2011-07-27 23 views
-1

警告:mysql_fetch_array()期望參數1是資源,布爾在C:\ public_html \ mars \ TaskC \ hitUrl.php中給出了行148我試圖用2「while while」循環運行2個不同的SQL表格

$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>"; 
    } 


} 
+3

你爲什麼不只是做一個加入一個單一的查詢? – Tremmors

回答

2

這似乎是的mysql_query失敗, 嘗試把這樣的事情

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

而一個SQL連接更原生的事,多業績爲導向。

0

始終本地連接:

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