Php Paypal IPN脚本/MySql插入查询问题
我的IPN脚本有问题。 它的罚款核实付款等,但我试图让它输入到一个数据库的数据。 之前的用户电子邮件和密码输入可以正常工作,但第二个不能Php Paypal IPN脚本/MySql插入查询问题,php,mysql,sql,paypal,Php,Mysql,Sql,Paypal,我的IPN脚本有问题。 它的罚款核实付款等,但我试图让它输入到一个数据库的数据。 之前的用户电子邮件和密码输入可以正常工作,但第二个不能 $item_name = $_POST['item_name']; $item_number = $_POST['item_number']; $payment_status = $_POST['payment_status']; $payment_amount = $_POST['mc_gross']; $payment_currency = $_POST[
$item_name = $_POST['item_name'];
$item_number = $_POST['item_number'];
$payment_status = $_POST['payment_status'];
$payment_amount = $_POST['mc_gross'];
$payment_currency = $_POST['mc_currency'];
$txn_id = $_POST['txn_id'];
$receiver_email = $_POST['receiver_email'];
$payer_email = $_POST['payer_email'];
$email = $_POST['item_name'];
$password = mt_rand(1000, 9999);
$Random = print_r($_POST);
mysql_query("INSERT INTO users (email, password) VALUES('". mysql_escape_string($email) ."', '".md5($password)."') ") or die(mysql_error());
mysql_query("INSERT INTO LNCH_Sales SET
item_name = '%s',
item_number = '%s',
payment_status = '%s',
payment_amount = '%s',
payment_currency = '%s',
txn_id = '%s',
receiver_email = '%s',
payer_email = '%s'",
mysql_real_escape_string($item['name']),
mysql_real_escape_string($item['number']),
mysql_real_escape_string($payment['status']),
mysql_real_escape_string($payment['amount']),
mysql_real_escape_string($payment['currency']),
mysql_real_escape_string($txn['id']),
mysql_real_escape_string($receiver['email']),
mysql_real_escape_string($payer['email'])
);
你知道这里怎么了吗?你的语法错了-而不是
INSERT INTO LNCH_SALES SET
(看起来像是插入和更新的混合体),您需要
INSERT INTO LNCH_SALES(<column_list>) VALUES (...
插入LNCH_SALES()值(。。。
您的语法错误-而不是
INSERT INTO LNCH_SALES SET
(看起来像是插入和更新的混合体),您需要
INSERT INTO LNCH_SALES(<column_list>) VALUES (...
插入LNCH_SALES()值(。。。
正如@FrankSchmitt所说,您可能缺少列列表
可以跳过列列表的唯一方法是插入列数=表列数,即:
jcho360> create table test (id int,name varchar(20));
Query OK, 0 rows affected (0.05 sec)
jcho360> insert into test values (1,'Jcho360');
Query OK, 1 row affected (0.00 sec)
jcho360> insert into test values (1);
ERROR 1136 (21S01): Column count doesn't match value count at row 1
jcho360> insert into test(id) values (1);
Query OK, 1 row affected (0.01 sec)
正如您所见,只有当我准备填写“插入”中的所有列时,您才可以跳过列列表。正如@FrankSchmitt所说,您可能缺少列列表 可以跳过列列表的唯一方法是插入列数=表列数,即:
jcho360> create table test (id int,name varchar(20));
Query OK, 0 rows affected (0.05 sec)
jcho360> insert into test values (1,'Jcho360');
Query OK, 1 row affected (0.00 sec)
jcho360> insert into test values (1);
ERROR 1136 (21S01): Column count doesn't match value count at row 1
jcho360> insert into test(id) values (1);
Query OK, 1 row affected (0.01 sec)
正如您所看到的,只有当我准备填写“插入”中的所有列时,您才能跳过列列表