HTML multiple select dropdown not returning multiple values in php,returnong only the last value instead of multiple

Question

I have a multiple select dropdown field in html which has to return multiple values but it is returning only single value.

 <select name="select[]"  multiple>
 <OPTGROUP label="cars">
 <option value="car1">car1</option>
 <option value="car2">car2</option>
 <option value="car3">car3</option>
 </OPTGROUP>
 <OPTGROUP label="bikes">
 <option value="bike1">bike1</option>
 <option value="bike2">bike2</option>
 <option value="bike3">bike3</option>

Below is the PHP to return the multiple values that are selected:

 $select = mysqli_real_escape_string($link, $_POST['select']);

Thanks in advance.

Check this link for more help... stackoverflow.com/questions/2407284/… — Gagan Upadhyay
– Gagan Upadhyay, Commented Feb 3, 2016 at 6:09
$_POST['select'] is an array. You can't do mysqli_real_escape_string() on an array, as it expects param 2 to be a string. You need to loop over each value to do escape, without overwriting the previous value. — Sean
– Sean, Commented Feb 3, 2016 at 6:23

Ravi Hirani · Accepted Answer · 2016-02-03 06:25:56Z

1

Your code looks ok. Write select tag as below:-

<select name="select[]" multiple="multiple">

Now print it:-

    if(isset($_POST['select'])){
        $selected = $_POST['select']; 
        // Here $selected is an array. So need a foreach loop            
        foreach ($selected as $option)
        {
             print "You are selected $option<br/>";
             //print mysqli_real_escape_string($link, $option);
        }
    }

edited Feb 3, 2016 at 6:25

answered Feb 3, 2016 at 6:12

Ravi Hirani

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4 Comments

user5358888 Over a year ago

when i use print or echo am able to get all the vlaues but when i assign mysqli_real_escape_string($link, $option); to $example to insert in db i am getting only one value.

Ravi Hirani Over a year ago

You're welcome :) You need an array to store all $option. Try this code :- $data[] = mysqli_real_escape_string($link, $option); and print_r($data) at end.

user5358888 Over a year ago

if(isset($_POST['select'])){ $selected = $_POST['select']; // Here $selected is an array. So need a foreach loop foreach ($selected as $option) { $data[] = mysqli_real_escape_string($link, $option); } } i have inserted this part of code but when iam inserting $data into the db column am not getting all the values.Please help in this

Ravi Hirani Over a year ago

You cannot store array in DB. So you need to use implode method. Write $result = implode(',',$data); Now store $result in DB

shyam sasi · Accepted Answer · 2016-02-03 06:15:20Z

0

there is two things you have to do one first check your form method if it is not POST then you can't call it like $_POST you have to use $_GET or $_REQUEST and change select[] to select

answered Feb 3, 2016 at 6:15

shyam sasi

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Comments

Junaid · Accepted Answer · 2016-02-03 06:18:21Z

0

<select name="select[]" multiple="multiple">

you can check it by

print_r($_POST['select']);

answered Feb 3, 2016 at 6:18

Junaid

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Comments

urmila manjarikar · Accepted Answer · 2016-02-23 11:51:03Z

0

<select name="select[]" multiple="multiple">
if(isset($_POST['select'])){
    $selected = $_POST['select'];
    $select = implode(",",$selected);
}

use implode method, to insert $select values in database ....

answered Feb 23, 2016 at 11:51

urmila manjarikar

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