PHP / MYSQL使用mysqli_query进行选择 [英] PHP / MYSQL select using mysqli_query

问题描述

我正在努力处理我的脚本，并且因为切换到mysqli而不再工作。我检查了PHP手册，但是看不到我做错了什么，肯定是一个初学者的错误。

以下是我的代码：

 <？php 
 
 // Connect到数据库
 include（'connection.php'）; 
 
 //检索表中的所有数据
 $ result = mysqli_query（SELECT * FROM gear）或die（mysqli_error（））; 
 
 echo< table border ='1'>; 
 echo< tr>制造商< th>模型< / th>< th>描述< / th>< / tr>; 
 //继续获取下一行，直到没有更多的元素获得
时（$ gear = mysqli_fetch_array（$ result））{
 
 //打印出每个元素的内容排成一张表
 echo< tr>< td>; 
 echo $ gear ['manu']; 
 echo< / td>< td>; 
 echo $ gear ['model']; 
 echo< / td>< td>; 
 echo $ gear ['desc']; 
回显< / td>< / tr>; 
 
} 
 echo< / table>; 
 
？> 
  

我想知道是否因为我正在使用另一个脚本进行连接，但它是抱怨我的mysqli_query，所以我得到这个错误：
$ b

[Wed Jan 01 21:14:54 2014] [error] [client： ：1] PHP警告：mysqli_error（）期望只有1个参数，0在第7行的/var/www/eml/includes/query_gear.php中给出

任何建议或建议，将不胜感激。

解决方案

c> mysqli_connect（），这是mysqli_ *所必需的，而不是mysql_ *所需的。假设你叫你 $ link ：

  $ result = mysqli_query（$链接，选择*从齿轮）或死（mysqli_error（$链接））; 
  

I am struggling with my script and since switching to mysqli it will no longer work.I have checked the PHP manual but just cannot see what I am doing wrong, surely a beginners mistake.

Here is my code:

<?php

//Connect to database
include ('connection.php');

// Retrieve all the data from the table
$result = mysqli_query("SELECT * FROM gear") or die(mysqli_error()); 

echo "<table border='1'>";
echo "<tr> <th>Manufacturer</th> <th>Model</th> <th>Description</th> </tr>";
// keeps getting the next row until there are no more to get
while($gear = mysqli_fetch_array( $result )) {

// Print out the contents of each row into a table
    echo "<tr><td>"; 
    echo $gear['manu'];
    echo "</td><td>"; 
    echo $gear['model'];
    echo "</td><td>"; 
    echo $gear['desc'];
    echo "</td></tr>"; 

} 
echo "</table>";

?>

I am wondering if it is because of the fact that I am using another script to connect but it is complaining about my mysqli_query so I am getting this error:

[Wed Jan 01 21:14:54 2014] [error] [client ::1] PHP Warning: mysqli_error() expects exactly 1 parameter, 0 given in /var/www/eml/includes/query_gear.php on line 7

Any advice or suggestions would be appreciated.

解决方案

You are missing your resource identifier obtained with mysqli_connect() which is required with mysqli_* and not with mysql_*. Assuming you called yours $link:

$result = mysqli_query($link, "SELECT * FROM gear") or die(mysqli_error($link)); 

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