带有自定义显示的 PHP/mysql 搜索

Question

我正在尝试使用 php 和 mysql 制作一个简单的搜索引擎。我知道如何使用 php 获取 mysql 结果。它的工作。

ex : - Mysql 数据库存储了这个(数据库表归档“meta”)

I am having some difficulty with my PHP/MySQL search. It worked great a year ago, but for some reason it has just stopped working. I have tried searching and cannot figure it out. I have it echo my query but it never adds in the WHERE clause. Any help is much appreciated. Form Code: PHP Code:

搜索词是搜索

我需要这样显示搜索结果

I have tried searching and cannot figure it out. I have it echo my query but it never adds in the WHERE clause. Any help is much appreciated. Form Code: PHP Code:

Answer 1

	
$query= mysql_query("SELECT meta FROM data WHERE LIKE '%searching%' ");

$count = mysqli_num_rows($query);
		
if($count >0){
		while($row = mysqli_fetch_array($query)) {

		
	    echo "$row[id]'>$row[name]</option>";
	       $count =1;

		}
		else{
		echo "<option value='' selected='selected'></option>";
		}
		mysqli_close($con);	

从上面的代码中，有几个突出的问题:

1. 该选项的回显没有 open for ，因此不会显示。
2. 当您从 $row 变量回显信息时，您正在引用 id 和 name，但查询仅输出元。
3. 查询未在字段中搜索 where 子句，因此失败

据我所知，代码应该更改为以下内容才能工作(假设没有其他周围问题):

$query = mysql_query("SELECT meta, id, name FROM data WHERE meta LIKE '%searching%'");
$count = mysqli_num_rows($query);

if ($count > 0)
{
    while ($row = $mysqli_fetch_array($query))
    {
        echo "<option value='" . $row['id'] . "'>" . $row['name'] . "</option>";
        $count = 1; // not sure why this needs to be here, but it's in your code so I'm keeping it there
    }
}
else
{
    echo "<option value='' selected='selected'></option>";
} 

mysqli_close($con);