how to return multiple rows using a function in php

Question

Look at the following code :

<?php

function getmangainfo($teamname)
{
    $rValue = "";
    $lValue = "";
    $query = ("SELECT pic, mn_title FROM table Where mn_team LIKE '%" . $teamname . "%' Limit 0,4 ");
    $row_data = mysql_query($query);
    while ($row_data = mysql_fetch_array($row_data)) {
        $rValue = $row['pic'];
        $lValue = $row['mn_title'];
        return "<a class='linksrepeated' href='" . $ABSPATH . "/" . $lValue . "/'> <img src='" . $rValue . "'/></a>";
    }
}

this function is not returning anything! I am thinking it is because that the return statement is inside the while loop. I tried many things hoping it will return the 4 results but nothing happened. the SQL query works 100%. the problem is with my function. please let me know what is wrong and how to fix it.

Why aren't you echoing the data out? Return will not work with multiple iterations (I don't believe) — Dylan Cross
– Dylan Cross, Commented Jan 4, 2013 at 4:25
the loop will always be executed once because of the return statement inside it. — John Woo
– John Woo, Commented Jan 4, 2013 at 4:25
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Mayank K. Swami
– Mayank K. Swami, Commented Jan 4, 2013 at 4:39

Netorica · Accepted Answer · 2013-01-04 05:07:55Z

2

change the $row_data to $row in your while statement

while($row = mysql_fetch_array($row_data))

because as I see the codes inside the while

$rValue = $row['pic'];
$lValue = $row['mn_title'];

you get your data as $row but your in while statement is $row_data

the problem is not on the while loop as the execution reaches the return statement, the execution pointer will exit the function(of course in the while statement)

but for me to make your code cleaner as i see you expect only one row in return pull out the return statement on your while

$rValue = "";
 $lValue = "";
 while ($row_data = mysql_fetch_array($row_data)) {
        $rValue = $row['pic'];
        $lValue = $row['mn_title'];
        break; //just to make sure for one row return
 }
 return "<a class='linksrepeated' href='" . $ABSPATH . "/" . $lValue . "/'> <img src='" . $rValue . "'/></a>";

but as the others says that you expect 4 row returns you can create a variable that would store all the returns in a single string

   $rValue = "";
     $lValue = "";
     $links = "";
     while ($row_data = mysql_fetch_array($row_data)) {
            $rValue = $row['pic'];
            $lValue = $row['mn_title'];
            $links .="<a class='linksrepeated' href='" . $ABSPATH . "/" . $lValue . "/'> <img src='" . $rValue . "'/></a>";
     }
     return $links

reference: http://php.net/manual/en/function.mysql-fetch-array.php

edited Jan 4, 2013 at 5:07

answered Jan 4, 2013 at 4:26

Netorica

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3 Comments

syrkull Over a year ago

best answer of them all. you are the chosen one :P

Netorica Over a year ago

you're the chosen one also! enjoy coding =)

Hanky Panky Over a year ago

@Mahan I think that "break" doesn't have to be there according to her question. In question it was said "I tried many things hoping it will return the 4 results but nothing happened. " So that appears like 4 results are expected and not 1

Hanky Panky · Accepted Answer · 2013-01-04 04:25:39Z

1

    function getmangainfo($teamname){
    $rValue = "";
    $lValue = "";
    $query = ("SELECT pic, mn_title FROM table Where mn_team LIKE '%".$teamname."%' Limit 0,4 ");
    $row_data = mysql_query($query);
    $output=array();
    while($row = mysql_fetch_array($row_data))
    {
    $rValue = $row['pic'];
    $lValue = $row['mn_title'];
    $output[]="<a class='linksrepeated' href='".$ABSPATH."/".$lValue."/'> <img src='".$rValue."'/></a>";
    }   
return $output;
}

Edit: Updated variable name

answered Jan 4, 2013 at 4:25

Hanky Panky

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2 Comments

syrkull Over a year ago

I get "Array" as a result no data at all!

Hanky Panky Over a year ago

@shnisaka yes, that array contains your data. run print_r() on it rather than echo

Bhupinder Singh · Accepted Answer · 2013-01-04 04:29:17Z

1

Its not giving full result set because you are 'returning' from within the loop. try the following it should help.

function getmangainfo($teamname){
    $rValue = "";
    $lValue = "";
    $query = ("SELECT pic, mn_title FROM table Where mn_team LIKE '%".$teamname."%' Limit 0,4 ");
    $row_data = mysql_query($query);
    $return = '';
    while($row_data = mysql_fetch_array($row_data))
    {
        $rValue = $row['pic'];
        $lValue = $row['mn_title'];
        $return .= "<a class='linksrepeated' href='".$ABSPATH."/".$lValue."/'> <img src='".$rValue."'/></a>";
    }
    return $return;
}

answered Jan 4, 2013 at 4:29

Bhupinder Singh

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