Php 注意：第23行中的数组到字符串转换

此处的这一行：

<?php
include 'db.inc.php';


//function that well return search results
//takes the keywords entered by user

function search_results ($keywords){

    //code here
    $returned_results= array();//array that will contain results
    $where= ""; // constract of query for results

    $keywords = preg_split ('/[\s]+/', $keywords);
         //function split values by space
     // (hello aya) is like ( hello  aya )

    $total_keywords = count($keywords);
         //count number of words entered ex(ayoy adel =2 , a b c s = 4)


    foreach ($keywords as $key=>$keyword){
    $where .="`keywords` LIKE '%$keywords%'";
               if ($key != ($total_keywords -1 )){
                 $where .= " AND " ;
               }

        }

       $results = "SELECT `title` , `subject` , `id` FROM `` WHERE $where ; ";
       }

    //array begins from 0 , this erro

    //$key=>$keyword key=0 , 1, 2   $keyword value in database
    //
    //
    //

    /*$sqlQuery= " SELECT `keywords` FROM `question` WHERE `keywords` LIKE `%hci%` AND  `keywords` LIKE `%what%` ";
*/

?>

---

包括

$keywords

变量，该变量是数组，而不是字符串。您可能想在那里使用

$keyword

变量。

$where.=“`keywords``类似于“%$keyword%”

$keywords

是您的数组

$keyword

是foreach循环中的单个关键字您从phpacadmy的视频教程中获得了这个脚本，我建议您将连接源更改为PDO，因为它不安全是的，是的，我仍然是php和mysql的初学者。PDO是什么意思？提前感谢汉克斯和皮尔塞涅克斯，这帮了我的忙那是我的错误^^

$where .="`keywords` LIKE '%$keywords%'";