使用php自定义函数将值插入mysql

我想在MySQL数据库中插入值，但这些值没有插入，我不知道出了什么问题

这是我的密码：

<?php


include("inc/connection.php");

function RegisterStep1() {

        $txtuser = $_POST['txtName'];
        $txtsurname = $_POST['txtSurname'];
        $txtemail = $_POST['txtEmail'];
        $txtpass = $_POST['txtPassword'];
        $txtconfirmpass = $_POST['txtConfirmPass'];
        $txtcontactperson = $_POST['txtcompanycontactname'];
        $txtCompAddress = $_POST['txtCompAddress'];
        $txtRegNo = $_POST['txtRegNo'];
        $txtuserpos = $_POST['txtuserpos'];
        $txtdepartment = $_POST['txtdepartment'];
        $txtcontacts = $_POST['txtcontacts'];
        /* check if values are posted */



        $txtuser = mysql_real_escape_string($_POST['txtName']);
        /* combo company type */
        $txtCompAddress = mysql_real_escape_string($_POST['txtCompAddress']);
        /* combo locations
         */
        $txtRegNo = mysql_real_escape_string($_POST['txtRegNo']);
        /* combo companysize */
        $txtcontactperson = mysql_real_escape_string($_POST['txtcompanycontactname']);
        $txtsurname = mysql_real_escape_string($_POST['txtSurname']);
        $txtuserpos = mysql_real_escape_string($_POST['txtuserpos']);

        $txtdepartment = mysql_real_escape_string($_POST['txtdepartment']);
        $txtcontacts = mysql_real_escape_string($_POST['txtcontacts']);
        $txtemail = mysql_real_escape_string($_POST['txtEmail']);
        $txtpass = mysql_real_escape_string($_POST['txtPassword']);
        $txtconfirmpass = mysql_real_escape_string($_POST['txtConfirmPass']);



        $q = "INSERT INTO company(Name,Type,Address,Location,RegisteredNumber,CompanySize,ContactPerson,Surname,Position,Department,Contacts,DateResgistered,AccountStatus,Email,Password)
      Values('" . $txtuser . "','" . $txtCompAddress . "','" . $txtRegNo . "','" . $txtcontactperson . "','" . $txtsurname . "','" . $txtuserpos . "','" . $txtRegNo . "','" . $txtdepartment . "','" . $txtcontacts . "','" . $txtemail . "','" . $txtpass . "')";

        $submitquery = mysql_query($q);
        if ($submitquery) {
            echo"<div id='results'>Error occured while creating account,please try again in few minutes</div>";
        } else {
            echo "<div id='results'>Thanks for signing up</div>";
        }

}

?>

---

有件有趣的事，请检查一下你的情况

$submitquery = mysql_query($q);
        if ($submitquery) {
            echo"<div id='results'>Error occured while creating account,please try again in few minutes</div>";
        } else {
            echo "<div id='results'>Thanks for signing up</div>";
        }

$submitquery=mysql\u查询（$q）；
如果（$submitquery）{
echo“创建帐户时出错，请几分钟后重试”；
}否则{
echo“感谢注册”；
}

应该是

$submitquery = mysql_query($q);
            if (!$submitquery) {
                echo"<div id='results'>Error occured while creating account,please try again in few minutes</div>";
            } else {
                echo "<div id='results'>Thanks for signing up</div>";
            }

$submitquery=mysql\u查询（$q）；
如果（！$submitquery）{
echo“创建帐户时出错，请几分钟后重试”；
}否则{
echo“感谢注册”；
}

---

我认为您的代码运行良好。：）

您得到的输出是什么？最后是否出现了错误？请使用

echo mysql\u Error（）

找出您遇到的错误。发布更少的代码。将您的代码减少到仍然产生错误的最小样本，并发布剩下的内容。同时，你自己也可能会发现这个问题。这是调试，在要求堆栈溢出为您执行此操作之前，您需要自己尝试。输出是创建帐户时出错的，请在几分钟后重试。我一直在尝试跟踪它，但没有得到任何粘合。让我尝试mysql\u Error（）以找出哪里出了问题。

mysql\u Error（）的输出是什么

+1和deceze感谢mysql_error（）问题出在查询本身，但我不知道如何调试它。它现在工作正常，谢谢：）。谢谢大家。如果遇到任何问题，我会再次发布