php - 使用 mysqli bind_result 未从数据库中获取值

Question

我正在尝试使用 mysqli 获取在我的数据库表中找到的 int。然后我使用 bind_result() 绑定(bind)结果。但是，当我尝试使用该值时，我只是得到一个 0。我该如何处理？

我的代码是这样的:

$sql = <<<EOF
        SELECT
        project_salary_amount
        FROM projects_set_salary
        WHERE project_id = ? ORDER BY project_salary_id DESC LIMIT 1
EOF;
    $stmt = $mysqli->prepare($sql) or die ("Feil i database<br>" . $sql . "<br><b>Feilmelding:</b> " . $mysqli->error);
    $stmt->bind_param("i", $project_id);
    $stmt->execute();
    $stmt->store_result();
    $stmt->bind_result($dbb_salary_amount);
    $db_salary_amount = $dbb_salary_amount;
    $num_salary_results = $stmt->num_rows;
    $stmt->free_result();
    $stmt->close();

    if($num_salary_results == 0){
        $sql = <<<EOF
        INSERT INTO
        projects_set_salary (project_id)
        VALUES ($project_id)
EOF;
        $stmt = $mysqli->prepare($sql) or die ("Feil i database<br>" . $sql . "<br><b>Feilmelding:</b> " . $mysqli->error);
        $stmt->execute();
        $stmt->close();
        $db_salary_amount = 10;
        $html_set_salary = "";
        $html_set_salary .= "<form id=\"form_send_salary\"method=\"post\" action=\"create_set_salary.php\">\n";
        $html_set_salary .= "<input type=\"number\" name=\"set_salary\" value=\"$db_salary_amount\">";
    }
    else{
        $html_set_salary = "";
        $html_set_salary .= "<form id=\"form_send_salary\"method=\"post\" action=\"create_set_salary.php\">\n";
        $html_set_salary .= "<input type=\"number\" name=\"set_salary\" value=\"$db_salary_amount\">";
    }

现在输入框内不显示10了。这意味着它不执行 if 语句。但是，为什么我没有从 $db_salary_amount 得到任何结果？我知道我可能不必将变量从 $db_salary_amount 更改为 $db_salary_amount。我只是想找出问题所在。

Answer 1

您在 bind_result() 之后缺少一个 ->fetch()

$stmt->execute();
$stmt->store_result();
$stmt->bind_result($dbb_salary_amount);
$stmt->fetch(); // <--- missing this ->fetch()
$db_salary_amount = $dbb_salary_amount;

来自 ->bind_result() 的文档-> 当调用 mysqli_stmt_fetch() 获取数据时，MySQL 客户端/服务器协议(protocol)将绑定(bind)列的数据放入指定变量 var1，....