Question

无法理解为什么经过多次修修补补后，我们将不胜感激：

<?php
// Fetch the user row from the query above
$userData = mysqli_fetch_array($query, MYSQL_ASSOC);

$password = $userData["password"];

    if(isset($_POST['password2'])){
        $p2 = $_POST['password2'];
        }


    $password2 = sha1($p2);

// Ajax calls this REGISTRATION code to execute
if($password2 == $password){
// GATHER THE POSTED DATA INTO LOCAL VARIABLES
$personname = preg_replace('#[^a-z ]#i', '', $_POST['personname']);
$email = mysqli_real_escape_string($db_conx, $_POST['email']);
$address_line1 = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_line1']);
$address_line2 = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_line2']);
$address_city = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_city']);
$address_state = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_state']);
$address_zip = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_zip']);
$address_country = preg_replace('#[^a-z ]#i', '',   $_POST['address_country']);

////////电子邮件的重复数据检查

    $sql1 = "SELECT email FROM users WHERE id='$id' AND activated='1' LIMIT 1";

    if($email =! email){
        $sql = "SELECT id FROM users WHERE email='$email' LIMIT 1";
        $query = mysqli_query($db_conx, $sql); 
        $email_check = mysqli_num_rows($query);

        if ($email_check > 0){ 
        echo "That email address is already in use in the system";
        exit();
}

//////////表格数据错误处理

else if($personname == "" || $email == "" || $password == "" ||   $address_country == "" || $address_line1 == "" || $address_line2 == "" || $address_city == "" || $address_state == "" || $address_zip == ""){
    echo "The form submission is missing values.";
    exit();
}

    else {

//////// END FORM DATA ERROR HANDLING ////////开始将数据替换为数据库

//////将用户信息添加到主站点表的数据库表中

$query = "UPDATE users (personname, email, address_line1, address_line2, address_city, address_state, address_zip, address_country)
            VALUES('$personname','$email','$address_line1','$address_line2','$address_city','$address_state','$address_zip','$address_country')
            WHERE id = '$id'";

            if(mysql_query($query)){
                echo "updated";
                header("location: user.php?id=".$_SESSION["id"]);
                } else{ echo "fail";}

    }
exit();
}
?>

Answer 1

首先，}没有if($password2 == $password){的右括号$address_country...，它可能属于}之后。但是，您的支撑有点偏离，exit;之后的if($email =! email){应该与其他几个一起提升。例如if(mysql_query($query))的那个。

您还要将API与if(mysqli_query($db_conx, $query))混合，需要MYSQL_ASSOC和MYSQLI_ASSOC到if($email =! email){。那些不同的MySQL函数不会相互混合。另外，请确保您已启动会话，因为您正在使用会话。

然后，此行email - if($email != email){被视为constant，并且也是错误的。它应显示为$sql1。但是，我不知道您想要使用/检查哪个变量（或字符串），甚至不知道您表中的一行。

然后这个查询的变量email - 它处于不确定状态。我不知道你想怎么做。

如果您想要对＃34;电子邮件＆＃34;匹配一行，那么您需要这样做，从DB中获取行。

咨询：http://php.net/manual/en/mysqli-result.fetch-row.php

参见脚注

尝试以下内容，我已经重新安排了你的支撑：

旁注： 关于<?php // Fetch the user row from the query above $userData = mysqli_fetch_array($query, MYSQLI_ASSOC); $password = $userData["password"]; if(isset($_POST['password2'])){ $p2 = $_POST['password2']; } $password2 = sha1($p2); // Ajax calls this REGISTRATION code to execute if($password2 == $password){ // GATHER THE POSTED DATA INTO LOCAL VARIABLES $personname = preg_replace('#[^a-z ]#i', '', $_POST['personname']); $email = mysqli_real_escape_string($db_conx, $_POST['email']); $address_line1 = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_line1']); $address_line2 = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_line2']); $address_city = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_city']); $address_state = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_state']); $address_zip = preg_replace('#[^a-z0-9 ]#i', '', $_POST['address_zip']); $address_country = preg_replace('#[^a-z ]#i', '', $_POST['address_country']); } //////// DUPLICATE DATA CHECKS FOR EMAIL $sql1 = "SELECT email FROM users WHERE id='$id' AND activated='1' LIMIT 1"; // unsure what you want to do here if($email != email){ $sql = "SELECT id FROM users WHERE email='$email' LIMIT 1"; $query = mysqli_query($db_conx, $sql); $email_check = mysqli_num_rows($query); if ($email_check > 0){ echo "That email address is already in use in the system"; exit(); } } ////////// FORM DATA ERROR HANDLING else if($personname == "" || $email == "" || $password == "" || $address_country == "" || $address_line1 == "" || $address_line2 == "" || $address_city == "" || $address_state == "" || $address_zip == ""){ echo "The form submission is missing values."; exit(); } else { //////// END FORM DATA ERROR HANDLING //////// Begin Replacement of data into the database //////Add user info into the database table for the main site table $query = "UPDATE users (personname, email, address_line1, address_line2, address_city, address_state, address_zip, address_country) VALUES('$personname','$email','$address_line1','$address_line2','$address_city','$address_state','$address_zip','$address_country') WHERE id = '$id'"; if(mysqli_query($db_conx, $query)) echo "updated"; header("location: user.php?id=".$_SESSION["id"]); } else{ echo "fail";} } exit(); ?>是一个常数，请检查您对上述内容的看法。它在下面没有改变。

UPDATE users...

旁注：

您使用的UPDATE table SET column_x = '$var'并不是UPDATE的工作方式。

语法为UPDATE [LOW_ PRIORITY] [IGNORE] table_name [, table_name...] SET column_name1 = expr1 [, column_name2=expr2 ...] [WHERE condition]

阅读更新：

http://www.mysqltutorial.org/mysql-update-data.aspx

即。从该页面：

INSERT INTO

如果您不想更新，请使用or die(mysqli_error($db_conx))，我更倾向于此。

将mysqli_query()添加到$sql1 = "SELECT email FROM users WHERE id='$id' AND activated='1' LIMIT 1"; $result = mysqli_query($db_conx, $sql1); $row = mysqli_fetch_row($result); $email_row = $row['email'];，您将看到错误。

<强>脚注：

以下是获取行的示例：

if($email != $email_row){...}

然后你可以这样做：

{{1}}

由于某种原因，在最后一行解析错误？

1 个答案: