How to properly pass an array from a select form to another PHP page

Question

Let me explain further. I have a select form on my first php page (lets call this page first.php). I do have a submit button. I am catching the array on the second page (lets cal this page sec.php) with $_POST and then setting it to a PHP variable. However, I can not get it to print. Here's what my code/mark up looks like on first.php

<label>Product:</label>
    <select name="arr[]">
        <option value="Mobile">Mobile</option>
        <option value="Social">Social</option>
        <option value="Online">Online</option>
    </select>

Note = I know I don't need to have an array for this. But I want to keep it this way.

Here's what my code looks like on sec.php:

<?php
    $arr= $_POST['arr'];
?>

I want it to print in this HTML table:

<tr>
    <td width="200"> <?php echo $url[0]; ?></td>
    <td width="200"> <?php echo $sMonth[0] . "/" . $sDay[0] . "/" . $sYear[0]; ?></td>
    <td width="200"> <?php echo $eMonth[0] . "/" . $eDay[0] . "/" . $eYear[0]; ?></td>
    <td> <?php echo $tBudget[0]; ?></td>
    <td> <?php echo $dBudget[0]; ?></td>
    <td> <?php echo $model[0]; ?></td>
    <td> <?php echo $bid[0]; ?></td>
    <td> <?php echo $target[0]; ?></td>
    <td> <?php echo $status[0]; ?></td>
    <td width="200"> <?php echo $arr[0]; ?></td>
    <td> <?php echo $tUnits[0]; ?></td>
</tr>

NOTE = all other variables and values print just FINE (by using the same method and concept as the $arr variable) But ONLY the $arr variable is unable to print).

What am I doing wrong?

What does var_dump($arr); show? How about var_dump($_POST);? — gen_Eric
– gen_Eric, Commented Dec 17, 2013 at 19:58
var_dump($arr) says this: dasarray(1) { [0]=> string(6) "Mobile" } — user3108505
– user3108505, Commented Dec 17, 2013 at 20:05

Leonardo · Accepted Answer · 2013-12-17 20:09:09Z

2

Change

<select name="arr[]">
...
<td width="200"> <?php echo $arr[0]; ?></td>

to

<select name="arr">
...
<td width="200"> <?php echo $arr; ?></td>

You'll only want to use array naming when you have a multiple select option

edited Dec 17, 2013 at 20:09

answered Dec 17, 2013 at 19:57

Leonardo

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10 Comments

user3108505 Over a year ago

Can you elaborate on that further? it's not multiple. User selects ONE option, that option will be put into the array.

Leonardo Over a year ago

@user3108505 exactly, since you only passing ONE value, you don't need to name your select arr[] and you only read it in your php like this: $arr instead of arr[0] since it's not an array. Read my updated answer.

user3108505 Over a year ago

My other variables are implemented the same way using the same concept and implementation. They all print except for arr

Leonardo Over a year ago

@user3108505 is your select into the <form> tags of your form?

user3108505 Over a year ago

Yes it is. So is my other select tags that I have used for my other variables/arrays.

|

zeropaper · Accepted Answer · 2013-12-17 20:12:04Z

0

Actually, it's probably something like that:

   <select name="arr[]" multiple>
      <option value="Mobile">Mobile</option>
      <option value="Social">Social</option>
      <option value="Online">Online</option>
   </select>

answered Dec 17, 2013 at 20:12

zeropaper

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1 Comment

user3108505 Over a year ago

but having "multiple" would cause all 3 values to be inserted. I don't want that. I want the selected option to be inserted into the first elemental position only.

Ahmed Magdy · Accepted Answer · 2016-02-14 18:21:14Z

0

u need to use for loop to get values in your array check this

<form action="<?php echo @$_SERVER['PHP_SELF'];?>" method="POST"> 
            <select name="arr[]" multiple>
                <option>Radio</option>
                <option>TV</option>
                <option>Keyboard</option>
                <option>DVD Player</option>
                <option>Screen</option>
            </select>
            <input type="submit" value="GO!">
        </form>
<?php
$data=@$_POST['arr'];
$len = count($data); // getting length of ur array that u need to condition ur loop
for($y=0;$y<$len;$y++){
    echo "$data[$y]"."<br />";
}
?>

edited Feb 14, 2016 at 18:21

answered Feb 14, 2016 at 4:28

Ahmed Magdy

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1 Comment

Ahmed Magdy Over a year ago

if u wont use loop don't use multiple then , multiple let user choose many value that's why you make select arr[] as (array) , best way to print array values = looping , in you'r code you just print the first index of arr[0] lemme know why ?!!

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