Question

我正在研究PHP中的一个小型搜索引擎应用程序，并且我被困在排名网站上。在代码中我有一个简单的查询，它应该匹配用户查询（输入）与网站的描述。

$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++;
        }

当我添加另一列以匹配时，每次出现错误

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

编辑查询代码：

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

Answer 1

首先在sql浏览器中首先使用数据库中的现有信息运行查询（如果查询正常）

检查输入是否已接收

似乎您的错误来自sql语法或输入已被提供给sql语法。

mysqli_fetch_array（） - 添加另一列以匹配时出错

1 个答案: