php mysqli_query() 需要至少 2 个参数，其中 1 个给定？
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mysqli_query() expects at least 2 parameters, 1 given in?
提问by user3606199
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?
每次运行这个 php 时，我都会遇到相同的 3 个错误。我不知道我做错了什么，有人可以帮忙吗？
Here are the errors:
以下是错误：
[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
[05-May-2014 19:20:50 America/Chicago] PHP 警告：mysqli_query() 需要至少 2 个参数，1 个在第 10 行的 /home/sagginev/public_html/Nutrifitness/search.php 中给出
[05-May-2014 19:20:50 America/Chicago] PHP 警告：mysqli_num_rows() 期望参数 1 为 mysqli_result，在第 11 行的 /home/sagginev/public_html/Nutrifitness/search.php 中给出 null
[05-May-2014 19:20:50 America/Chicago] PHP 警告：mysqli_num_rows() 期望参数 1 为 mysqli_result，在第 16 行的 /home/sagginev/public_html/Nutrifitness/search.php 中给出 null
here is my code
这是我的代码
enter code here
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>
回答by John Conde
The error message is quite clear. mysqli_query()requires twoparameters. You only provide one. When you see an error message like this the first thingyou need to do is go to the manual. If you did you would see you must provide your MySQLi link as the first parameter:
错误信息非常清楚。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，您需要做的第一件事就是查看手册。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：
$search_query=mysqli_query($con, $search_sql);
回答by Jesse Hockenbury
You need to add the connection variable as the first argument which in this case is $con:
您需要添加连接变量作为第一个参数，在这种情况下是$con：
$search_query=mysqli_query($con, $search_sql);
回答by kimbarcelona
You need to change
你需要改变
$search_query=mysqli_query($search_sql);
to:
到：
$search_query=mysqli_query($con, $search_sql);
You need the connection string in the first parameter and the second is your query. Manual says:
您需要第一个参数中的连接字符串，第二个参数是您的查询。手册说：
mysqli_query(connection,query,resultmode);
Read more.
阅读更多。
