Need Some Help Regarding Fetching Data from Mysql using explode function

Question

In the below script I want to fetch data from mysql using a explode function and also a variable within an explode function. Here's how I want to get

<?php
include ('config.php');
$track = "1,2,3";
$i = 1
$trackcount = explode(",",$track);
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>

This is the code

$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";

I want sql to fetch data from tracks table where id = $trackcount[$i] Whatever the value of $trackcount[$i] mysql should fetch but it shows a blank screen. If I put this

$sql = "SELECT * FROM tracks WHERE id='$trackcount[1]'";

It works perfectly

Vigneswaran S · Accepted Answer · 2015-11-21 18:25:17Z

1

save your $trackcount[$i] in one variable and then pass it in the query as given below

<?php
include ('config.php');
$track = "1,2,3";
$i = 1;
$trackcount = explode(",",$track);
$id=$trackcount[$i];
$sql = "SELECT * FROM tracks WHERE id='$id'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>

and one more thing check your previous code with echo of your query and see what is passing ok.

echo $sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//like this

problem is with your query

$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//change

to

$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";

edited Nov 21, 2015 at 18:25

answered Nov 21, 2015 at 18:11

Vigneswaran S

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Comments

Professor Abronsius · Accepted Answer · 2015-11-21 17:51:01Z

0

Generally you would want to use the IN operator with this type of query, so for you this would be:-

$sql="SELECT * FROM `tracks` WHERE `id` in (".$track.");";

or, if the $ids are in an array,

$sql="SELECT * FROM `tracks` WHERE `id` in (".implode( ',', $array ) .");";

answered Nov 21, 2015 at 17:51

Professor Abronsius

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