed.."; 
    } 
} 

也只是爲了讓你知道，PDO有來獲取最後一個ID的方法，所以不是 mysqli_insert_id 你可以使用 $ stmt-> lastInsertId（）

if(isset($_POST['user_id']) && isset($_POST['content']) && isset($_POST['date']) && isset($_POST['category_id'])){ 

    $content = $_POST['content']; 
    $date = $_POST['date']; 
    $user_id = $_POST['user_id']; 
    $category_id = $_POST['category_id']; 

    $stmt = $db->prepare("INSERT INTO post_items(`content`,`date`,`user_id`,`category_id`) 
     VALUES (?,?,?,?)"); 
    $stmt->bindParam(1, $content); 
    ... 
    $stmt->bindParam(4, $category_id); 

    if($stmt->execute()) { 
     echo $stmt->lastInsertId(); 

    }else{ 
     echo "Something is wrong. Insert failed.."; 
    } 
}