mysqli_query() 需要至少 2 个参数，其中 1 个给定? [英] mysqli_query() expects at least 2 parameters, 1 given in?

问题描述

每次运行这个 php 时，我都会遇到同样的 3 个错误.我不知道我做错了什么，有人可以帮忙吗?

I keep getting the same 3 errors every time I run this php. I have no clue what I am doing wrong, can anyone help?

以下是错误:

[05-May-2014 19:20:50 America/Chicago] PHP 警告:mysqli_query()预计至少有 2 个参数，其中 1 个在/home/sagginev/public_html/Nutrifitness/search.php 第 10 行

[05-May-2014 19:20:50 America/Chicago] PHP Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/sagginev/public_html/Nutrifitness/search.php on line 10

[05-May-2014 19:20:50 America/Chicago] PHP 警告:mysqli_num_rows()期望参数 1 是 mysqli_result, null 在/home/sagginev/public_html/Nutrifitness/search.php 第 11 行

[05-May-2014 19:20:50 America/Chicago] PHP Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /home/sagginev/public_html/Nutrifitness/search.php on line 11

[05-May-2014 19:20:50 America/Chicago] PHP 警告:mysqli_num_rows()期望参数 1 是 mysqli_result, null 在/home/sagginev/public_html/Nutrifitness/search.php 第 16 行

[05-May-2014 19:20:50 America/Chicago] PHP Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /home/sagginev/public_html/Nutrifitness/search.php on line 16

这是我的代码

在这里输入代码

    <?php
    $con=mysqli_connect('localhost','sagginev_rob','122989','sagginev_Nutrifitness');
    if (mysqli_connect_errno()) // Check connection
      {   echo "Failed to connect to MySQL: " . mysqli_connect_error();  }

        if(!isset($_POST['search'])) {
    header("Location:home.php");
    }
    $search_sql="Select * FROM Questions WHERE username LIKE '%".$_POST['search']."%' OR feedback LIKE '%".$_POST['search']."%'";
    $search_query=mysqli_query($search_sql);
    if(mysqli_num_rows($search_query)!=0) {
    $search_rs=mysqli_fetch_assoc($search_query);
    }
    ?>
    <H2> Search Results</H2>
    <?php if(mysqli_num_rows($search_query)!=0) {
     do { ?>
     <p><?php echo $search_rs['name']; ?> </p>
    <?php } while ($search_rs=mysqli_fetch_assoc($search_query));
    } else {
       echo "No results found";
    } ?>
    <form>
    <br>
    <input type="button" value="Go Back Home" onClick="parent.location='http://sagginevo.com/Nutrifitness/home.php'">
    </form>

推荐答案

错误信息很清楚.mysqli_query() 需要两个参数.你只提供一个.当您看到这样的错误消息时，第一件事您需要做的是go到手册.如果你这样做了，你会看到你必须提供你的 MySQLi 链接作为第一个参数:

The error message is quite clear. mysqli_query() requires two parameters. You only provide one. When you see an error message like this the first thing you need to do is go to the manual. If you did you would see you must provide your MySQLi link as the first parameter:

$search_query=mysqli_query($con, $search_sql);

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