Mysqli查询无结果

Question

我在mysqli准备查询中遇到问题，我无法弄清楚出了什么问题。

在phpmyadmin中，我正在编写此查询

SELECT `country_id` FROM `oa_country` WHERE `name` = "Ελλάδα"

我得到了成功的结果。

当我尝试使用mysqli prepare语句执行此操作时，出现错误。

function getCountryId($country) {
    print($country); \\prints "Ελλάδα" or anything else, even countries in latin letters.
    $db = mysqli_connect(DB_HOSTNAME,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
    $result = mysqli_prepare($db, "SELECT `country_id` FROM `oa_country` WHERE `name` = ? ;");
    mysqli_stmt_bind_param($result, 's', $country);
    mysqli_stmt_execute($result);
    mysqli_stmt_bind_result($result, $countcountry);
    print_r($result);\\ see below.
    while(mysqli_stmt_fetch($result))
      {
            $countryid = $countcountry;

      }
    mysqli_stmt_close($result);
    mysqli_close($db);

    echo $countryid; \\undefined variable countryid
    return $countryid;
}

Print $ result打印这些数据。

mysqli_stmt Object
(
    [affected_rows] => -1
    [insert_id] => 0
    [num_rows] => 0
    [param_count] => 1
    [field_count] => 1
    [errno] => 0
    [error] => 
    [error_list] => Array
        (
        )

    [sqlstate] => 00000
    [id] => 1
)

奇怪的是，它曾经可以工作，但是突然间停了下来，我无法解释为什么！

PS所有插入都在使用mysqli + prepare语句。 诸如此类的查询停止工作。

更新1。

---

我在班级上方记录了此错误

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

现在我得到这个错误。

Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'Column 'country_id' cannot be null' in /some/url/lib/registerClass.php:49
Stack trace:
#0 /some/url/lib/registerClass.php(49): mysqli_stmt_execute(Object(mysqli_stmt))
#1 /some/url/register.php(34): RegisterLibrary->Address(21267, 'kapa\n', '\xCE\x9D\xCE\xAC\xCF\x84\xCF\x83\xCE\xB9\xCE\xBF\xCF\x82', 'oriste', 'pws', '123123', 'asdasd', '4121sadsa', 'sadas', '13322', '\xCE\x95\xCE\xBB\xCE\xBB\xCE\xAC\xCE\xB4\xCE\xB1', '\xCE\x91\xCF\x84\xCF\x84\xCE\xB9\xCE\xBA\xCE\xAE')

在$db变量下面，我将字符集设置为utf8。

function getCountryId($country) {
    $db = mysqli_connect(DB_HOSTNAME,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
    mysqli_set_charset( $db, 'utf8' );
    if ($stmt = mysqli_prepare($db, "SELECT country_id FROM oa_country WHERE name=?")) {

        /* bind parameters for markers */
        mysqli_stmt_bind_param($stmt, "s", $country);

        /* execute query */
        mysqli_stmt_execute($stmt);

        /* bind result variables */
        mysqli_stmt_bind_result($stmt, $countryidr);
        $thecountryid = $countryidr;
        echo $thecountryid;  //no echo.
        /* fetch value */
        mysqli_stmt_fetch($stmt);
        /* close statement */
        mysqli_stmt_close($stmt);

    } 
    return $thecountryid;  //no return.
}

Answer 1

它看起来像字符集问题

尝试设置mysqli_set_charset