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()绑定结果。但是，当我尝试使用该值时，我只得到一个0。我怎么处理这个？

我的代码看起来像这样：

$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，它不会在输入框中显示10。这意味着它不执行if语句。但是，为什么我没有从$db_salary_amount获得任何结果？我知道我可能不必将变量从$db_salary_amount更改为$db_salary_amount。我只是想找出问题所在。

Answer 1

在->fetch()

之后，你错过了bind_result()

$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 - ＆gt; When mysqli_stmt_fetch() is called to fetch data, the MySQL client/server protocol places the data for the bound columns into the specified variables var1, ....

使用mysqli bind_result不会从数据库中获取值

1 个答案: