我正在尝试使用ajax更新数据库,但如果在控制台中看到,则会收到一些错误,如果有错误。
INSERT INTO customers(customerName,contactLastName,contactFirstName, phone, addressLine1,addressLine2, city, state, postalCode,country,salesRepEmployeeNumber,creditLimit) VALUES (,,,, ,,,,,,,)
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ',,, ,,,,,,,)' at line 1
我已经高兴了包含原型ajax代码的JS代码,然后是包含插入细节的php代码
你能帮我解决这个问题吗?我的网站链接
http://localhost/fashionsite/customer.php
我的js代码enter code here
$('submit_btn').observe('click', function(ev) {
$('customerdetails').request({
method: 'get',
onFailure: function() {
alert("failed");
},
onComplete: function(details){
console.log(details.responseText);
//alert("inserted success fully");
//$("content_updated").update(details.responseText);
}
});
ev.preventDefault();
});
我的PHP代码
<?php
include 'config.php';
include 'opendb.php';
//$sNO = $_POST["sNO"];
//$customerNumber = $_POST["customerNumber"];
$customerNames = $_POST["customerName"];
$contactLastName = $_POST["contactLastName"];
$contactFirstName = $_POST["contactFirstName"];
$phone = $_POST["phone"];
$addressLine1 = $_POST["addressLine1"];
$addressLine2 = $_POST["addressLine2"];
$city = $_POST["city"];
$state = $_POST["state"];
$postalCode = $_POST["postalCode"];
$country = $_POST["countryText"];
$salesRepEmployeeNumber = $_POST["salesRepEmployeeNumber"];
$creditLimit = $_POST["creditLimit"];
/*echo $customerNumber .'<br/>'. $customerName .'<br/>'. $contactLastName .'<br/>'. $contactFirstName .'<br/>'. $phone .'<br/>'. $addressLine1 .'<br/>'. $addressLine2 .'<br/>'. $city .'<br/>'. $state .'<br/>'. $postalCode .'<br/>'. $country .'<br/>'. $salesRepEmployeeNumber .'<br/>'. $creditLimit;*/
$sql = "INSERT INTO customers(customerName,contactLastName,contactFirstName, phone, addressLine1,addressLine2, city, state, postalCode,country,salesRepEmployeeNumber,creditLimit) VALUES ($customerNames,$contactLastName,$contactFirstName,$phone, $addressLine1,$addressLine2,$city,$state,$postalCode,$country,$salesRepEmployeeNumber,$creditLimit)";
print $sql;
//echo $sql .'<br/><br/><br/><br/>';
if (!mysql_query($sql,$conn))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
?>
我已经高兴了包含原型ajax代码的JS代码,然后是包含插入细节的php代码
答案 0 :(得分:2)
您发送GET
请求,但尝试使用php从$_POST
数组中获取值。那么在sql中你得到'VALUES (,,,, ,,,,,,,)'
并且它是错误原因。