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Home/ Questions/Q 561893
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T12:29:52+00:00 2026-05-13T12:29:52+00:00

$result = mysql_query(SELECT avg(r.rate) FROM rate r where ImgName=’1′); this php is not working.

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$result = mysql_query("SELECT avg(r.rate) FROM rate r where ImgName='1'");

this php is not working.

Originally my code is

<?php
$con = mysql_connect("localhost","root","sql");

if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("photogallery", $con);

$result = mysql_query("SELECT avg(r.rate) FROM rate r ");
echo "<table border='0' cellspacing='5'>";
echo "<th> Average Rating </td>";

while($row = mysql_fetch_array($result))
{
    echo "<tr>";
    echo "<td> " . $row['rate'] . "</td>";
    echo "</tr>";
}

echo "</table>";

mysql_close($con);
?> 

the above is not showing any out put.

but modify code i.e. then its workin.

$result = mysql_query(“SELECT r.rate FROM rate r “);

but i want to aggregate function

thanks in advance

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  1. Editorial Team
    Editorial Team
    2026-05-13T12:29:53+00:00Added an answer on May 13, 2026 at 12:29 pm

    Your query is producing a scalar rather than a set of rows. If you want to get the average rate per item then you should do something like:

    SELECT avg(r.rate) FROM rate r GROUP BY ItemIdColumn
    

    And yes, if you want to fetch the value by column name, you should use an alias, like knittl mentioned.

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