警告:mysql_fetch_array()要求参数1为资源,布尔值在第148行的C:\ public_html \ mars \ TaskC \ hitUrl.php中给出
$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>";
}
}
答案 0 :(得分:2)
这似乎是mysql_query失败了, 尝试放这样的东西
$tableQuery = mysql_query($query) or die("Query failed with error: ".mysql_error());
SQL连接更具原生性,更具性能导向性。
答案 1 :(得分:0)
始终本地加入:
$newQuery="SELECT urls,date FROM urls_info as info INNER JOIN urls ON info.url = urls.url";