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up vote 1 down vote favorite

Ok, I have the following code:

$array = mysql_query("SELECT artist FROM directory WHERE artist LIKE 'a%' 
        OR artist LIKE 'b%' 
        OR artist LIKE 'c%'");
    $array_result= mysql_fetch_array($array);

Then, when I try to echo the contents, I can only echo $array_result[0];, which outputs the first item, but if I try to echo $array_result[1]; I get an undefined offset.

Yet if I run the above query through PHPMyAdmin it returns a list of 10 items. Why is this not recognized as an array of 10 items, allowing me to echo 0-9?

Thanks for the help.

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I know that % is the usual wildcard, but have you tried using * instead? –  JMichelB Aug 9 '11 at 18:52
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4 Answers 4

up vote 11 down vote accepted

That's because the array represents a single row in the returned result set. You need to execute the mysql_fetch_array() function again to get the next record. Example:

while($data = mysql_fetch_array($array)) {
  //will output all data on each loop.
  var_dump($data);
}
share|improve this answer
    
Thanks a bunch. Oddly enough, when I use this, it doesn't get the value for the data in first row. I.e. Say first artist is Apple, the second is Banana, the third is Cantaloupe. When I var dump, it just does Banana -> Cantaloupe -> etc, no apple. –  Ando Aug 9 '11 at 19:09
    
@Ando: Does it get 'apple' when you use PMA (PhpMyAdmin)? Could be: (1) Your php-generated SQL query doesn't filter the records appropriately. (2) You don't ORDER BY, but you LIMIT, thus maybe the record with 'apple' is left behind while filtering. –  Dor Aug 9 '11 at 19:14
    
My SQL query does bring back Apple, and I do ORDER BY artist ASC. What is peculiar is if I ORDER BY desc, then it shows Apple at the bottom, but then it is missing the item that should be first...very odd –  Ando Aug 9 '11 at 19:20
    
Also, if I use while(list($artist) = mysql_fetch_array($result)) { echo $artist; } the list shows only items 1-9 instead of 0-9. –  Ando Aug 9 '11 at 19:23
    
Try posting the exact code, table description (SHOW TABLE STATUS LIKE "tbl_name"; DESCRIBE tbl_name;) and output, we'll try and see where's the problem. –  Dor Aug 9 '11 at 19:32
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up vote 0 down vote 
$array = mysql_query("SELECT artist FROM directory WHERE artist LIKE 'a%' 
    OR artist LIKE 'b%' 
    OR artist LIKE 'c%'") or die(mysql_error());
while($array_result= mysql_fetch_array($array))
{
  echo $array_result['artist'].'<br>';
}

Reference: mysql_fetch_array

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up vote 1 down vote

I prefer to use this code instead:

$query  = "SELECT artist FROM directory WHERE artist LIKE 'a%' 
        OR artist LIKE 'b%' 
        OR artist LIKE 'c%'";           
$result = mysql_query($query) or die(mysql_error());


while(list($artist) = mysql_fetch_array($result))
{
echo $artist;
}

If you add more fields to your query just add more variables to list($artist,$field1,$field2, etc...)
I hope it helps :)

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Or: while($data = mysql_fetch_assoc($result)) { extract($data); /*...*/ }. See: php.net/manual/en/function.extract.php –  Dor Aug 9 '11 at 19:09
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up vote 4 down vote

You should be using while to get all data. 

$array_result = array();
while ($row = mysql_fetch_array($array, MYSQL_NUM)) {
    $array_result[] = $row;
}
echo $array_result[4];
share|improve this answer
1  
The data is stored in PHP's memory space anyway. Copying that data is usually useless. –  Dor Aug 9 '11 at 19:16
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