Not getting in PHP Array in required format

Question

The array I want is given below;

Array ( [0] => John [1] => toby [2] => hrithik )

I tried below code for

$r = array();
$queryx = mysql_query("SELECT `name` FROM `student` WHERE `status` = '1'") or die ('Query is invalid: ' . mysql_error());

while ($row = mysql_fetch_assoc($queryx)) {
$r[] = $row;
}


print_r($r);

The output I am getting is given below;

Array ( [0] => Array ( [name] => John ) [1] => Array ( [name] => toby ) [2] => Array ( [name] => hrithik ) )

@Deep I tried but its fetching only one row and same array which I was getting earlier. — Riya Patel
– Riya Patel, Commented Feb 28, 2017 at 6:12
then no need to declare as array , just use as in the answer given below — Deep Kakkar
– Deep Kakkar, Commented Feb 28, 2017 at 6:20

Mayank Pandeyz · Accepted Answer · 2017-02-28 06:10:58Z

3

Try this:

while ($row = mysql_fetch_assoc($queryx)) {
    $r[] = $row['name'];
}

It will put name in $r array on different index like 0, 1, 2 and so on

answered Feb 28, 2017 at 6:10

Mayank Pandeyz

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Tonmoy · Accepted Answer · 2017-02-28 06:20:53Z

1

try the following code below,

$r = array();
$i = 0;
$queryx = mysql_query("SELECT `name` FROM `student` WHERE `status` = '1'") or die ('Query is invalid: ' . mysql_error());
while ($row = mysql_fetch_assoc($queryx)) {
$r[$i] = row["name"];
$i++;
}
print_r($r);

answered Feb 28, 2017 at 6:20

Tonmoy

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Comments

Josh Lin · Accepted Answer · 2017-02-28 06:26:32Z

0

what you want is mysql-fetch-row http://php.net/manual/en/function.mysql-fetch-row.php

you can refer api to check how these function works mysql_fetch_array http://php.net/manual/en/function.mysql-fetch-array.php mysql_fetch_assoc http://php.net/manual/en/function.mysql-fetch-assoc.php

answered Feb 28, 2017 at 6:26

Josh Lin

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