通过PDO将数组插入带有PHP的MySQL数据库中对我不起作用。即使数组有41个值,它也会显示“无效的参数编号”;PDO命名为41列;并且准备好的查询有41个问号。数组具有41个值(我用var_dump检查),如下所示:$data = [ 'countrycode' => $_GET["id"], 'country' => $_GET["country"], 'region' => $_GET["region"], 'law_db_region' => $_GET["law_db_region"], [... etc ...] ]我的PDO代码如下(是的,这些是41列和41个问号):$sql = "INSERT INTO awards_country (countrycode, country, region, law_db_region, law_db_national, law_db_national2, const_url, const_version, const_date, const_dummy, const_article, const_wording, const_article2, const_wording2, const_article3, const_wording3, law_dummy, law_act, law_article, law_source, law_date, law_act2, law_article2, law_source2, law_date2, law_act3, law_article3, law_source3, law_date3, web_dummy, web_url, web_organ, web_language, web_date, web_url2, web_organ2, web_language2, web_date2, wikipedia_url, wikipedia_date, comment) VALUES ( ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ? )";$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);$stmt = $conn->prepare($sql);$stmt->execute($data);不过,我得到的错误是:致命错误:未捕获的PDOException:SQLSTATE [HY093]:无效的参数编号:在国家/地区插入代码.php:99中未定义参数堆栈跟踪:#0 [...] country-insert.php(99):PDOStatement-> execute(Array)#1 {main}放在/data/web/e96483/html/awards/country-insert.php中99行
2 回答
哔哔one
TA贡献1854条经验 获得超8个赞
为了 ?参数,您不应使用关联数组,而应将值作为位置传递
$data = [
$_GET["id"],
$_GET["country"],
$_GET["region"],
$_GET["law_db_region"],
[... etc ...]
]
- 2 回答
- 0 关注
- 169 浏览
添加回答
举报
0/150
提交
取消