Question

我想创建搜索页面（php），我希望在“div”中显示我从数据库中搜索到的数据？我与数据库建立了连接并在一个phppage中搜索数据并在另一个phppage中创建了div标签。如何显示一个php页面的搜索数据，以显示在另一个php页面的“div”

中

的search.php：

<?php 
include 'Searchdata.php';
include 'connect.php';
if(isset($_POST['submit'])){
   $searchkey= $_POST['search'];    
   $searchkey=preg_replace("#[^0-9a-z]#i", "", $searchkey); 
   $query = mysqli_query($conn, "SELECT * FROM newentry WHERE Date LIKE '%$searchkey%'")or die("Could not search!");
   $count = mysqli_num_rows($query); 
   if(!($count == 0)) {    
      while($row=mysqli_fetch_array($query)){
         $Date=$row['Date'];
         $Entry=$row['Entry'];
         echo'<div>'.$Date.'<br>'.$Entry.'</div>';
       }
    } else {echo "There was no search result!";} 
}?>

Searchdata.php：

<div>
<form action="Search.php" method="post">  
   <input type="text" name="search" placeholder="Search"> 
   <input type="submit" value="Search" />
</form>

Answer 1

只需创建一个变量来存储结果，在这种情况下是$ data。将include searchdata.php移动到代码的底部，以便它可以重新计算$ data。然后在你的html页面上回复它。

html page

<div>
<form action="Searchdata.php" method="post">  
<input type="text" name="search" placeholder="Search"> 
<input type="submit" value="Search">
</form>
<div><?php echo $data ?></div>
</div>

PHP代码。

<?php
include 'connect.php';
$data = '';
if(isset($_POST['submit'])){
$searchkey= $_POST['search'];   
$searchkey=preg_replace("#[^0-9a-z]#i", "", $searchkey); 
$query = mysqli_query($conn, "SELECT * FROM newentry WHERE Date LIKE '%$searchkey%'")or die("Could not search!");
$count = mysqli_num_rows($query); 
if(!($count == 0)) {    
while($row=mysqli_fetch_array($query)){ 
$Date=$row['Date'];
$Entry=$row['Entry'];
$data = '<div>'.$Date.'<br>'.$Entry.'</div>';
}
} else {
$data = "There was no search result!";}}
include 'Search.php';
?>

我想创建搜索页面，我希望在“div”中显示数据库中的搜索数据？

1 个答案: