Trying to insert array into database... (implode() [function.implode]: Invalid arguments passed )

Question

I have a select input where you can choose multiple options. Unfortunately, I have no clue how to insert the values selected into my database. Here is the code I am trying to use, which is giving me an invalid argument passed on my attempt to implode it.

Thanks for any help...

Here is my php:

  $distribs = implode("|", $_POST["distributors"]); 
$distribs  = mysql_real_escape_string($distribs ); 


$distribs = mysql_real_escape_string($_POST['distributors']);
$sql="UPDATE customers SET distributors = '$distribs'
                                            WHERE id='$id'";

Here is my select:

  echo "<select name='distributors' multiple='multiple'>";

what datatype you are using for distributors in DB as implode return strings ?? — swapnesh
– swapnesh, Commented Dec 6, 2012 at 3:56
@alex please enlighten me. my select name is supplosed to be distributors[] rather than just distributors, correct? — Alex
– Alex, Commented Dec 6, 2012 at 3:59
are you getting the selected value in print_r($_POST['distributors']) — DeDav
– DeDav, Commented Dec 6, 2012 at 3:59

DeDav · Accepted Answer · 2012-12-06 04:01:14Z

1

try like this

echo "<select name='distributors[]' multiple='multiple'>";

answered Dec 6, 2012 at 4:01

DeDav

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1 Comment

Alex Over a year ago

Okay, this is what I had at one point, but for whatever reason, it wasn't working earlier. Now it works! Go figure...I must have had my code wrong elsewhere at the time. Thanks!

Community · Accepted Answer · 2023-11-17 20:24:57Z

1

$_POST["distributors"]; is not an array, implode(); require an array to function,

PHP.net Function for implode();

If your $_POST is a correct array format; then implode will convert the array into a single string.

If your $_POST contains information from a <select> </select> Then the submission would be a single string which can be inserted into a database without the need for your implode(); function.

if you are certain that your $_POST is being submitted in an array format. Use as DeDav Has suggested.

Run this command when your $_POST is populated.

print_r($_POST['distributors'])

If your using a multiple select. Use as again DeDav suggested:

echo "<select name='distributors[]' multiple='multiple'>";

edited Nov 17, 2023 at 20:24

CommunityBot

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answered Dec 6, 2012 at 4:02

Daryl Gill

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2 Comments

Mayank K. Swami Over a year ago

i suspect explode doesnot require an array. here explode — Split a string by string no mention of array

Daryl Gill Over a year ago

Simple mistake; shouldn't have mentioned explode(); This has now been removed from my post.

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