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php - 无法将新用户插入 mysql 数据库

转载 作者:行者123 更新时间:2023-11-29 01:23:17 25 4
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我有一个表单,允许我将用户输入到我的数据库中。但是,每当我单击提交时,我都会收到 query failed 错误消息。以下是我构建的表单:

注册-admin.php

<form id="resgisterform" name="registerform" method="post" action="register-admin-exec.php">
<table width="300" border="0" align="center" cellpadding="2" cellspacing="0">
<tr>
<th>Username </th>
<td><input name="username" type="text" class="textfield" id="username" /></td>
</tr>
<tr>
<th>First Name </th>
<td><input name="first_name" type="text" class="textfield" id="first_name" /></td>
</tr>
<tr>
<th>Last Name </th>
<td><input name="last_name" type="text" class="textfield" id="last_name" /></td>
</tr>
<tr>
<th>Muvdigital Email </th>
<td><input name="muvdigital_email" type="text" class="textfield" id="muvdigital_email" /></td>
</tr>
<tr>
<th>Personal Email </th>
<td><input name="personal_email" type="text" class="textfield" id="personal_email" /></td>
</tr>
<tr>
<th>Title </th>
<td><input name="title" type="text" class="textfield" id="title" /></td>
</tr>
<tr>
<th>Address 1 </th>
<td><input name="address_1" type="text" class="textfield" id="address_1" /></td>
</tr>
<tr>
<th>Address 2 </th>
<td><input name="address_2" type="text" class="textfield" id="address_2" /></td>
</tr>
<tr>
<th>City </th>
<td><input name="city" type="text" class="textfield" id="city" /></td>
</tr>
<tr>
<th>State </th>
<td><input name="state" type="text" class="textfield" id="state" /></td>
</tr>
<tr>
<th>Zip Code </th>
<td><input name="zip" type="text" class="textfield" id="zip" /></td>
</tr>
<tr>
<th>Phone </th>
<td><input name="phone" type="text" class="textfield" id="phone" /></td>
</tr>
<tr>
<th>Password </th>
<td><input name="password" type="password" class="textfield" id="password" /></td>
</tr>
<tr>
<th>Confirm Password </th>
<td><input name="cpassword" type="password" class="textfield" id="cpassword" /></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input type="submit" name="Submit" value="Register" /></td>
</tr>
</table>
</form>

然后将此表单中的值带到下面的 register-admin-exec.php 页面

<?php
//Start session
session_start();

//Include database connection details
require_once('config.php');

//Array to store validation errors
$errmsg_arr = array();

//Validation error flag
$errflag = false;



//Function to sanitize values received from the form. Prevents SQL injection
function clean($str) {
$str = @trim($str);
if(get_magic_quotes_gpc()) {
$str = stripslashes($str);
}
return mysql_real_escape_string($str);
}

//define and validate the post values
/*if (isset ($_POST['admin_id']) && !empty ($_POST['admin_id'])) {
$admin_id = $_POST['admin_id'];
} else {
echo 'Error: admin id not provided!';
}*/

if (isset ($_POST['username']) && !empty ($_POST['username'])) {
$username = clean($_POST['username']);
} else {
echo 'Error: username not provided!';
}

if (isset ($_POST['first_name']) && !empty ($_POST['first_name'])) {
$first_name = clean($_POST['first_name']);
} else {
echo 'Error: first name not provided!';
}

if (isset ($_POST['last_name']) && !empty ($_POST['last_name'])) {
$last_name = clean($_POST['last_name']);
} else {
echo 'Error: last name not provided!';
}

if (isset ($_POST['muvdigital_email']) && !empty ($_POST['muvdigital_email'])) {
$muvdigital_email = clean($_POST['muvdigital_email']);
} else {
echo 'Error: muvdigital email not provided!';
}

if (isset ($_POST['personal_email']) && !empty ($_POST['personal_email'])) {
$personal_email = clean($_POST['personal_email']);
} else {
echo 'Error: personal email not provided!';
}

if (isset ($_POST['title']) && !empty ($_POST['title'])) {
$title = clean($_POST['title']);
} else {
echo 'Error: title not provided!';
}

if (isset ($_POST['phone']) && !empty ($_POST['phone'])) {
$phone = clean($_POST['phone']);
} else {
echo 'Error: phone not provided!';
}

if (isset ($_POST['address_1']) && !empty ($_POST['address_1'])) {
$address_1 = clean($_POST['address_1']);
} else {
echo 'Error: address 1 not provided!';
}

