php - 从多个表中选择-PHP

Question

我正在尝试执行此查询

SELECT category_request.request_id
FROM category_request, event, user
WHERE category_request.category_name = event.category_name 
AND user.user_id = event.moderator_id

当我尝试在我的数据库上运行时，它工作正常，如下所示:

但出于某种原因(也许是因为我在做一个连接？)我得到一个错误:

mysqli_stmt_bind_param(): Number of variables doesn't match number of parameters in prepared statement

而且我确定只需要表中的一列这是 php 文件

<?php
define('__ROOT__', dirname(dirname(__FILE__))); 
require_once(__ROOT__.'/config.php'); 


/*  $event_name = $_POST["event_name"];
    $event_description = $_POST["event_description"];
     $event_date = $_POST["event_date"];
    $event_time = $_POST["event_time"];
    $max_members = $_POST["max_members"]; 
    $ack_needed = $_POST["ack_needed"];
    $category_name = $_POST["category_name"];
    $event_location = $_POST["event_location"]; */

    $event_name = (isset($_POST['event_name']) ? $_POST['event_name'] : '');
    $event_description = (isset($_POST['event_description']) ? $_POST['event_description'] : '');
    $event_date = (isset($_POST['event_date']) ? $_POST['event_date'] : '');
    $event_time = (isset($_POST['event_time']) ? $_POST['event_time'] : '');
    $max_members = (isset($_POST['max_members']) ? $_POST['max_members'] : '');
    $ack_needed = (isset($_POST['ack_needed']) ? $_POST['ack_needed'] : '');
    $category_name = (isset($_POST['category_name']) ? $_POST['category_name'] : '');
    $event_location = (isset($_POST['event_location']) ? $_POST['event_location'] : '');
    $moderator_id = (isset($_POST['moderator_id']) ? $_POST['moderator_id'] : '');



    $result = mysqli_prepare( $con, "SELECT category_request.request_id FROM category_request, event, user
        WHERE category_request.category_name = event.category_name AND user.user_id = event.moderator_id")                                              

    or die(mysqli_error($con));

    mysqli_stmt_bind_param($result,"i", $category_id)
    or die(mysqli_error($con)); 


    mysqli_stmt_execute($result)
    or die(mysqli_error($con));







   /* $statement = mysqli_prepare($con,     
    "INSERT INTO event (event_name, event_description, event_date
                                    , event_time, max_members, ack_needed, 
                                    category_name, moderator_id,event_location)     VALUES (?, ?, ?, ? ,?, ?, ?, ?, ?)")                                                                                    
    or die(mysqli_error($con));

    mysqli_stmt_bind_param($statement, "ssssssisis", $event_name, $event_description, $event_date,   $event_time, $max_members, $ack_needed,                                                                                    
                                                                            , $category_name, $moderator_id,$event_location)
    or die(mysqli_error($con));
    mysqli_stmt_execute($statement)
    or die(mysqli_error($con)); */

    $response = array(); 
    $response["success"] = true;  

    echo json_encode($response);

    mysqli_close($con)
    or die(mysqli_error($con));
?>

Answer 1

您正在尝试制作一个准备好的语句，而不是仅仅使用 mysqli_query。如果您确实打算使用准备好的语句，例如为您的 mysqli_stmt_bind_param 函数使用占位符，那么您需要在准备好的语句中包含 ? 作为占位符。

Buuuut，因为我没有在您的查询中的任何地方看到引用的 category_id，也许您需要使用 mysqli_query 并获取所有结果？例如:

$result = mysqli_query( $con, "SELECT category_request.request_id FROM category_request, event, user WHERE category_request.category_name = event.category_name AND user.user_id = event.moderator_id");
if(mysqli_num_rows($result)) {
  while($row = mysqli_fetch_row($result)) {
    // do something with $row
  }
} else {
  echo "no results";
}