How to insert the string data from PHP into MySql table

Question

How to insert these string data from PHP into MySQL table, I can get the code running but it not inserting into database.

The php code:

<?php

//mysql connection
$connect_error = 'Sorry, the database server is down. Please try again later.';
mysql_connect('xxxxxxxxx.com', 'xxxxxxx', 'xxxxxxxxx') or die($connect_error);
mysql_select_db('checkbox_test') or die($connect_error);

if (isset($_POST['submit'])) {
    if (isset($_POST['subject'])) {
        $strsubject = implode(",", ($_POST['subject']));
    } else {
        $strsubject = "";
    }

    echo $strsubject;
}

mysql_query("INSERT INTO levels (level_1, level_2, level_3, level_4, level_5, level_6) VALUES ($strsubject");
?>

The html:

<form action="" method="post" name="form_subject" id="form_subject">

    <input type="checkbox" name="subject[]" value="level_1"><label>Level 1</label><p>    
    <input type="checkbox" name="subject[]" value="level_2"></label>Level 2</label><p>  
    <input type="checkbox" name="subject[]" value="level_3"></label>Level 3</label><p>   
    <input type="checkbox" name="subject[]" value="level_4"><label>Level 4</label><p>    
    <input type="checkbox" name="subject[]" value="level_5"></label>Level 5</label><p>       
    <input type="checkbox" name="subject[]" value="level_6"></label>Level 6</label><p>

    <input type="submit" action= "" name="submit"  value="Submit" />

</form>

The database table has 7 columns, 1 for id (set to A_I) and the rest named for each level. I'm new to web developing, thanks guys

[How to include a PHP variable inside a MySQL insert statement](http://stackoverflow.com/a/7537500/285587) — Your Common Sense, Jul 31 '13 at 13:00
@igasparetto this tutorial was outdated already ten years ago. — Your Common Sense, Jul 31 '13 at 13:06
possible duplicate of [How to include a PHP variable inside a MySQL insert statement](http://stackoverflow.com/questions/7537377/how-to-include-a-php-variable-inside-a-mysql-insert-statement) — Fabio Antunes, Jul 31 '13 at 13:18

score 0 · Answer 1 · answered Jul 31 '13 at 13:04

You're missing a closing parenthesis in your INSERT statement, right after $strsubject and before the ":

mysql_query("INSERT INTO levels (level_1, level_2, level_3, level_4, level_5, level_6) 
                    VALUES ($strsubject")

This may not be the only problem.

score -1 · Answer 2 · answered Jul 31 '13 at 13:43

In PHP, like in JS, C++ and Java, one does not simply put a variable into another. Use the right concat character .:

.

$what      = "lazr!";
$variable1 = "firin " . $what;

$sql = "....SET column1='" . $data1 . "')";

.

Don't use mysql_* functions.
Use prepared statements within PDO: http://php.net/manual/de/pdo.prepare.php

.

<?php
/* Execute a prepared statement by passing an array of values */
$sth = $dbh->prepare('SELECT name, colour, calories
    FROM fruit
    WHERE calories < ? AND colour = ?');
$sth->execute(array(150, 'red'));
$red = $sth->fetchAll();
$sth->execute(array(175, 'yellow'));
$yellow = $sth->fetchAll();

$strsubject = implode(",", ($_POST['subject'])); should be:

.

$strsubject = "'" . implode("', '", ($_POST['subject'])) . "'";

My brain explodes when I try to comprehend this answer. What does Java to do here? What is "to put one variable into another" thing? What do these fruits and calories to do here? Why does it came the to the same dangerous code at the end? — Your Common Sense, Jul 31 '13 at 13:57

How to insert the string data from PHP into MySql table

2 Answers2