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Error at line 25


b0y

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I can't fix the error
fhw6HVY.png

 

This is line 25 code: 

if (mysql_num_rows($results)==0)

This is full code in case you need it: 

 

<?php
$db = "user";
$tb = "clients";
$link = mysql_connect('localhost', 'root', ''); 

if (!$link) die (mysql_error()) ;
else echo 'Connect to MySql Server - Success! <br>';

// Open Database - USE SEARCH ENGINE
mysql_select_db($db , $link) or die(mysql_error());

// Retrieve table
if (isset($_POST['Search']))
{
$keywoord = $_POST['keyword'];
$keywoord = trim($keywoord);
$results = mysql_query($result = "SELECT * FROM clients WHERE name like '%".$keywoord. "%' \n"
    . "OR\n"
    . " address like '%". $keywoord. "%' \n"
    . "OR\n"
    . " contact like '%". $keywoord. "%' ;");

echo "<br><h2>Search Results:</h2><hr>"; 

if (mysql_num_rows($results)==0)
    {
    echo "Nothing was found! Please try Joogle! Again";
    }
else 
    {while($row = mysql_fetch_array( $results, MYSQL_NUM))
        {
        
        echo "Name: "            . $row[1]        .    "<br>";
        echo "Address: "         . $row[2]        .    "<br>";
        echo "Contact: "         . $row[3]        .    "<br>";
// printf(" %s    %s    %s", $row[1], $row[2], $row[3]); echo "<br>";
        
        echo "<hr>";
        }
    }
}
mysql_close($link);
?>

 


I don't know what's the problem. 

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99% of the time that error means your query failed. Use mysql_error to find out why.

 

And by the way, don't include semicolons in queries.

And also by the way, please try to switch to the mysqli or PDO functions. The mysql extension and its mysql_* functions are deprecated because they're old and slow and inefficient.

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99% of the time that error means your query failed. Use mysql_error to find out why.

 

And by the way, don't include semicolons in queries.

And also by the way, please try to switch to the mysqli or PDO functions. The mysql extension and its mysql_* functions are deprecated because they're old and slow and inefficient.

 

But I have to use MySQL because of assessment I'm doing (TAFE) 

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You need to echo mysql_error to see what the error is. Post the full error message here and we'll try to help solve it for you.

$results = mysql_query($result = "SELECT * FROM clients WHERE name like '%".$keywoord. "%' \n"
    . "OR\n"
    . " address like '%". $keywoord. "%' \n"
    . "OR\n"
    . " contact like '%". $keywoord. "%' ;");

// output mysql error
if(!$results)
   echo 'DB Error: ' . mysql_error();
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You need to echo mysql_error to see what the error is. Post the full error message here and we'll try to help solve it for you.

$results = mysql_query($result = "SELECT * FROM clients WHERE name like '%".$keywoord. "%' \n"
    . "OR\n"
    . " address like '%". $keywoord. "%' \n"
    . "OR\n"
    . " contact like '%". $keywoord. "%' ;");

// output mysql error
if(!$results)
   echo 'DB Error: ' . mysql_error();

 

Okay I just did that and this is what I got o.O

http://prntscr.com/35pzmx

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Okay guys thanks for the help. I tried it at TAFE and it was fully working. I don't know it's not working at home on my very own computer confused.gif but anyway, sorry for wasting your time, guys. Thanks again. Really appreciated! <3

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