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[HELP]Warning:mysql_fetch_array(): supplied argument is not a valid MySQL result

alanlaw87

By alanlaw87
in PHP Coding Help

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alanlaw87 Newbie

alanlaw87

Posted
Posted

Hi,

I am having problem with mysql and php.... when i test the database on my local host it didt produce any error but when i put it on my web host it gave me error messages... "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource" Please help me as i am new to php.

 

 

 

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">

<html>

    <head>

        <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">

        <link rel="stylesheet" type="text/css" href="http://www.hingloong.com/pricesearch/style.css" />

        <title>Price Search Query</title>

    </head>

    <body>

        <form method=GET action="search.php">

            <div id=container>

               

                <select name=general>

                    <option value = all>Select</option>

                    <option value = laptop>Laptop</option>

                    <option value = mobile>Mobile</option>

                    <option value = games>Games</option>

                    <option value = camera>Camera</option>

                </select>

               

                <label><b>Brand:</b></label>

                <input type=text name=brand>

                <label><b>Model:</b></label>

                <input type=text name=model>

                <input type=submit name=search value=Search>

               

               

        </form>

        <br><br>

        <input type=button name=gamelist value=View Games onclick="location.href='gamelist.php'">

        <input type=button name=mobilelist value=View Mobile onclick="location.href='mobilelist.php'">

        <input type=button name=cameralist value=View Cameras onclick="location.href='cameralist.php'">

        <input type=button name=laptoplist value=View Laptops onclick="location.href='laptoplist.php'">

        </div>

    </body>

</html>

 

<?php

include ("connect.php");

 

$general = $_GET['general'];

$brand = $_GET['brand'];

$model = $_GET['model'];

 

 

 

 

if($general == 'mobile')

{

   

    echo "<br>";

    echo "<h3>Mobile Phone Prices</h3>";

    echo "<div id=container>";

    echo "<table>";

    echo "<tr>";

    echo "<td class=head>Phone Brand</td>";

    echo "<td class=head>Phone Model</td>";

    echo "<td class=head>Phone Loan</td>";

    echo "<td class=head>Phone Buy</td>";

    echo "<td class=head>Phone Sell</td>";

    echo "<td class=head>Phone Rrp</td>";

    echo "</tr>";

   

   

$query = "select * from mobilephones where phone_brand LIKE ('$brand%') AND phone_model LIKE ('$model%') ";

$temp = mysql_query($query) or die(mysql_error());

 

 

 

 

 

}

 

 

while ($row = mysql_fetch_array($temp))

{

 

    echo "<tr>";

    echo "<td>" .$row['phone_brand']. "</td>";

    echo "<td>" .$row['phone_model']. "</td>";

    echo "<td>" .$row['phone_loan']. "</td>";

    echo "<td>" .$row['phone_buy']. "</td>";

    echo "<td>" .$row['phone_sell']. "</td>";

    echo "<td>" .$row['phone_rrp']. "</td>";

 

 

   

   

}

 

    echo "</table>";

    echo "</div>";

 

 

 

?>

PFMaBiSmAd Advanced Member

PFMaBiSmAd

Posted
Posted

Your query is inside of the conditional logic - if($general == 'mobile'){ your query is in here} but you are attempting to fetch the results after and outside of that conditional logic.

 

Your code will produce that error every time it is executed and $general is not equal to 'mobile'.

 

You need to fix your logic so that you only fetch the results from the query when the query is executed.

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