php - mysqli_fetch_array() - 添加另一列来匹配时出错

Question

我正在用 PHP 开发一个小型搜索引擎应用程序，但我陷入了对网站排名的困境。在代码中，我有一个简单的查询，它应该将用户查询(输入)与网站的描述相匹配。

$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

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";

Answer 1

如果查询正常，首先在 SQL 浏览器中使用数据库中的现有信息运行查询

检查收到的输入是否正常

看来您的错误来自 sql 语法或输入已馈送到 sql 语法。