問題描述

可捕獲的致命錯誤:第 20 行 C:xampphtdocsxxxdash.php 中的類 mysqli_result 的對象無法轉換為字符串

Catchable fatal error: Object of class mysqli_result could not be converted to string in C:xampphtdocsxxxdash.php on line 20

我是一個相當新的人，作為一個老派的編碼員，只是使用 mysql_result 來獲取這些數據，我不知道如何去做.我有一個類-> 功能設置.

I am quite fairly new, and being a old-school coder, simply using mysql_result to grab such data, I am unaware of how to go about this. I have a class->function setup.

dash.php 的第 20 行包含:

Line 20 of dash.php contains:

echo $user->GetVar('rank', 'Liam', $mysqli);

雖然，功能是:

function GetVar($var, $username, $mysqli)
    {
        $result = $mysqli->query("SELECT " . $var . " FROM users WHERE username = '" . $username . "' LIMIT 1");
        return $result;
        $result->close();
    }

現在，據我所知，我打算將 $result 轉換為字符串，但我不完全了解如何執行此操作.我嘗試了幾種方法，但都無濟于事.所以我來到社區希望得到答案，我也環顧四周，但注意到所有其他線程都要求 num_rows，而我只想從查詢選擇中獲取字符串.

Now, to my understanding, I am meant to convert $result into a string, but I am not fully aware of how to do so. I've tried using a few methods, but to no avail. So I've come to the community to hopefully get a answer, I've also looked around but noticed that all other threads are asking for num_rows, while I just want to grab the string from the query select.

推薦答案

在回顯結果之前，您必須先獲取它.粗略示例:

You have to fetch it first before echoing the results. Rough Example:

function GetVar($var, $username, $mysqli) {
    // make the query
    $query = $mysqli->query("SELECT ".$var." FROM users WHERE username = '".$username."' LIMIT 1");
    $result = $query->fetch_assoc(); // fetch it first
    return $result[$var];
}

然后使用你的函數:

echo $user->GetVar('rank', 'Liam', $mysqli);

重要提示:既然你剛開始，請檢查準備好的語句.不要直接在您的查詢中附加用戶輸入.

Important Note: Since you're starting out, kindly check about prepared statements. Don't directly append user input on your query.

這篇關于對象無法在 MySQLi PHP 中轉換為字符串的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！