试图保存用户信息，错误
试图保存用户信息，错误
提问于 2017-08-23 06:49:45
我试图保存存储在SQL中的信息，但是这个错误不断出现：“错误保存数据。您的MySQL语法出现了错误；检查与您的MySQL服务器版本对应的手册，以获得正确的语法，以便使用”company= 'GlobalTop Inc.‘；在第6行“”中的regid = 1’其中，什么似乎是错误？
以下是完整的代码：
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<?php
include "db.php";
$gresult = ''; //declare global variable
//Start of edit contact read
if(isset($_POST["action"]) and $_POST["action"]=="edit"){
$id = (isset($_POST["ci"])? $_POST["ci"] : '');
$sql = "select regid, regname,
address, phone,
email,company from tblregistrants
where regid = $id";
$result = mysqli_query($link, $sql);
if(!$result)
{
echo mysqli_error($link);
exit();
}
$gresult = mysqli_fetch_array($result);
include 'update.php';
exit();
}
//Insert or Update contact information
if(isset($_POST['action_type']))
{
if ($_POST['action_type'] == 'add' or $_POST['action_type'] == 'edit')
{
//Sanitize the data and assign to variables
$regid = mysqli_real_escape_string($link, strip_tags($_POST['regid']));
$regname = mysqli_real_escape_string($link, strip_tags($_POST['regname']));
$phone = mysqli_real_escape_string($link, strip_tags($_POST['phone']));
$address = mysqli_real_escape_string($link, strip_tags($_POST['address']));
$email = mysqli_real_escape_string($link, strip_tags($_POST['email']));
$company = mysqli_real_escape_string($link, strip_tags($_POST['company']));
if ($_POST['action_type'] == 'add')
{
$sql = "insert into tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email'
company = '$company'";
}else{
$sql = "update tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email'
company = '$company'
where regid = $regid";
}
if (!mysqli_query($link, $sql))
{
echo 'Error Saving Data. ' . mysqli_error($link);
exit();
}
}
header('Location: view.php');
exit();
}
//Read registrants information from database : Stage 1
$sql = "select * from tblregistrants";
$result = mysqli_query($link, $sql);
if(!$result)
{
echo mysqli_error($link);
exit();
}
//Loop through each row on array and store the data to $reg_list[] : Stage 2
while($rows = mysqli_fetch_array($result))
{
$reg_list[] = array('regid' => $rows['regid'],
'regname' => $rows['regname'],
'address' => $rows['address'],
'phone' => $rows['phone'],
'email' => $rows['email'],
'company' => $rows['company']);
}
include 'view.php';
exit();
?>回答 3
Stack Overflow用户
发布于 2017-08-23 06:56:40
在,之后的if语句和else语句中，您都错过了email = '$email'
if ($_POST['action_type'] == 'add')
{
$sql = "insert into tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email',
company = '$company'";
}else{
$sql = "update tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email',
company = '$company'
where regid = $regid";
}还可以使用准备语句防止SQL注入。
Stack Overflow用户
发布于 2017-08-23 06:59:48
就像劳伦斯说你失踪了一样，在你的查询中
试试这个：
$sql = "insert into tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email',
company = '$company'";Stack Overflow用户
发布于 2017-08-23 06:58:34
改变这个，
$sql = "update tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email',
company = '$company'
where regid = $regid";到这个
$sql = "update tblregistrants set
name = '$regname',
phone = '$phone',
address = '$address',
email = '$email',
company = '$company'
where regid = '$regid'";页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/45832529
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