When I run the script below I get an "undefined offset 2" error in php

Question

if (isset($_POST['sub']))
{
    $grpm_phno = $_POST['grpm_phno'];
    $grmid = $_POST['groupid'];
    $grpm_name = $_POST['grpm_name'];
    $name = explode(',', "$grpm_name");
    $phone = explode(',', "$grpm_phno");
    $countt = count($name);
    for ($i = 0; $i <= $countt; $i++)
    {
        $x = $name[$i];
        $y = $phone[$i];
        $dt = date('Y-m-d h:i:s');

        // Insert Query of SQL

        $query = mysql_query("insert into grp_mst(grpm_name, grpm_phno, grpm_grpcatm_id,grpm_typ,grpm_crtdon) values ('$x', '$y', '$grmid', 'b', '$dt')");
    }
}

input given:

name:raj,mohan

number:61231,3618372

output:

raj :61231

mohan:3618372

and empty row any help

instead of `for` loop try `foreach` loop http://php.net/manual/en/control-structures.foreach.php — Pragnesh Chauhan, Dec 15 '17 at 06:44
`mysql_*` is deprecated as of [tag:php-5.5] and removed in [tag:php-7]. So instead use `mysqli_*` or `PDO`. https://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php/14110189#14110189 — mega6382, Dec 15 '17 at 06:44
Possible duplicate of [Why shouldn't I use mysql\_\* functions in PHP?](https://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) — Mathews Sunny, Dec 15 '17 at 06:49
@While youre perfectly right that mysql_* is deprecated and should not be used for more then two reasons this isnt a duplicate of that question you linked at all. — Fabian S., Dec 15 '17 at 07:10

Alive to die - Anant · Answer 1 · 2017-12-15T06:56:51.403

Why won't you use foreach()(as it take care of indexes itself):-

foreach($name as $key=>$value){
 $x = $value;
 $y = $phone[$key];
 $dt = date('Y-m-d h:i:s');

 $query = mysql_query("insert into grp_mst(grpm_name, grpm_phno, grpm_grpcatm_id,grpm_typ,grpm_crtdon) values ('$x', '$y', '$grmid', 'b', '$dt')");
}

Important notes:-

1.mysql_* library is deprecated in php-5.5 and removed in php-7. Turn towards mysqli_* OR PDO along with latest php 7 version.

2.Always use prepared statements of mysqli_* or PDO library to prevent Your code from SQL INJECTION.

REFERENCE:-

php mysqli prepare

php PDO prepare

score 1 · Answer 2 · answered Dec 15 '17 at 06:45

1

You have 2 items in your array, index 2 is the third one (and does not exists given your input)

$countt = count($name);
for ($i = 0; $i <= $countt; $i++)
/* => */
for ($i = 0; $i < $countt; $i++)

answered Dec 15 '17 at 06:45

Oulalahakabu

504
3
6

thanks Oulalahakabu , its working fine – jayaram Dec 15 '17 at 07:04
@jayaram it's not about working fine and go. Its about upgrading your knowledge and code as well. Please check once my answer and read notes section carefully – Alive to die - Anant Dec 15 '17 at 07:12

score 0 · Answer 3 · edited Dec 15 '17 at 06:45

0

$name = explode(',', "$grpm_name");
$phone = explode(',', "$grpm_phno");

rewrite it as

$name = explode(',', $grpm_name);
$phone = explode(',', $grpm_phno);

edited Dec 15 '17 at 06:45

mega6382

9,211
17
48
69

answered Dec 15 '17 at 06:45

Satish51

119
6

it getting Parse error: syntax error, unexpected '"' in C:\xampp\htdocs\ – jayaram Dec 15 '17 at 06:47
when u pass a data in double quotes it act's like a string not as a variable – Satish51 Dec 15 '17 at 06:49

score 0 · Answer 4 · answered Dec 15 '17 at 07:30

0

either use foreach loop or modify your for loop like this:

for ($i = 0; $i < $countt; $i++)

answered Dec 15 '17 at 07:30

mehrdadep

972
1
15
37

When I run the script below I get an "undefined offset 2" error in php

4 Answers4