<?php include ("db.php"); $result = mysql

<?php
include ("db.php");
$result = mysql_query("SELECT ques_id FROM questionbank
order by ques_id limit 5 ");

while($obj=mysql_fetch_array($result))
{
$ad1[$obj['ques_id']]++;//Used an array and inserted the
database query results into it.
}
$rand_keys=array_rand($ad1,1); //Did a random array function
echo "First random id = ".$ad1[$rand_keys[0]];
echo "<br>Second random id = ".$ad1[$rand_keys[1]];

?>

Question Posted / rabeeshm

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<?php include ("db.php"); $result = mysql_query("SELECT ques_id FROM questionban..

Answer / bibhu

The following is your program. you need some little bit
change on this program.
<?php
include ("db.php");
$result = mysql_query("SELECT ques_id FROM questionbank
order by ques_id limit 5 ");

while($obj=mysql_fetch_array($result))
{
$ad1[$obj['ques_id']]++;//
}
$rand_keys=array_rand($ad1,1); //Did a random array function
echo "First random id = ".$ad1[$rand_keys[0]];
echo "<br>Second random id = ".$ad1[$rand_keys[1]];

?>

Your Mistake
1. in while statement you add $ad1[$obj['ques_id']]++; which
is not assignment statement. It is a increment statement.
The correct answer is $ad1[$obj['ques_id']] = $obj['ques_id'];

2.in array random function You use 1 key but you display random
array value using 2 keys.

$rand_keys=array_rand($ad1,1); //Did a random array function
echo "First random id = ".$ad1[$rand_keys[0]];
echo "<br>Second random id = ".$ad1[$rand_keys[1]];

just change 1 to 2 in array_rand($ad1,2);

The Changes are from while statement.

while($obj=mysql_fetch_array($select))
{
$ad1[$obj['ques_id']] = $obj['ques_id'];
}
$rand_keys=array_rand($ad1,2); //Did a random array function
echo "First random id = ".$ad1[$rand_keys[0]];
echo "<br>Second random id = ".$ad1[$rand_keys[1]];

Is This Answer Correct ?

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