检查在mysql db中是否找不到结果 [英] Check if no result found in mysql db

问题描述

我写了一个phpi在mysqli中搜索一个表，它工作正常，但是如果没有找到结果，我想向用户显示正确的消息.

I wrote a php search for a table in mysqli and it works fine but i want to show the correct message to user if no result were found.

这是我当前的代码:

$search_string=$_GET["design"];

$connect= mysqli_connect("mysql.myhost.com","abc","123456","mydb_db");
$query="select * from product where product_design like '%$search_string%'";
$rows= @mysqli_query($connect,$query) or die("Error: ".mysqli_error($connect));
if ($rows!=null)//I put this if to check if there is any result or not but its not working
{

while(($record=mysqli_fetch_row($rows))!=null)
{
    .
            .//i have working code for showing the result here
            .
}   
mysqli_close($connect);
}
else{
echo"no result found";
}

您能帮我解决什么问题吗，即使我搜索数据库中不存在的内容，程序仍不会显示未找到结果"

could you please help me what is wrong , even when i search for something which is not exist in the db still the program not displaying "no result found"

谢谢

推荐答案

您需要的是 mysqli_num_rows ，特别是mysqli_result::num_rows位.这会将num_rows属性添加到mysqli结果集中.这意味着您可以

What you need is mysqli_num_rows specifically the mysqli_result::num_rows bit. This adds a num_rows property to mysqli result sets. This means you can do

$rowCount = $rows->num_rows

还有一个非OO等效项...

There's also a non-OO equivalent ...

$rowCount = mysqli_num_rows($rows);

(区别纯粹是编码风格之一)

(The difference is purely one of coding style)

使用其中之一来确定返回多少记录并输出适当的消息.

Use one of these to determine how many records are returned and output the appropriate messages.

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