dongtiannan0367 2015-04-26 10:14
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表单写入mysql数据库

I'm currently working on writing a form to a database, still a beginner. I attempted a form that i will show below but it seems to give me errors. If anyone could help me find out what is wrong with my PHP code then it would be great. Thanks in advance:

My form.php

<form method="post" action="db.php" name="OverrideForm" id="OverrideForm" autocomplete="on">
    <fieldset>
        <legend>Contact Details</legend>
        <div>
            <label for="name" accesskey="N">First Name</label>
            <input name="name" type="text" id="name" required />
        </div>
        <div>
            <label for="mname" accesskey="M">Middle Name</label>
            <input name="mname" type="text" id="mname" required />
        </div>
        <div>
            <label for="fname" accesskey="F">Last Name</label>
            <input name="fname" type="text" id="fname" required />
        </div>
        <div>
            <label for="sid" accesskey="i">Student ID</label>
            <input name="sid" type="text" id="sid" size="10"  required />
        </div>
<div>
            <label for="email" accesskey="E">Email</label>
            <input name="email" type="email" id="email" pattern="^[A-Za-z0-9](([_\.\-]?[a-zA-Z0-9]+)*)@([A-Za-z0-9]+)(([\.\-]?[a-zA-Z0-9]+)*)\.([A-Za-z]{2,})$" required />
        </div>
        <div>
            <label for="phone" accesskey="p">Phone Number</label>
            <input name="phone" type="text" id="phone" size="8"  required />
        </div>
<div>
            <label for="sc" accesskey="s">Scolarship</label>
            <select name="sc" id="sc" required="required">
                <option value="0">Yes</option>
                <option value="1">No</option>
            </select>
        </div>            
   </fieldset>

    <fieldset>
        <legend>Subject Details</legend>
<div>
            <label for="class" accesskey="c">Class</label>
            <input name="class" type="text" id="class" size="50" required />
        </div>
        <div>
            <label for="section" accesskey="o">Section</label>
            <input name="section" type="text" id="section" size="1" required />
        </div>
        <div>
            <label for="semester" accesskey="S">Semester</label>
            <select name="semester" id="semester" required="required">
                <option value="F15">Fall 2015</option>
                <option value="S15">Summer 2015</option>
                <option value="SP16">Spring 2016</option>
            </select>
        </div>
    </fieldset>

    <input type="submit" class="submit" id="submit" value="Submit" />
</form>

my db.php

    <?php
$mysql_host     = "localhost";
$mysql_username = "blahblah";
$mysql_password = "blahblah";
$mysql_database = "blahblah";

$mysqli  = new Mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database) or die(mysql_error());
$prepare = $mysqli->prepare("INSERT INTO `Overrides`(`name`,`mname`,`fname`,`sid`,`email`,`phone`,`sc`,`class`,`section`,`semester`) VALUES ('$name','$mname','$fname','$sid','$email','$phone','$sc','$class','$section','$semester')");
$prepare->bind_param("ssssssssss", $_POST['name'], $_POST['mname'], $_POST['fname'], $_POST['sid'], $_POST['email'], $_POST['phone'], $_POST['sc'], $_POST['class'], $_POST['section'], $_POST['semester']);
$prepare->execute();
print_r($_POST)
?>

Error i'm getting is:

Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement in /home/aukwizcq/public_html/db.php on line 9
Array ( [name] => aaa [mname] => aaa [fname] => aaaa [sid] => 123456 [email] => fgfg@hotmail.com [phone] => 45454 [sc] => 1 [class] => Cpeg 340 [section] => 1 [semester] => S15 )

My db structure:

Name Type Collation Attributes Null Default Extra Action

1   name    varchar(30) latin1_swedish_ci       No  None        Change 
2   mname   varchar(30) latin1_swedish_ci       No  None        Change 
3   fname   varchar(30) latin1_swedish_ci       No  None        Change 
4   sid varchar(11) latin1_swedish_ci       No  None        Change Change   
5   email   varchar(50) latin1_swedish_ci       No  None        Change 
6   phone   int(8)          No  None        Change Change   Drop Drop   
7   sc  bit(1)          No  None        Change Change   Drop Drop   
8   class   varchar(10) latin1_swedish_ci       No  None        Change 
9   section int(1)          No  None        Change Change   Drop Drop   
10  semester    varchar(11) latin1_swedish_ci       No  None        Change 
  • 写回答

1条回答 默认 最新

  • duanlingzei0170 2015-04-26 10:16
    关注

    The syntax of the prepared statement is not right, the variables should be question marks. It should be like this:

    $prepare = $mysqli->prepare("INSERT INTO `Overrides`(`name`,`mname`,`fname`,`sid`,`email`,`phone`,`sc`,`class`,`section`,`semester`) VALUES (?,?,?,?,?,?,?,?,?,?)");
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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