我无法让php postgre使用以下代码从php数组new_address
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
);
$sth->execute();
表格中的最后一列是id
,它是serial
所以我省略了它,认为它会自动递增,但请告诉我,如果我错了。
我收到错误:
致命错误:带有消息的未捕获异常'PDOException' 'SQLSTATE [42601]:语法错误:7错误:语法错误在或附近 “'storeid'”第3行:('storeid',^'in
print_r($new_address);
告诉我:
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
)
)
感谢您的任何建议!
答案 0 :(得分:1)
根据4.1. Lexical Structure,您必须使用双引号"
还有第二种标识符:分隔标识符或带引号的标识符。它是通过用双引号(“)括起任意字符序列而形成的。分隔标识符始终是标识符,绝不是关键字。因此,”select“可用于引用名为”select“的列或表,而不带引号的选择将被视为关键字,因此在预期使用表或列名称时会引发解析错误。
INSERT INTO address
("storeid",
"classtypeid",
...
此外,如果您将列的默认值设置为NULL
,则可以从列列表中省略它们,并仅使用您真正需要的列
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,不按名称。
如果您可以将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;
}
}