CODEWITHSUNDEEP
php
The Power in the Right Hand
The Power in the Right Hand

Reputation: 15

My PHP Implode function does not work

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.

Upvotes: 0

Views: 4008

Answers (8)

user1647981
user1647981

Reputation:

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];

Upvotes: 2

Basith
Basith

Reputation: 1075

<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...

Upvotes: 3

Sibu
Sibu

Reputation: 4617

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>

Upvotes: 0

Vinay
Vinay

Reputation: 2594

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;
}

Upvotes: 2

Kichu
Kichu

Reputation: 3267

Just look up this:

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

And $ is used with variable.

Upvotes: 0

nageeb
nageeb

Reputation: 2042

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.

Upvotes: 0

alex
alex

Reputation: 490253

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

Upvotes: 0

Pramod Kumar Sharma
Pramod Kumar Sharma

Reputation: 8012

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

Upvotes: 0

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