Question

if (isset($_GET['k']))
{

    $k=$_GET['k'];
    $query="SELECT * FROM `upload` WHERE `keywords` LIKE '%$k%'  ";

 @mysql_connect("localhost","root","") or die("error");
 mysql_select_db("lol") or die ("error");

 $query= mysql_query($query);
  $numrows=mysql_num_rows($query);
 if($numrows >0)
 {
    while ($row =mysql_fetch_assoc($query))
    {

    $keywords=$row['keywords'];
    $name2=$row['name2'];
    $URL=$row['URL'];
    echo "<h2><a href='$URL'>$name2</a></h2><br/><br/>";

    }
 }
 else
 echo "no result found";
 }
 else
 {
    echo "please enter some value in search";
 }

?>

我正在制作一个简单的搜索引擎问题是，如果不在搜索栏中传递一个值，它会显示数据库中的所有结果。为了使其正确我使用ISSET但它不起作用PLZ帮我这个。这是附加的代码

Answer 1

这是Html：

<form method="post" action="#">
    <input type="text" name="k" value="">
    <input type="submit" name="submit">
</form>

其次是PHP代码：

<?php
if (isset($_POST['submit']) && isset($_POST['k']) && ($_POST['k'] != ""))
{

    $k=$_POST['k'];
    $query="SELECT * FROM `upload` WHERE `keywords` LIKE '%$k%'  ";

 @mysql_connect("localhost","root","") or die("error");
 mysql_select_db("lol") or die ("error");

 $query= mysql_query($query);
  $numrows=mysql_num_rows($query);
 if($numrows >0)
 {
    while ($row =mysql_fetch_assoc($query))
    {

    $keywords=$row['keywords'];
    $name2=$row['name2'];
    $URL=$row['URL'];
    echo "<h2><a href='$URL'>$name2</a></h2><br/><br/>";

    }
 }
 else
 echo "no result found";
 }
 else
 {
    echo "please enter some value in search";
 }

?>

使用php搜索isset无法正常工作

1 个答案: