空白/空搜索表單字段

Question

我已經爲我的數據庫設置了搜索表單。當我搜索並找到結果時，消息會在搜索表單下方回顯。例如，找到10條記錄，找到0條記錄。

如果搜索表單字段爲空白/空白，我怎樣才能讓該消息消失？目前它顯示15個記錄找到爲空白/空搜索字段。這是所有的數據庫記錄。

感謝您的任何幫助。

表：

<form action="" method="post"> 
    <input type="text" name="search_box" value="<?php if (isset($_POST['search_box'])) echo $_POST['search_box']; ?>" placeholder="Search here ..."/> 
    <input value="Search" name="search" type="submit" /><br> 
</form> 

PHP：

<?php 
     $count = mysqli_num_rows($result); 
     if($count > 0){ 
      echo $count . " Records Found"; 
     }if($count == 0){ 
      echo "0 Records Found"; 
     }if($count == ""){ 
      echo ""; 
     } 
     ?> 

查詢：

//Retrieve the practice posts from the database table 
$query = "SELECT * FROM practice"; 

//check if search... button clicked, if so query fields 
if(isset($_POST['search'])){ 
    $search_term = trim($_POST['search_box']); 
    $query .= " WHERE title = '{$search_term}'"; 
    $query .= " or subject LIKE '%{$search_term}%'";} 

Answer 1

<?php 

//Retrieve the practice posts from the database table 
$query = "SELECT * FROM practice"; 

//check if search... button clicked, if so query fields 
if(isset($_POST['search'])){ 
    $search_term = trim($_POST['search_box']); 
    $query .= " WHERE title = '{$search_term}'"; 
    $query .= " or subject LIKE '%{$search_term}%'"; 

    //execute your query 
    $result = $dbconnect->query($query); 

    $count = mysqli_num_rows($result); 
    if($count > 0){ 
     echo $count . " Records Found"; 
    } 
    if($count == 0){ 
     echo "0 Records Found"; 
    } 
} 
else { 
    // it is mean your search box value($_POST['search']) is empty, so it will echo null value 
    echo $_POST['search']; 
} 
?> 

請嘗試，希望將節省您的一天：d

Answer 2

請與這些更換的PHP代碼： -

<?php 
      if(isset($_REQUEST['search_box'])){ 
     $count = mysqli_num_rows($result); 
     } else { 
     $count = ''; 
     } 
     if($count > 0){ 
      echo $count . " Records Found"; 
     }if($count == 0){ 
      echo "0 Records Found"; 
     }if($count == ""){ 
      echo ""; 
     } 
     ?> 

Answer 3

只要把你的代碼在POST方法

if ((isset($_POST['search_box'])) && ($_POST['search_box']!="")) 
{ 

    $count = mysqli_num_rows($result); 
    if($count > 0){ 
     echo $count . " Records Found"; 
    }if($count == 0){ 
     echo "0 Records Found"; 
    }if($count == ""){ 
     echo ""; 
    } 
}