PHP Multiple Dropdown Box Form Submit To MySQL (Part 2)

Question

This is a continuation of the discussion at

PHP Multiple Dropdown Box Form Submit To MySQL

which ended with the words: "Once you have the variables, it is trivial to create new rows." No doubt that's generally true, but apparently not for this learner... :-D

Given the following form:

    <form action="form.php" method="POST">
    <select name="colors[]" multiple="yes" size="2">
    <option>Red</option>
    <option>Blue</option>
    </select>
    <input type="submit" value="Go!"> 
    </form>

how do I create new rows? The following script

foreach($_POST['colors[]'] as $color) 
    {
        $id = mysqli_real_escape_string($link, $color);
        $sql = "INSERT INTO colors SET id = '$id'";
    }

raises the error

 Warning: Invalid argument supplied for foreach() in form.php on line ...

whereas the following

    $colors = $_POST['colors[]']; 
    for ($i = 0; $i < count($colors); $i++) 
        {
            $color = $colors[$i];
            $sql = "INSERT INTO colors SET id = '$color'";
        }

raises no errors but does no row creation.

What triviality am I missing here?

Sarfraz · Accepted Answer · 2010-02-08 16:06:51Z

2

Use:

foreach($_POST['colors'] as $color)

No need to specify [] here, php knows that it is an array.

answered Feb 8, 2010 at 16:06

Sarfraz

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5 Comments

PeterC Over a year ago

Yes, however I'm still not getting the database to create rows. I created a table like so: CREATE TABLE colors ( id INTEGER NOT NULL AUTO_INCREMENT PRIMARY KEY, color TEXT ); --- I can't see what the problem is...

Sarfraz Over a year ago

@Peterc: I could not understand your comment completely but why don't you use a sql client software such as sqlyog to create your table?

PeterC Over a year ago

@Sarfraz - good suggestion. I think I found the problem. Thanks again for your help!

Anthony Forloney Over a year ago

What was the problem? It helps for others who encounter the problem to know where you found the problem and an alternative solution for it.

PeterC Over a year ago

@Anthony - it was precisely what you said in your answer below - I had left out the mysql_query statement. Doh! :-D

Anthony Forloney · Accepted Answer · 2010-02-08 16:49:49Z

1

Inside of your foreach loop I did not see that you are executing the query. You need to execute mysql_query with your INSERT statement.

foreach($_POST['colors'] as $color) {
 $id = mysqli_real_escape_string($link, $color);
 $sql = "INSERT INTO colors SET id = '$id'";
  if (!mysql_query($sql)) {  // if the query encountered a problem.
     die('Invalid query: ' . mysql_error());
  } 
}

edited Feb 8, 2010 at 16:49

answered Feb 8, 2010 at 16:30

Anthony Forloney

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1 Comment

PeterC Over a year ago

@Anthony - That's correct - as I just discovered (to my chagrin)! Thanks for the help.

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