php mysql query if $num_results > 0

I want to make a php search query.

First put a sentence and explode every word get $name.

Then I put $name make a query to match all the name which exist in my database.

How to use a if ($num_results > 0) if I am not sure how many $name are matched and echo out? (may be $row['name'] is null, maybe $row['name'] is 1 or 2, I want to get them one by one)

$query = mysql_query("SELECT * FROM books 
  WHERE name LIKE '$name[0]' OR name LIKE '$name[1]' OR name LIKE '$name[2]' 
        OR name l开发者_如何学JAVAike '$name[3]' ");
while ($row = mysql_fetch_array($query)) {
  if ($num_results > 0) {
    echo $row['name'][0] ;
  }
  if ($num_results > 1) {
    echo $row['name'][1] ;
  }
  if ($num_results > 2) {
    echo $row['name'][2] ;
  }
  if ($num_results > 3) {
    echo $row['name'][3] ;
  }
}

use:

mysql_num_rows($query);

to get the amount of rows you get from your mysql query

<?php if (mysql_num_rows($query)>0){?>
Do stuff
<? }else{ ?>
We didn't find anything!
<?php }?>

How about:

$i = 0
while($row = mysql_fetch_array($query)) {
    echo $row['name'][$i];
    $i++;
}

I didn't actually understood what you really want but your code should make more sense like this:

<?php
$searchphrase="alex nick george";
// this:
$names = explode(" ",$searchphrase);
// produces the following:
$names = array("alex","nick","george");
// so you can make the query:
if(is_array($names)){
    $query = "SELECT * FROM books WHERE name IN ('".implode(",",$names)."') ORDER BY FIELD (name,".implode(",",$names).")";
    $result = mysql_query($query) or die(mysql_error());
    if (mysql_num_rows($result) == 0){
        // no results
    }else{
        while ($row = mysql_fetch_array($result)) {
            echo $row['name'];
        }
    }
}
?>

Perhaps if you describe it better we can help more efficiently.