使用 PDO 将行插入 SQL

Question

我无法让 php postgre 使用以下代码从 php 数组 new_address 中插入一行：

    $customer_id = '2319';
    $use_frequency = 1;

    $sth = $dbh->prepare("
    INSERT INTO address 
            ('storeid', 
             'classtypeid', 
             'modifiedbyuser', 
             'modifiedbycomputer', 
             'modifieddate', 
             'seqid', 
             'issystem', 
             'isactive', 
             'streetaddress1', 
             'streetaddress2', 
             'city', 
             'state', 
             'county', 
             'postalcode', 
             'country', 
             'formattedtext', 
             'taxclassid', 
             'isvalidated', 
             'validatedaddress', 
             'hasvalidationerror', 
             'validationerror', 
             'customer_id', 
             'use_frequency') 
   VALUES   ( NULL, 
              NULL, 
              NULL, 
              NULL, 
              NULL, 
              NULL, 
              NULL, 
              NULL, 
             :address_1, 
             :address_2, 
             :city, 
             :state, 
              NULL, 
             :zip, 
             :country, 
             :formatted_text, 
              NULL, 
              NULL, 
              NULL, 
              NULL, 
              NULL, 
             :customer_id, 
             :use_frequency");



$sth->execute(array(
    ':address_1' => $new_address['address_1'],
    ':address_2' => $new_address['address_2'],
    ':city' => $new_address['city'],
    ':state' => $new_address['state'],
    ':zip' => $new_address['zip'],
    ':country' =>$new_address['country'],
    ':formatted_text' => $formatted_text,
    ':customer_id' => $customer_id, 
    ':use_frequency' => $use_frequency
    );           


$sth->execute();

表格中的最后一列是id，它是serial，所以我省略了它，认为它会自动递增，但如果我错了，请告诉我。

我收到错误：

致命错误：带有消息的未捕获异常“PDOException” 'SQLSTATE [42601]：语法错误：7 错误：在或附近出现语法错误 "'storeid'" LINE 3: ('storeid', ^' in

print_r($new_address); 给我看：

Array (
[0] => stdClass Object (
[customer_id] => 9319
)
[1] => stdClass Object (
[address_1] => 1515 example st
)
[2] => stdClass Object (
[address_2] => box 1
)
[3] => stdClass Object (
[city] => town
)
[4] => stdClass Object (
[state] => ST
)
[5] => stdClass Object (
[zip] => 12345
)
[6] => stdClass Object (
[country] => US
)
)

感谢您的建议！

Answer 1

根据4.1. Lexical Structure，必须用双引号"转义列名

还有第二种标识符：分隔标识符或引用标识符。它是通过将任意字符序列括在双引号 (") 中而形成的。分隔标识符始终是标识符，而不是关键字。因此“select”可用于引用名为“select”的列或表，而未加引号的选择将被视为关键字，因此在需要表或列名的地方使用时会引发解析错误。

INSERT INTO address 
        ("storeid", 
         "classtypeid", 
         ...

另外，如果您将列的默认值设置为NULL，您可以从列列表中省略它们并仅使用您真正需要的那些

insert into address
    ("streetaddress1", 
     "streetaddress2", 
     "city", 
     "state", 
     ...)
values (:address_1, 
       :address_2, 
       :city, 
       :state, 
       ...)

根据您的评论，您必须修改 $new_address 数组。它按数字索引，不是按名称。

如果可以将 JSON 更改为

{ "customer_id": 9319,
  "address_1": "1515 example trail",
  "address_2": "box 1",
  "city": "town city",
  "state": "MI",
  "zip": "12345",
  "country": "US" }

你可以使用

$new_address = json_decode($json, true);

获取关联数组。

如果无法更改 JSON，则必须将其映射到关联数组

$json = json_decode('[ { "customer_id": 9319 }, { "address_1": "1515 example trail" }, { "address_2": "box 1" }, { "city": "town city" }, { "state": "MI" }, { "zip": "12345" }, { "country": "US" } ]');

foreach ($json as $element) {
    foreach ($element as $key => $val) {
        $new_address[$key] = $val;
    }
}