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Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\public_html\mars\TaskC\hitUrl.php on line 148

$query  = "SELECT url FROM urls";
$tableQuery = mysql_query($query);


while($row = mysql_fetch_array($tableQuery)){
    $newQuery="SELECT urls,date FROM urls_info WHERE url='{$row['url']}'";
    $newTableQuery=mysql_query($newQuery);

    $newGroupedQuery = $newQuery . "GROUP BY date";

    while($newRow = mysql_fetch_array($newTableQuery)){
        $displayStr .=    "<tr>";
        $displayStr .=    "<td>".$newRow['date']."</td>";
        $displayStr .=    "<td>".$newRow['url']."</td>";
        $displayStr .=    "<td>".num_rows($newQuery)."</td>";
        $displayStr .=    "</tr>";   
    }


}
hakre
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2 Answers2

1

this seems like the mysql_query is failing, try putting something like this

$tableQuery = mysql_query($query) or die("Query failed with error: ".mysql_error());

And an SQL join is more native to do, and more performance oriented.

fatnjazzy
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0

Always natively join:

$newQuery="SELECT urls,date FROM urls_info as info INNER JOIN urls ON info.url = urls.url";
Martijn
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