调用一个成员函数bind_param（）布尔，mysqli_stmt :: store_result和mysqli_stmt :: close（）方法

Question

这里是我的代码...它是关于使用PHP mysqli扩展

<?php 
error_reporting(E_ALL); 
$db = new mysqli("localhost","root","","dee"); 

if ($db->connect_errno) 
{ 
    die('Unable to connect to database'); 
} 
mysqli_set_charset($db,"utf8"); 


$storeid=4; 
$categoryid=6; 

$statement_store = $db->prepare('SELECT * FROM tbl_store WHERE store_id=?'); 
$statement_store->bind_param('i',$storeid); 
$statement_store->execute(); 

$statement_store->store_result();//---------------(1) 

$statement_store->bind_result($store_id,$store_name,$store_description,$store_image,$store_open,$store_close,$store_foldername); 
$statement_store->fetch(); 
$store = $store_name; 

//$statement_store->close();//--------------(2) 


$statement_category = $db->prepare('SELECT * FROM tbl_category WHERE category_id=?'); 
$statement_category->bind_param('i',$categoryid); 
$statement_category->execute(); 
$statement_category->bind_result($category_id,$category_name); 
$statement_category->fetch(); 
$category = $category_name; 

echo $store; 
echo '<br>'; 
echo $category; 

?> 

1. 致命错误：调用一个成员函数bind_param（）上布尔错误不使用时给出 两者（1）和（2）
2. 当使用（1）或（2）不给出错误
   
3. 当同时使用（1）和（2）不会给出错误

有人可以告诉我这里发生了什么吗？

Answer 1

当您不使用store_result()或close()时，您的第一个准备好的语句（或其结果）仍然是“活动的”。这意味着您必须以某种方式读取数据，然后才能发出新的预备声明。因为你的第二个prepare()语句将失败，它返回布尔值false。

检查$db->error字段，你会看到你得到“命令不同步;你现在不能运行这个命令”的错误信息。来自MySQL手册B.5.2.14 Commands out of sync：

If you get Commands out of sync; you can't run this command now in your client code, you are calling client functions in the wrong order.

This can happen, for example, if you are using mysql_use_result() and try to execute a new query before you have called mysql_free_result() . It can also happen if you try to execute two queries that return data without calling mysql_use_result() or mysql_store_result() in between.