每次运行此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?
以下是错误:
[05-May-2014 19:20:50 America / Chicago] PHP警告:mysqli_query() 期望至少有2个参数,1个给出 第10行/home/sagginev/public_html/Nutrifitness/search.php
[05-May-2014 19:20:50 America / Chicago] PHP警告:mysqli_num_rows() 期望参数1为mysqli_result,null中给出null 第11行/home/sagginev/public_html/Nutrifitness/search.php
[05-May-2014 19:20:50 America / Chicago] PHP警告:mysqli_num_rows() 期望参数1为mysqli_result,null中给出null 第16行/home/sagginev/public_html/Nutrifitness/search.php
这是我的代码
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>
答案 0 :(得分:5)
错误信息非常清楚。 mysqli_query()
需要两个参数。你只提供一个。当您看到类似这样的错误消息时,您需要做的是 go to the manual {{3}} 。如果你这样做,你会发现你必须提供你的MySQLi链接作为第一个参数:
$search_query=mysqli_query($con, $search_sql);
答案 1 :(得分:1)
您需要添加连接变量作为第一个参数,在本例中为$con
:
$search_query=mysqli_query($con, $search_sql);
答案 2 :(得分:0)
您需要更改
$search_query=mysqli_query($search_sql);
为:
$search_query=mysqli_query($con, $search_sql);
您需要第一个参数中的连接字符串,第二个是您的查询。手册说:
mysqli_query(connection,query,resultmode);
阅读more.