Mysql extracting array in Php

Question

What is the problem with my mysql array extraction below:

        $mysql_array = mysql_query("SELECT * FROM table1 WHERE uid1='$uid1'"); 
        $array = array();
        while($row = mysql_fetch_assoc($mysql_array)){
            $array[] = $row;
        }

        $array = array_unique($array);
        $array = array_reverse($array);
        $emails = array();
        $numbers = array();

        foreach($array as $row){
            $uid2 = $row['uid2'];
            $number = number($uid2);
            if(strlen($number) > 9){
                $numbers[] = array('uid2' => $uid2, 'number' => $number);
            }
            else{
                $email = email($uid2);
                $emails[] = array('uid2' => $uid2, 'email' => $email);
            }
        }
        $numbers = array_unique($numbers);
        $emails = array_unique($emails);            
        var_dump($numbers);
        var_dump($emails);

There must be something I need to do to convert the "Resource" from mysql into an array. There must be a problem with the above code. This is what I am getting on the var_dumps: array(0) { } and array(1) { [0]=> array(2) { ["uid2"]=> NULL ["email"]=> NULL } }

What the purpose of this code? $number = number($yfbid);if(strlen($number) > 9){ , What? — Ofir Baruch
– Ofir Baruch, Commented Feb 29, 2012 at 19:30
possible duplicate of php - create assciative array while loop — Vyktor
– Vyktor, Commented Feb 29, 2012 at 19:32
what are you expecting to accomplish? I see several issues but I'm not even sure where to begin. — Cfreak
– Cfreak, Commented Feb 29, 2012 at 19:34
no it is not a possible duplicate, there the question was about the loop, now that that is covered, the question is about extracting the array from mysql. regarding the variables - you are right i had some of them wrong when i copied the code, now it's editted — Lucy Weatherford
– Lucy Weatherford, Commented Feb 29, 2012 at 19:34

jribeiro · Accepted Answer · 2012-02-29 19:32:48Z

2

Look at this foreach:

        foreach($array as $uid2){
            $uid2 = $array['uid2'];
            $number = number($yfbid);
            if(strlen($number) > 9){
                $numbers[] = array('uid2' => $uid2, 'number' => $number);
            }
            else{
                $email = email($uid2);
                $yfbid[] = array('uid2' => $uid2, 'email' => $email);
            }
        }

You iterate all positions of array and call them $uid2. Then, in the first line you do $uid2 = $array['uid2'];. You loose your array position.

Maybe you wanted something like: $var = $uid2['uid2'];

answered Feb 29, 2012 at 19:32

jribeiro

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Comments

Travesty3 · Accepted Answer · 2012-03-01 13:28:01Z

1

I think this is what you're trying to do:

$result = mysql_query("SELECT DISTINCT * FROM table1 WHERE uid1 = '{$uid1}' ORDER BY /* insert column name here, followed by ASC or DESC */");
$table = array();
while ($row = mysql_fetch_assoc($result))
    $table[] = $row;

// at this point, each entry in the $table array is one row from 'table1'.
// shouldn't need to do array_unique or array_reverse with the modified SQL query above.

$numbers = $emails = array();
foreach ($table as $row)
{
    $number = number($row["uid2"]);
    if (strlen($number) > 9 && !in_array($numbers[$row["uid2"]], $number, true))
        $numbers[$row["uid2"]][] = $number;
    else
    {
        $email = email($row["uid2"]);
        if (!in_array($emails[$row["uid2"]], $email, true))
            $emails[$row["uid2"]][] = $email;
    }
}

// shouldn't need to do array_unique with the if-statements above

var_dump($numbers);
var_dump($emails);

EDIT Answering question from comments:

Based on the logic you are using, the result will probably be the same. My example above will allow this scenario for $numbers, while your example will not:

array(2)
{
    [123] => array(2)
    {
        [0] => 1234567890,    // same as $numbers[456][0]
        [1] => 9876543210
    },
    [456] => array(1)
    {
        [0] => 1234567890    // same as $numbers[123][0]
    }
}

But based on the way that $number is generated based on $uid2, you probably won't see any difference. What I mean is that, if number(123) returns 1234567890, then number(456) probably will not return 1234567890, so you probably wouldn't ever run into a scenario where you would see a difference.

EDIT 2 After thinking about this some more, I'm betting that your code can be greatly simplified to this:

// only selecting uid2 from the table
$result = mysql_query("SELECT DISTINCT uid2 FROM table1 WHERE uid1 = '{$uid1}' ORDER BY /* insert column name here, followed by ASC or DESC */");

$output = array();
while (list($uid2) = mysql_fetch_row($result))
{
    $number = number($uid2);
    if (strlen($number) > 9)
        $output[$uid2]["number"] = $number;
    else
        $output[$uid2]["email"] = email($uid2);
}

var_dump($output);

LAST EDIT (Hopefully):

// only selecting uid2 from the table
$result = mysql_query("SELECT DISTINCT uid2 FROM table1 WHERE uid1 = '{$uid1}' ORDER BY /* insert column name here, followed by ASC or DESC */");

$numbers = $emails = array();
while (list($uid2) = mysql_fetch_row($result))
{
    $number = number($uid2);
    if (strlen($number) > 9)
        $numbers[$uid2] = $number;
    else
        $emails[$uid2] = email($uid2);
}

var_dump($numbers);
var_dump($emails);

edited Mar 1, 2012 at 13:28

answered Feb 29, 2012 at 19:52

Travesty3

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6 Comments

Lucy Weatherford Over a year ago

this looks great! thanks. but shouldn't the in array function look like this: !in_array($number, $numbers)

Lucy Weatherford Over a year ago

oh awesome!! I was hoping/ thinking that this may indeed can be simplified directly from the mysql. thanks!

Travesty3 Over a year ago

mysql_fetch_row($result) will return an array. If you use list($uid2) = mysql_fetch_row($result), it gets the first value from the numeric array and assigns it to the variable $uid2. If you wanted the first two values from an array, you could use list($value1, $value2) = $array;, or if you just wanted the third value from an array, you could do list( , , $value3) = $array;. See the PHP documentation for further explanation.

Lucy Weatherford Over a year ago

why do my arrays look like this { [345678]=> array(1) { ["email"]=> string(19) "[email protected]" }, how do I get it to be just: { [345678]=> "[email protected]" }

Travesty3 Over a year ago

I had both numbers and emails going into one array. See updated answer again. This will give you two separate arrays in the structure you want.

|

Marc B · Accepted Answer · 2012-02-29 19:34:26Z

0

foreach($array as $uid2){
    $uid2 = $array['uid2'];

makes no sense. Loop over each element in $array and assign the elements to $uid2 sequentially, then kill that assigned and pull a fixed value from the same array?

As well, where is $yfbid set? From your code sample, it's undefined, and you try to convert this undefined value to a number, then check the string length of that number. Why not just strlen($yfbid) and let PHP handle the conversion?

answered Feb 29, 2012 at 19:34

Marc B

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