mysql将值插入表中？

Question

我正在尝试将我的注册表单数据插入到我的表中，但由于某种原因没有插入任何内容并且我没有收到错误。

有人可以告诉我我哪里出了问题，因为我对 php 和 MySQL 真的很陌生，在此先感谢。

我的 config.php 文件保存了我的连接：

<?php
$host="localhost";
$username="mark";
$password="password";
$db_name="hewden1";
$conn = mysql_connect("$host", "$username", "$password") or die("Could Not Connect to Server");
$db = mysql_select_db("$db_name")or die("Cannot Connect the Database"); 
?>

这是我的注册表单html：

<form name="test" action="validation/signup_process.php" method="post">
<input type="text" name="compname" class="login_form" placeholder="Company or Trading Name">
<br>
<input type="text" name="contactname" class="login_form" placeholder="Contact Name"><br><br><br>
<input type="text" name="emailaddress" class="login_form" placeholder="Email Address">
<br>
 <input type="text" name="password1" class="login_form" placeholder="Password">
 <input type="text" name="password2" class="login_form" placeholder="Confirm Password"><br>

 <input type="submit" name="submit" value="Register" class="buttons">
</form> 

我的 php/MySQL 代码：

<?php 
session_start();
include("config.php");
//retrieve our data from POST
$compname = $_POST['compname'];
$contactname = $_POST['contactname'];
$username = $_POST['emailaddress'];
$password1 = $_POST['password1'];
$password2 = $_POST['password2'];

if($password1 != $password2) {
$_SESSION['message2'] = '<div id="message_box2"><div class="boxclose" id="boxclose" onclick="this.parentNode.parentNode.removeChild(this.parentNode);"></div><h23>Ooops!</h23><p>The Password&#39;s did not match.</p> </div>';
header("location:..\sign-up.php");

}else{
if(strlen($username) > 30) {
$_SESSION['message2'] = '<div id="message_box2"><div class="boxclose" id="boxclose" onclick="this.parentNode.parentNode.removeChild(this.parentNode);"></div><h23>Ooops!</h23><p>The Username you have selected is incorrect.</p> </div>';
header("location:..\sign-up.php");

}else{

$hash = hash('sha256', $password1);

function createSalt(){
$text = md5(uniqid(rand(), true));
return substr($text, 0, 3); }
$salt = createSalt();
$password = hash('sha256', $salt . $hash);

$username = mysql_real_escape_string($username);
$query = "INSERT INTO supplier_pre_sign (contactname, company_name, supplier_email, password, date, user_type) VALUES ('$contactname','$compname','$username', '$salt', now(), 'visitor');" or die(mysql_error());

echo "eveything ok";

} }?>

Answer 1

您忘记执行查询。您需要执行查询才能插入数据库。

试试这个：

    <?php 
    session_start();
    include("config.php");
    //retrieve our data from POST
    $compname = $_POST['compname'];
    $contactname = $_POST['contactname'];
    $username = $_POST['emailaddress'];
    $password1 = $_POST['password1'];
    $password2 = $_POST['password2'];

    if($password1 != $password2) {
    $_SESSION['message2'] = '<div id="message_box2"><div class="boxclose" id="boxclose" onclick="this.parentNode.parentNode.removeChild(this.parentNode);"></div><h23>Ooops!</h23><p>The Password&#39;s did not match.</p> </div>';
    header("location:..\sign-up.php");

    }else{
    if(strlen($username) > 30) {
    $_SESSION['message2'] = '<div id="message_box2"><div class="boxclose" id="boxclose" onclick="this.parentNode.parentNode.removeChild(this.parentNode);"></div><h23>Ooops!</h23><p>The Username you have selected is incorrect.</p> </div>';
    header("location:..\sign-up.php");

    }else{

    $hash = hash('sha256', $password1);

    function createSalt(){
    $text = md5(uniqid(rand(), true));
    return substr($text, 0, 3); }
    $salt = createSalt();
    $password = hash('sha256', $salt . $hash);

    $username = mysql_real_escape_string($username);
    $query = "INSERT INTO supplier_pre_sign (contactname, company_name, supplier_email, password, date, user_type) VALUES ('$contactname','$compname','$username', '$salt', now(), 'visitor')";
    mysql_query($query); //You forgot to add this line
    echo "eveything ok";

    } }?>

Answer 2

使用这个

$host="localhost";
$username="mark";
$password="password";
$db_name="hewden1";
$conn = mysql_connect($host, $username, $password) or die("Could Not Connect to Server");
$db = mysql_select_db($db_name)or die("Cannot Connect the Database"); 

//插入查询

$query = "INSERT INTO supplier_pre_sign (contactname, company_name, supplier_email, password, date, user_type) VALUES ('$contactname','$compname','$username', '$salt', now(), 'visitor');" or die(mysql_error());
$res = mysql_query($query);

Answer 3

您的插入查询在一个函数中，我没有看到它被调用。
然后，您的$query 也不会被执行。
在您创建查询的行之后添加mysql_query($query);。

Answer 4

Hey Only 像这样更改查询

$query = "INSERT INTO supplier_pre_sign (contactname, company_name, supplier_email, password, date, user_type) VALUES ($contactname,$compname,$username, $salt, now(), 'visitor');" or die(mysql_error());

我认为它会正常工作

然后查看类似的答案

Inserting $variable or $_POST value into mysql table

Answer 5

试试这个：

<?php 
session_start();
include("config.php");
//retrieve our data from POST
$compname = $_POST['compname'];
$contactname = $_POST['contactname'];
$username = $_POST['emailaddress'];
$password1 = $_POST['password1'];
$password2 = $_POST['password2'];

if($password1 != $password2) {
$_SESSION['message2'] = '<div id="message_box2"><div class="boxclose" id="boxclose" onclick="this.parentNode.parentNode.removeChild(this.parentNode);"></div><h23>Ooops!</h23><p>The Password&#39;s did not match.</p> </div>';
header("location:..\sign-up.php");

}elseif($username != "something") {  //Write a code in the (). 
if(strlen($username) > 30) {
$_SESSION['message2'] = '<div id="message_box2"><div class="boxclose" id="boxclose" onclick="this.parentNode.parentNode.removeChild(this.parentNode);"></div><h23>Ooops!</h23><p>The Username you have selected is incorrect.</p> </div>';
header("location:..\sign-up.php");

}else{

$hash = hash('sha256', $password1);

function createSalt(){
$text = md5(uniqid(rand(), true));
return substr($text, 0, 3); }
$salt = createSalt();
$password = hash('sha256', $salt . $hash);

$username = mysql_real_escape_string($username);
$query = "INSERT INTO supplier_pre_sign (contactname, company_name, supplier_email, password, date, user_type) VALUES ($contactname,$compname,$username, $salt, now(), 'visitor');" or die(mysql_error());

$res = mysql_query($query);

if($res){echo "inserted";} else{ mysql_error($res) }

    echo "eveything ok";

} }?>