CODEWITHSUNDEEP
phpjavascriptmysqlajax
Richard Lewis
Richard Lewis

Reputation: 11

Live Ajax Search not printing all results

I have a HTML page with a text box in which a user would type a country into, and then AJAX and PHP would query a mysql database of countries and print them in a DIV below. The code works just fine apart from one thing. If I type 'A' in the text box, it only prints one result on screen, 'Afghanistan', whereas it should be printing every country starting with 'A'. I imagine the fault lies with the response text, but how would I print out multiple responses, or does that even make sense :P Below is the relevant code.

The PHP database query

<?php
$country=$_REQUEST['country']; //get the country sent by Ajax
$connection = mysql_connect("localhost","MYUSERNAME","MYPASSWORD");
mysql_select_db("countrydb", $connection);  
$result=mysql_query("SELECT * FROM countries WHERE country LIKE '%$country%'",$connection);
if($row = mysql_fetch_array($result))

echo $row['country'];
else
  echo "noresults";
mysql_close($connection); 
?>

This is the relevant javascript

    function showcountryStatus() {
  if (request.readyState == 4) {
    if (request.status == 200) {
              if (request.responseText == "noresults") {
    document.getElementById('results').innerHTML="No Results";
      } else {
        document.getElementById('results').innerHTML=request.responseText;
      }
    }
  }
}

So basically my search is returning only the first result, whereas it should be returning a whole list. Thank you in advance!

Upvotes: 1

Views: 306

Answers (1)

Marcel Korpel
Marcel Korpel

Reputation: 21763

As you can read on the page about mysql_fetch_array you should use it in a loop, e.g.

while ($row = mysql_fetch_array($result)) {
    echo $row['country'];
}

Now you're only fetching the first result.

Upvotes: 1

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