php - mysqli_num_rows()期望参数1为mysqli_result php

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mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result
                                
                                    （32个答案）
                                
                        
                                2年前关闭。
            
                    
我创建了一个搜索功能来搜索和检索数据库中的数据，但出现此错误


  mysqli_num_rows（）期望参数1为mysqli_result php in
  16行


这是我的PHP代码

<?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";
            }
        }
    }
?>

我的代码有什么问题吗？？？

最佳答案

您的代码中有2个错误


使用'%$searchq%'代替'%searchq%'。
将mysqli_query($conn, $query)替换为$queryresult = mysqli_query($conn, $query)，并将$queryresult用于计数和结果数组。


以此替换您的代码

<?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";
            }
        }
    }
?>

关于php - mysqli_num_rows()期望参数1为mysqli_result php ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40560034/

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Answer 1

您的代码中有2个错误


使用'%$searchq%'代替'%searchq%'。
将mysqli_query($conn, $query)替换为$queryresult = mysqli_query($conn, $query)，并将$queryresult用于计数和结果数组。


以此替换您的代码

<?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";
            }
        }
    }
?>