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

问题描述

每次运行这个php时，我都会收到相同的3个错误。

以下是错误：

[05-May-2014 19:20:50 America / Chicago] PHP警告：mysqli_query（）
至少需要2个参数，1给出
/ home / sagginev / public_html /Nutrifitness/search.php on line 10

[05-May-2014 19:20:50 America / Chicago] PHP警告：mysqli_num_rows（）
预计参数1为mysqli_result，null在
/home/sagginev/public_html/Nutrifitness/search.php中给出的第11行

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

这是我的代码

code $
$ p $ <？php
$ con = mysqli_connect（'localhost'，'sagginev_rob'， '122989'，'sagginev_Nut rifitness’）;
if（mysqli_connect_errno（））//检查连接
{echo无法连接到MySQL：。 mysqli_connect_error（）; }
$ b $ if（！isset（$ _ POST ['search']））{
header（Location：home.php）;

$ search_sql =Select * FROM WHERE USER 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>搜寻结果< / H2>
<？php if（mysqli_num_rows（$ search_query）！= 0）{
do {？>
< p><？php echo $ search_rs ['name']; ？> < / p为H.
<？php} while（$ search_rs = mysqli_fetch_assoc（$ search_query））;
} else {
echo未找到结果;
}？>
< form>
< br>
< input type =buttonvalue =Go Back HomeonClick =parent.location ='http：//sagginevo.com/Nutrifitness/home.php'>
< / form>





解决方案

错误信息非常清晰。 mysqli_query（）需要两个参数。你只提供一个。当您看到像这样的错误消息时，您需要做的第一件事： 去到手册 。如果你确实看到过，你必须提供MySQLi链接作为第一个参数：



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




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?

Here are the errors:

[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 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 Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /home/sagginev/public_html/Nutrifitness/search.php on line 16

here is my code

enter code here

    <?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>

解决方案

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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