Question

请帮我解决指定的问题：

代码部分：

$result = mysql_query("SELECT *, UNIX_TIMESTAMP(eventdate) AS eventdate,    
  UNIX_TIMESTAMP(throughdate) AS throughdate FROM events ORDER BY eventdate where 
  id='$_GET[id];'");

// the above query is not working   

if ( mysql_num_rows($result) == 0 ) {  
   print "<p>No events right now.</p>\n";
   }
else {

    $lasteventmonth = '';
    while ($row = mysql_fetch_array($result)) { 
        $eventmonth="";
        $eventmonth = date("F Y",$row['eventdate']);

        if ($lasteventmonth != $eventmonth) {
            print "<p style='font-size: 18px;'><b>$eventmonth</b></p>";
        }
        $lasteventmonth = $eventmonth;

        showEvent($row);    
        }
    }
   ?>
........................
........................//other codes

当代码评估如下：

警告：mysql_num_rows（）：提供的参数不是第122行的C：\ Users \ Fagun \ Desktop \ UsbWebserver \ Root \ mapcal \ mapcal.php中的有效MySQL结果资源现在没有活动。+++++++++++++++++

Answer 1

致电mysql_query后，请使用此身份：

if (! $result) {
    echo mysql_errno() . ": " . mysql_error(). "\n";
}

这将告诉您究竟为什么MySQL不会运行您的查询。

Answer 2

是ID字符串还是int？无论哪种方式，我猜你不应该包括一个尾随的分号？

尝试更改如下... $ result = mysql_query（“SELECT *，UNIX_TIMESTAMP（eventdate）AS eventdate，
UNIX_TIMESTAMP（throughdate）AS throughdate FROM events ORDER BY eventdate where ID = '$ _ GET [ID]'“）;

Answer 3

我认为这是一个关于如何使用构建查询和连接id的问题。请尝试这样做（注意ID的连接方式）：

$query = "SELECT *, UNIX_TIMESTAMP(eventdate) AS eventdate,    
    UNIX_TIMESTAMP(throughdate) AS throughdate FROM events
    ORDER BY eventdate where 
    id='".$_GET[id]."'";

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

您不必将其分为两部分，但是 - 这应该更容易阅读和理解。您甚至可以在运行查询之前回显查询以查看实际创建的查询，并在数据库上手动尝试。

或die(mysql_error())部分会详细说明问题所在（如果不是ID问题）。

Answer 4

尝试：

$result = mysql_query("SELECT   *,
                                UNIX_TIMESTAMP(eventdate) AS eventdate,    
                                UNIX_TIMESTAMP(throughdate) AS throughdate
                       FROM     events
                       ORDER BY eventdate
                       where    id= '" . intval($_GET['id']) . "'");

if($result)
{
    //Do code
}

使用intval（）确保$_GET['id']是一个整数使用if语句确保查询已正确执行。

Answer 5

试试这个

 $result = mysql_query("SELECT *, UNIX_TIMESTAMP(eventdate) AS eventdate,    
 UNIX_TIMESTAMP(throughdate) AS throughdate FROM events where 
 id='".$_GET['id']."' ORDER BY eventdate");

Answer 6

正确引用值：

$_GET[id]应为$_GET['id']

尝试以下：

$result = mysql_query("SELECT *, UNIX_TIMESTAMP(eventdate) AS eventdate,    
  UNIX_TIMESTAMP(throughdate) AS throughdate FROM events ORDER BY eventdate where 
  id='".$_GET['id']."');

Answer 7

试试这个：

"SELECT *, UNIX_TIMESTAMP(eventdate) AS eventdate, 
    UNIX_TIMESTAMP(throughdate) AS throughdate FROM events ORDER BY eventdate where 
    id='".$_GET['id'].";'"

我假设id 不来自用户输入。如果是这样，则容易受到SQL注入攻击。

为什么这个mysql查询不起作用？

7 个答案: