【发布时间】:2014-05-25 19:23:49
【问题描述】:
大家下午好。运行以下 INSERT INTO 语句时出现以下错误。
错误:您的 SQL 语法有错误;检查手册 对应于您的 MySQL 服务器版本,以便使用正确的语法 在第 2 行的 ' , , , , , , , , , )' 附近
我完全被难住了。这可能与我在表中名为“id”的增量值有关吗?这是代码。非常感谢提前!
<?php
$con=mysqli_connect("localhost","root","root","wexdemo");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name = mysqli_real_escape_string($_POST['name']);
$address = mysqli_real_escape_string($_POST['address']);
$area = mysqli_real_escape_string($_POST['area']);
$postcode = mysqli_real_escape_string($_POST['postcode']);
$sector = mysqli_real_escape_string($_POST['sector']);
$subsector = mysqli_real_escape_string($_POST['subsector']);
$contact = mysqli_real_escape_string($_POST['contact']);
$position = mysqli_real_escape_string($_POST['position']);
$email = mysqli_real_escape_string($_POST['email']);
$telephone = mysqli_real_escape_string($_POST['telephone']);
$sql="INSERT INTO employers (name, address, area, postcode, sector, subsector, contact, position, email, telephone)
VALUES ($name, $address, $area, $postcode, $sector, $subsector, $contact, $position, $email, $telephone)";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
【问题讨论】:
-
总是尝试调试你的代码,你可以简单地打印你的查询并检查正在执行的实际查询是什么
-
了解prepared statements,真的很聪明!
-
@PhilHowell:如果与
mysqli_prepare合作,那将是最好的