search script

pinehead18

if ($_POST['search']) {
             $con = mysql_connect("localhost","pi","") or die(mysql_error());
        $db = mysql_select_db("pinehead",$con) or die(mysql_error());
       // $sql = "SELECT * FROM users WHERE (age > $agemin) AND (age < $agemax) AND city like '%$city%' limit 0,5";
        $sql = "SELECT * from users WHERE (age > $agemin) AND (age < $agemax) AND city like '%$city%'";   

        $result = mysql_query($sql,$con) or die(mysql_error());
        while ($row = mysql_fetch_array($result)) {
        $fname = $row['fname'];
        $uname = $row['user'];

 echo $uname

}

This is my script. It only returns one result even though i know for a fact their are more.

Thank you for any help that you can provide.

leatherback

Hey Pinehead,

From the query you showed .. It looks like it should display more then one record.. IF there are more records!

Have you checked wether there are more hits? Say:

if($result)
 {
 $count = mysql_num_rows($result);
 echo $count;
 }

And, perhaps it is an idea to add echo $query; too, and past it directly in your sql processor and see what it outputs there?

LordShryku

This clearly isn't your whole script, since you never close your if ($POST['search']) bracket. Try echoing mysql_num_rows to see how many rows it thinks were returned

if(isset($_POST['search'])) {
   $con = mysql_connect("localhost","pi","") or die(mysql_error());
   $db = mysql_select_db("pinehead",$con) or die(mysql_error());
   $sql = "SELECT * 
             FROM users 
            WHERE age > ".$agemin." 
              AND age < ".$agemax." 
              AND city LIKE '%$".city."%'";   

   $result = mysql_query($sql) or die(mysql_error());
   echo "Found ".mysql_num_rows($result)." results<br />";
   while ($row = mysql_fetch_array($result)) {
      $fname = $row['fname'];
      $uname = $row['user'];
      echo $uname."<br />";
   }
}

pinehead18

it says i have 333 results... Yet their are only 10 rows in the table.
Any ideas?

leatherback

A hunge: you have the echo statement within the mysql_fetch_array() loop? So that would give you three results, (Displayed three times, leading to 333) which is more likely.

Besides that.. Coud it be that the $uname is "" for two of the three results?

pinehead18

Ok i echoed $count after the loop works fine now. Says i'm suppose to have 3 results which is right. However i only get one.

Any ideas?
Thank you
Anthony

leatherback

Is theer actual content in the database for all of the results? Or are they filled with empty space??

You could test that with placing:

echo "Usernames fund: ".$uname."<br>";

in the loop. This will show something for every loop.

Besides.. <br /> is not part of the standard html.. Are you using xml? Else use <br>

pinehead18

No they all of the right content.

leatherback

Without trying to be annoying, or smart-ass..
Recheck? Did you place the little bit of text with your echo statement/ Did it give you multiple times the same text?

pinehead18

Ok, what i did was i took out the fancy html in which echoed the $name and i changed it to echo $fname and that was it.

It echoed all 3 results. This is my entire code.. Maby we can figure out why it wont post with my html.


<?PHP
// Add the next and previous buttons to this page as soon as you fix them 12-31-03

if ($_POST['search']) {

    $con = mysql_connect("localhost","pinehead","") or die(mysql_error());
    $db = mysql_select_db("pinehead",$con) or die(mysql_error());
   // $sql = "SELECT * FROM users WHERE (age > $agemin) AND (age < $agemax) AND city like '%$city%' limit 0,5";
    $sql = "SELECT * from users WHERE (age > $agemin) AND (age < $agemax) AND city like '%$city%'";
    $result = mysql_query($sql,$con) or die(mysql_error());
    while ($row = mysql_fetch_array($result)) {
    $fname = $row['fname'];
    $uname = $row['user'];
    $picurl = $row['pic1'];
    $count =  mysql_num_rows($result);



    $output = "<table border=1 bordercolor=475575 cellpadding=0 cellspacing=0 align=center width=80%>    
    <tr><td>
    <table border=0 cellpadding=0 cellspacing=0 width=100% bgcolor=475575>
            <tr><td><font size=-1 face=arial>» $uname</a></td><td align=right><a
href=http://lifeinkc.com/myprofile/$uname><font
size=-1 face=arial>View
Profile</a></font></td></tr>
        </table>
        <table border=0 cellpadding=0 cellspacing=0 bgcolor=3D4A67 width=100%>
        <tr><td width=20%><img src=/pics/$picurl width=75 height=56></td><td
valign=top><font size=-1 face=arial>Name: $fname<br>
           Age: $age<br>
           Location: $location</font></tr><td>
        </table>
        <table border=0 cellpadding=0 cellspacing=0 width=100% bgcolor=475575>
                <tr><td><img src=images/sendafriend_1.gif></td></a><td align=right><img
src=images/sendamessage_1.gif></td></tr>
        </table>
        </tr></td></table>
        <table border=0 cellpadding=1 cellspacing=1 width=50% height=5%>
                <tr><Td>&nbsp;</td></tr>
        </table>";


}
}

include('html.inc');  

start_header($output);


?>

leatherback

Hi,

easy:

Your $output = ..
should be
$output .= ..

Next time: Please tell that you replaced such abig chunk of code:

There is a big difference between echo-ing something, and storing it in a var!

Goodluck with the rest!

PS: That goes of course for all var's in the for-loop.
Only leve vars in the for-loop which get new values in the loop. Otherwise leave them outside the loop (Such as the count() ): Why waste time redefining variables when they don't change, right?

pinehead18

O my goodness it fixed it.. why?

leatherback

Haha, I like your "Oh my goodness! It is always worth reading all the non-sense post for a responce like this! You just made my day! I'd better stop (That, and the fact that it is 1.20 am here..)"

Last post today.. Off to bed.. But just a short explanation:

Say you have var $a. You give it a value:

$a = "Testing ";
Now you have a variable $b :
$b = "combining strings";

If you now would state:
$a = $b, you'd know what it would do, right? It would assign the value of $a (Testing ) to $b, so echo $a would give "comining strings "But that is not what you want, you want to combine:0
$a = $a . $b -> Quicker way: $a .= $b;

So, just the = sign will overwrite the existing $a, and put the value of $b instead. Using the .= is like combining the two.

Happy coding.

J.
Here is a test-snippet for you:

<?php

$a = "Testing "; 
$b = "combining strings"; 
$c = $a . $b; 
$d = $a; //backup of $a; 

echo "a is now: ".$a; 
echo "<br>b is now: ".$b; 
echo "<br>c is now: ".$c; 
$a = $b; 
echo "<br>a = b is now: ".$a; 
$a = $d;


 $a .= $b;


echo "<br>a: .= b is now: ".$a;
?>