mysqli_num_rows（）預計參數1是

Question

我創建的搜索和檢索在數據庫中的數據的搜索功能，但得到這個錯誤

mysqli_num_rows（）預計參數1至 線16

可以mysqli_result PHP mysqli_result PHP

這裏是我的PHP代碼

<?php 
    include 'database_conn.php'; 

    //collect search title 
    if (isset($_GET['keywords'])){ 

     $searchq = $_GET['keywords']; 
     $searchq = preg_replace("#[^a-z0-9]#i" , "", $searchq); 

     $query = "SELECT eventTitle FROM te_events WHERE eventTitle LIKE '%searchq%'"; 

     mysqli_query($conn, $query) 
     or die ("SQL error:" .mysqli_error($conn)); 


     $count = mysqli_num_rows($query); 

     if($count==0){ 
      echo "<p>There was no search result!</p>\n"; 
     } 

     else{ 
      while ($row = mysqli_fetch_array($query)){ 
       $title = $row['eventTitle']; 
       $id = $row['eventID']; 

       echo "<p>title</p>\n"; 
      } 
     } 
    } 
?> 

有沒有在我的代碼的任何問題???

Answer 1

2誤區在你的代碼

1. 使用'%$searchq%'而不是'%searchq%'。

2. 將mysqli_query($conn, $query)替換爲$queryresult = mysqli_query($conn, $query)並將$queryresult替換爲計數和結果數組。

本

<?php 
    include 'database_conn.php'; 

    //collect search title 
    if (isset($_GET['keywords'])){ 

     $searchq = $_GET['keywords']; 
     $searchq = preg_replace("#[^a-z0-9]#i" , "", $searchq); 

     $query = "SELECT eventTitle FROM te_events WHERE eventTitle LIKE '%$searchq%'"; 

     $queryresult = mysqli_query($conn, $query) 
     or die ("SQL error:" .mysqli_error($conn)); 


     $count = mysqli_num_rows($queryresult); 

     if($count==0){ 
      echo "<p>There was no search result!</p>\n"; 
     } 

     else{ 
      while ($row = mysqli_fetch_array($queryresult)){ 
       $title = $row['eventTitle']; 
       $id = $row['eventID']; 

       echo "<p>title</p>\n"; 
      } 
     } 
    } 
?>