Question

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
  <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
      <head>
          <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
          <title>Aliens Abducted Me - Report an Abduction</title>
      </head>
      <body>
      <h2>Aliens Abducted Me - Report an Abduction</h2>

          <?php
              $firstname = $_POST['firstname'] ;
              $lastname= $_POST['lastname'];
              $when_it_happened = $_POST['whenithappened'];
              $how_long = $_POST['howlong'];
              $how_many = $_POST['howmany'];
              $alien_description = $_POST['howlook'];
              $what_they_did = $_POST['whattheydid'];
              $fang_spotted = $_POST['fangspotted'];
              $email = $_POST['email'];
              $other = $_POST['anythingelse'];

                /* $to = "random@gmail.com";

                $subject = "Aliens Abducted Me - Abduction Report";
                $msg ="$name was abducted $when_it_happened and was gone for $how_long.\n" .
                    "Number of aliens: $how_many\n" .
                    "Alien description: $alien_description\n" .
                    "What they did: $what_they_did\n" .
                    "Fang spotted: $fang_spotted\n" .
                    "Other comments: $other";
                 mail($to, $subject, $msg, 'From:' . $email); */


             $connect=mysqli_connect("localhost","gamenewton","random","aliensdata") or die( 'THE MYSQL CONNECTION FAILED TO CONNECT');
             $query="INSERT INTO alien_form "."(firstname,lastname,email,whenit,whattheydid,howlong,fang_spotted,anythingelse) ".
                 "VALUES('$firstname','$lastname','$email','$when_it_happened','$what_they_did,'$how_long','$fang_spotted','$other')" or die('Error in wirting query');

             $result= mysqli_query($connect, $query) or die('Error in contacting the database for query');

此错误发生在此消息上方的行

             mysqli_close($connect);

            echo 'Thanks for submitting the form.<br />';
            echo 'You were abducted ' . $when_it_happened;
            echo ' and were gone for ' . $how_long . '<br />';
            echo 'Number of aliens: ' . $how_many . '<br />';
            echo 'Describe them: ' . $alien_description . '<br />';
            echo 'The aliens did this: ' . $what_they_did . '<br />';
            echo 'Was Fang there? ' . $fang_spotted . '<br />';
            echo 'Other comments: ' . $other . '<br />';
            echo 'Your email address is ' . $email;
         ?>

我正在使用wampp处理php并收到此错误，因为我已经添加了die（）以了解此错误发生的位置。

Answer 1

在您的SQL插入中，您在what_they_did

之后缺少单引号

$query="INSERT INTO alien_form "."(...) ".
    "VALUES(...,'$what_they_did,'$how_long',...)"

mysqli_query（）中有MYSQL错误的PHP？我通过评论指出了错误的地方

1 个答案: