Php mysqli_fetch_array（）-添加另一列以匹配时出错

我正在用PHP开发一个小的搜索引擎应用程序，我被网站排名所困扰。在代码中我有一个简单的查询，它应该匹配用户查询（输入）和网站描述

$row2 = [];
$search_word = false;
if(isset($_POST["submit"])) {
    //edit this with your credentials
    $con = mysqli_connect("localhost", "root", "uThx6wuf", "search");

    if(mysqli_connect_error()) echo "Connection Fail";
    else {
        $search_word = true;
        $input = $_POST["s_input"];

        // tokenize input
        $tokens = tokenize($input);

        //compute weight of every token
        $token_weight = compute_weight($tokens, $con);

        $sql2 = "SELECT *, match(description) against('". $input ."') as score FROM search where match(description) against('".$input."') order by score desc";

        $result2 = mysqli_query($con, $sql2);

        $sql3 = "SELECT * FROM search";
        $numDocs = (mysqli_num_rows(mysqli_query($con, $sql3)));

        $maxOverlap = sizeof($tokens);
        $ctr2 = 0;
        while($ctr2 != $maxOverlap){
            //compute inverse_document_frequency of term
            $sql3 = "SELECT *, match(description) against('".$tokens[$ctr2]."') FROM search where match(description) against('".$tokens[$ctr2]."')";
            $docFreq = (mysqli_num_rows(mysqli_query($con, $sql3)));
            $idf[$tokens[$ctr2]] = idf($numDocs, $docFreq);
            $ctr2++;
        }

当我添加另一列进行匹配时，每次都会出现错误

mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in

已编辑查询的代码：

$sql2 = "SELECT *, match(description, title) against('". $input ."') as score FROM search where match(description, title) against('".$input."') order by score desc";

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首先，如果查询正常，请首先在sql浏览器中使用数据库中的现有信息运行查询

如果正常，则检查是否接收到输入

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您的错误似乎来自sql语法，或者输入被输入到sql语法

请尝试将该错误粘贴到搜索框中。它已经被问了几十次了。只有一个错误——我把它粘贴到了问题中。我可以毫无问题地运行旧的查询（sql browser，php），但一旦我在match（标题）中添加了另一列，它就会给我一个错误——在sql browser中，它是#1191——找不到与列列表匹配的全文索引。