向MySQL提交复选框时出错 [英] Error when submitting checkbox to MySQL

问题描述

我有以下php代码提交表单到数据库唯一的问题是与复选框...提交表单上显示

I have the following php code for submitting the form to the database the only problem is with the checkboxes ... on submitting the form this shows up

警告：join（）[function.join]：在第16行的/srv/disk6/1662822/www/website.co.nf/connect-mysql.php中传递的参数无效：

<?php
    $host="host.com" ;
    $username="1662822_db1" ;
    $password="awesomepassword" ;
    $db_name="1662822_db1" ;
    $tbl_name="courses" ;
        $dbcon = mysqli_connect("$host","$username","$password","$db_name") ;           

        if (!$dbcon) {
        die('error connecting to database'); }

        echo 'Courses successfully registerd , ' ;  

    // escape variables for security
    $studentid = mysqli_real_escape_string($dbcon, $_POST['studentid']); echo $studentid;
    **$ckb = join (', ', $_POST['ckb']);** 


    **$sql="INSERT INTO courses (studentid, ckb)
    VALUES ('$studentid', '$ckb')";**

    if (!mysqli_query($dbcon,$sql)) {
    die('Error: ' . mysqli_error($dbcon));
}
echo "  Thank you for using IME Virtual Registeration  ";   
        mysqli_close($dbcon);
?>

错误是
警告： join .join]：在第16行的/srv/disk6/1662822/www/website.com/connect-mysql.php中传递的参数无效

The error is Warning: join() [function.join]: Invalid arguments passed in /srv/disk6/1662822/www/website.com/connect-mysql.php on line 16

我理解它的事情（显然），但我不明白是什么...

I understand its something to do with join function (obviously) but I don't understand what it is...

复选框的HTML代码

<input type="checkbox" name="ckb" value="strenthofmaterials";>
<label for="StrengthofMaterials"> Strength Of Materials </label>
<input type="checkbox" name="ckb" value="dynamics";>
<label for="StrengthofMaterials"> dynamics </label>

对所有其他选项进行更改，只更改每个复选框的值

it goes on for all other choices with only changing the value of each checkbox

另一条信息，mysql数据库中的ckb字段类型在tinyint中，默认值为0 ...我猜它的不是我正在寻找的类型..？ / p>

another piece of information , the ckb field in mysql database type in tinyint with a default value of 0 ...and I'm guessing its not the type I'm looking for ..?

推荐答案

$ckb = array();
foreach($_POST['checkbox'] as $val){
    $ckb[] = (int) $val;
}
$ckb = implode(',', $ckb);

尝试这个。 $ ckb应该是一个数组。
为了安全起见，$ val被转换为整数。

Try this one. $ckb should be an array. For security purpose $val is converted in integer.

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