$address_2 = clean($_POST['address_2']);

if (isset ($_POST['city']) && !empty ($_POST['city'])) {
$city = clean($_POST['city']);
} else {
echo 'Error: city not provided!';
}

if (isset ($_POST['state']) && !empty ($_POST['state'])) {
$state = clean($_POST['state']);
} else {
echo 'Error: state not provided!';
}

if (isset ($_POST['zip']) && !empty ($_POST['zip'])) {
$zip = clean($_POST['zip']);
} else {
echo 'Error: zip not provided!';
}

if (isset ($_POST['password']) && !empty ($_POST['password'])) {
$password = clean($_POST['password']);
} else {
echo 'Error: password not provided!';
}

if (isset ($_POST['cpassword']) && !empty ($_POST['cpassword'])) {
$cpassword = clean($_POST['cpassword']);
} else {
echo 'Error: confirm password not provided!';
}



//encrypt the password
$salt = sha1($username);
$password = sha1($salt.$password);


//Check for duplicate login ID
if($username != '') {
$qry = "SELECT * FROM members WHERE username='".$username."'";
$result = mysql_query($qry);
if($result) {
if(mysql_num_rows($result) > 0) {
$errmsg_arr[] = 'Login ID already in use';
$errflag = true;
}
@mysql_free_result($result);
}
else {
die("Query failed");
}
}

//If there are input validations, redirect back to the registration form
if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
session_write_close();
header("location: register-admin.php");
exit();
}

//Create INSERT query
$qry = "INSERT INTO admins (
'username',
'password',
'first_name',
'last_name',
'muvdigital_email',
'personal_email',
'titles',
'phone',
'address_1',
'address_2',
'city',
'state',
'zip')
VALUES (
'$username',
'$password',
'$first_name',
'$last_name',
'$muvdigital_email',
'$personal_email',
'$title',
'$phone',
'$address_1',
'$address_2',
'$city',
'$state',
'$zip')";
$result = mysql_query($qry);

//Check whether the query was successful or not
if($result) {
header("location: register-success.php");
exit();
}else {
die("Query failed $qry");
}
?>

我知道它在我的插入语句中失败了,因为我已经尝试注释掉之前对重复登录 ID 的验证检查,但它仍然失败。我无法弄清楚为什么我的插入语句不起作用。在回显 $qry 之后,我得到了

INSERT INTO admins ( 'username', 'password', 'first_name', 'last_name', 'muvdigital_email', 'personal_email', 'titles', 'phone', 'address_1', 'address_2', 'city', 'state', 'zip') VALUES ( 'johndoe', '7afbb2186cf26d85bdfe948d367fb6baa6739283', 'john', 'doe', 'john.doe@muvdigital.com', 'jdoe@gmail.com', 'intern ', '6024013776', '18b main st', 'apt 12', 'Hooksett', 'NH', '03106')

因此 $_POST 函数正在运行。我尝试在命令行手动输入插入语句,但收到 ERROR 1054 (42S22): Unknown column 'johndoe' in 'field list'

admin_id 是一个 auto_increment 字段,这就是我将其注释掉的原因(我尝试取消注释并手动创建一个 admin_id,但仍然不起作用)

有人知道为什么会这样吗?

最佳答案

您用单引号引用了所有列名称,这是不正确的。它们应该不带引号,除非您使用了 MySQL reserved keyword。 (你没有)

// Column names are unquoted, but VALUES() should be quoted.
$qry = "INSERT INTO admins (
username,
password,
first_name,
last_name,
muvdigital_email,
personal_email,
titles,
phone,
address_1,
address_2,
city,
state,
zip)
VALUES (
'$username',
'$password',
'$first_name',
'$last_name',
'$muvdigital_email',
'$personal_email',
'$title',
'$phone',
'$address_1',
'$address_2',
'$city',
'$state',
'$zip')";
$result = mysql_query($qry);

关于php - 无法将新用户插入 mysql 数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11146448/

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