Php Mysqli查询没有结果
我在mysqli prepare查询中遇到了一个问题,我无法找出问题所在 在phpmyadmin中,我正在编写这个查询Php Mysqli查询没有结果,php,mysql,mysqli,prepared-statement,Php,Mysql,Mysqli,Prepared Statement,我在mysqli prepare查询中遇到了一个问题,我无法找出问题所在 在phpmyadmin中,我正在编写这个查询 SELECT `country_id` FROM `oa_country` WHERE `name` = "Ελλάδα" 我得到了一个成功的结果 当我尝试使用mysqli prepare语句执行此操作时,我得到一个错误 function getCountryId($country) { print($country); \\prints "Ελλάδα" or a
SELECT `country_id` FROM `oa_country` WHERE `name` = "Ελλάδα"
我得到了一个成功的结果
当我尝试使用mysqli prepare语句执行此操作时,我得到一个错误
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,它曾经工作过,但突然停止了,我无法解释为什么
另外,所有插入都与mysqli+prepare语句一起使用。
像这样的查询停止了工作
更新1。
我在我的类上面记录了这个错误
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
现在我得到了这个错误
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.
}
看起来像是charset问题
尝试设置mysqli\u set\u字符集可能尝试设置mysqli\u set_charset@bxN5我想这和charset有关。查看更新的问题@bxN5好的,我正确设置了字符集,它工作了。如果你想回答,我会接受:)那太酷了:)