Conflict between mysql_fetch_row and mysql_fetch_array

Question

I have a function in PHP that give me some values from DB

$select = mysql_query("SELECT * FROM sfide WHERE accepted = 0");

if($select){
    if(mysql_fetch_row($select) != 0){
        $sep = "~";
        $return = "";
        $i = 0;
        while($row = mysql_fetch_array($select)){
            if($i > 0){
                $return = $return.",";
            };
            $return = $return.$row['id'].$sep.$row['from_'];
            $i += 1;
        }
        return $return;
    }else{
        return "NO";
    }
}else{
return "E, 000, SELECT, ".mysql_error();        
}

The problem is that if I do like in the code above it returns me nothing, instead, if I duplicate the select variable it works:

$select = mysql_query("SELECT * FROM sfide WHERE accepted = 0");
$num_select = mysql_query("SELECT * FROM sfide WHERE accepted = 0");
if($select){
    if(mysql_fetch_row($num_select) != 0){
        $sep = "~";
        $return = "";
        $i = 0;
        while($row = mysql_fetch_array($select)){
            if($i > 0){
                $return = $return.",";
            };
            $return = $return.$row['id'].$sep.$row['from_'];
            $i += 1;
        }
        return $return;
    }else{
        return "NO";
    }
}else{
return "E, 000, SELECT, ".mysql_error();        
}

Why ?

mysql_fetch_row returns first row of the resultset and move internal pointer forward so mysql_fetch_array read the second row (or return false if it does not exists). You may take a look on mysql_num_rows function to see how many rows are in the resultset. — barell
– barell, Commented Mar 11, 2014 at 22:58

Funk Forty Niner · Accepted Answer · 2014-03-11 23:10:00Z

1

Instead of

mysql_fetch_row($select) != 0

use

mysql_num_rows($select) != 0

that will return an int for you to compare.

edited Mar 11, 2014 at 23:10

Funk Forty Niner

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answered Mar 11, 2014 at 23:01

Byron Claiborne

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