My PHP Implode function does not work

Question

I'm trying to make this implode function work.

This is the form part, assume all the item has been selected.

<form method="post">
<select name="test1" multiple="multiple" id="test1">
   <option value="1">item1</option>
   <option value="2">item2</option>
   <option value="3">item3</option>
   <option value="4">item4</option>
   <option value="5">item5</option>
</select>
</form>

The PHP part

<?php
$var1 = array();
$joinedString = array();
$var1 = $_POST['test1'];
$joinedString = implode(',', $var1);
?>

But The echoing part doesn't work, it gives me error, and only displaying only the first array value.

<?php
$echo $joinedString[0];
$echo $joinedString[1];
$echo $joinedString[2];
$echo $joinedString[3];
$echo $joinedString[4];
?>

Thank You guys, I'm quite new in programming. I always forgot that the code executed line by line, and always confused with variable and values, and yes, in real world I am also a clumsy & insignificant person.

what are you trying to do? JOIN an array, or SPLIT a string? — Andreas Grapentin
– Andreas Grapentin, Commented Sep 5, 2012 at 5:39

Basith · Accepted Answer · 2012-09-05 05:53:58Z

3

<form method="post" action="sear.php">
 <select name="test1[]" multiple="multiple" id="test1">
   <option value="1">item1</option>
   <option value="2">item2</option>
   <option value="3">item3</option>
   <option value="4">item4</option>
   <option value="5">item5</option>
   <input type="submit" name="submit" value="submit" />
  </select>
</form>
<?php
   $var1 = array();
   $joinedString = array();
   $var1 = $_POST['test1'];
   $joinedString = implode(',', $var1);
   echo $joinedString;
?>

After getting the post values it definitely works.... Try it...

edited Sep 5, 2012 at 5:53

answered Sep 5, 2012 at 5:39

Basith

1,0757 silver badges22 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

The Power in the Right Hand Over a year ago

yes, It's working now, how insignificant I am. thank you sir. next stop, sql left join() function.

Vinay · Accepted Answer · 2012-09-05 05:45:24Z

2

Use

<select name="test1[]" multiple="multiple" id="test1">

In php file.

$var1 = isset($_POST['test1']) ? $_POST['test1']: 0 ;
print_r($var1); //gives array
foreach($var1 as $var) {
    echo $var;
}

answered Sep 5, 2012 at 5:45

Vinay

2,5944 gold badges27 silver badges35 bronze badges

Comments

Sebass van Boxel · Accepted Answer · 2012-09-30 17:41:26Z

2

Change

<select name="test1" multiple="multiple" id="test1">

to

<select name="test1[]" multiple="multiple" id="test1">

And it's already an array

$var1 = $_POST['test1'];
    $imploded = implode(",", $var1);
    echo $imploded;
    //FOR GETTING INDIVIDUAL ITEMS FROM array
    echo $var1[0];

edited Sep 30, 2012 at 17:41

Sebass van Boxel

2,6221 gold badge24 silver badges37 bronze badges

answered Sep 5, 2012 at 5:41

user1647981

2 Comments

The Power in the Right Hand Over a year ago

so i just can call it like this : $val=$_post['test1[0]'];?

The Power in the Right Hand Over a year ago

Thank You Madame, now I just need to add the foreach loop to identify array values.

Pramod Kumar Sharma · Accepted Answer · 2012-09-05 05:36:44Z

0

Implode is not getting any array. Its works on Array.

answered Sep 5, 2012 at 5:36

Pramod Kumar Sharma

8,0325 gold badges30 silver badges53 bronze badges

1 Comment

The Power in the Right Hand Over a year ago

so, what function on getting array of values?

alex · Accepted Answer · 2012-09-05 05:37:36Z

0

You are passing a string instead of an array. implode() turns an array into a string.

answered Sep 5, 2012 at 5:37

alex

492k205 gold badges890 silver badges992 bronze badges

Comments

nageeb · Accepted Answer · 2012-09-05 05:39:08Z

0

In your script, $_POST['test1'] will only contain the submitted value of the select box, not the entire set of values. Since $var1 contains only a string, implode() will error out.

answered Sep 5, 2012 at 5:39

nageeb

2,0421 gold badge13 silver badges26 bronze badges

Comments

Kichu · Accepted Answer · 2012-09-05 05:39:38Z

0

Just look up this:

http://www.tizag.com/phpT/php-string-implode.php

And $ is used with variable.

answered Sep 5, 2012 at 5:39

Kichu

3,27916 gold badges73 silver badges136 bronze badges

Comments

Sibu · Accepted Answer · 2012-09-05 05:47:06Z

0

You are trying to use implode your array using (,) but post array values does not contain values seperated by comma, therefore you will have to use foreach.

<?php
$var1 = array();
$joinedString = array();
$var1 = $_POST['test1'];
foreach($var1 as $values)
echo $values."<br/>";
?>

<form method="post">
<select name="test1[]" multiple="multiple" id="test1">
   <option value="1">item1</option>
   <option value="2">item2</option>
   <option value="3">item3</option>
   <option value="4">item4</option>
   <option value="5">item5</option>
</select>
<input type="submit" >
</form>

answered Sep 5, 2012 at 5:47

Sibu

4,6272 gold badges30 silver badges38 bronze badges

4 Comments

The Power in the Right Hand Over a year ago

OMG, very straightforward.. Thank You Sibu.

Sibu Over a year ago

@ThePowerintheRightHand i guess for getting selected values you don't have to follow such long procedure as others have mentioned :-)

The Power in the Right Hand Over a year ago

I hope not, lol. my brain having a meltdown right now.. probably need to rest for awhile to avoid making simple mistakes.

Sibu Over a year ago

@ThePowerintheRightHand just rest, it will come with practice

Collectives™ on Stack Overflow

My PHP Implode function does not work

8 Answers 8

1 Comment

Comments

2 Comments

1 Comment

Comments

Comments

Comments

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

8 Answers 8

1 Comment

Comments

2 Comments

1 Comment

Comments

Comments

Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related