php - MySQL 如果可用数据转到页面并显示结果，如果不可用则返回检查号

Question

我需要一个我无法到达的小逻辑

我需要从数据库获取MySQL数据并在.php文件上回显，然后我需要检查数据是否可用(如果可用)转到one.php文件以显示结果(如果数据不可用)返回到我使用的上一个页面下面的代码但没有用请帮助

<?php

$connection = mysql_connect('localhost','asd_asd','13wda121asSS') or die ("Couldn't connect to server."); 
$db = mysql_select_db('asdasd_asds', $connection) or die ("Couldn't select database."); 

$warno=$_POST['search'];

$data = 'SELECT * FROM `cases` WHERE `warno` = "'.$warno.'"';

if (mysql_num_rows($result)==0) { header('Location: http://asdas.in/wcdsa.php')
} 
else
    {
    //We have no results and can't find the warranty, return a different page saying no warranty results exist
     header('Location: http://asdsd.in/wcasd.php'); //redirect
}

?>

这里我需要

$data = 'SELECT * FROM cases WHERE warno = "'.$warno.'"'; $result = mysql_query($data,$connection); //execute query 

我需要在其他页面(例如 results.php)中执行查询，我将在其中显示结果

如果数据库中不存在数据，则在执行查询的同一页面中显示“NO DATA FOUND”，请对此提供帮助

Answer 1

您没有执行查询。您需要先执行查询。您的 $result 未定义。

试试这个:

<?php

$connection = mysql_connect('localhost','asd_asd','13wda121asSS') or die ("Couldn't connect to server."); 
mysql_select_db('asdasd_asds', $connection) or die ("Couldn't select database."); 

$warno=$_POST['search'];

$data = 'SELECT * FROM `cases` WHERE `warno` = "'.$warno.'"';
$result = mysql_query($data,$connection); //execute query
if (mysql_num_rows($result)==0) { header('Location: http://asdas.in/wcdsa.php')
} 
else
    {
    //We have no results and can't find the warranty, return a different page saying no warranty results exist
     header('Location: http://asdsd.in/wcasd.php'); //redirect
}

?>