Question

我正在尝试从2个表中提取数据，当我在$ query中添加它时，当我点击我的链接ex时，结果都搞砸了。 /leads.php?contactstatus=Hot

当我只有“Hot”“Warm”“Cold”作为表格行contactstatus中的选项时，我如何默认显示所有的contactstatus类型？

<?
mysql_connect ("xxxx","xxxx","xxxx") or die ('Error: ' .mysql_error());
mysql_select_db ("xxxx");

if(isset($_GET['contactstatus'])
&& in_array($_GET['contactstatus'], array('Hot', 'Warm', 'Cold'))){     
$status = $_GET['contactstatus']; 
} 
else {      
$status = ''; // what do I put here so by default users for all contactstatus types show in results 
}  

 $query = "SELECT * FROM contacts,contacttodo WHERE contacts.contactstatus =    `'".$status."' OR contacttodo.type = 'Appointment'";`  

$result=mysql_query($query);

while($row = mysql_fetch_array($result)){

?>

更新：

这有效：

<div class="nav">
<table width="100%">
<tr>
<td>
<li><a href="/dbs/a.php?type=Appointment">Appointments</a></li> 
<li><a href="/dbs/a.php?contactstatus=Hot">Hot</a></li> 
<li><a href="/dbs/a.php?contactstatus=Warm">Warm</a></li> 
<li><a href="/dbs/a.php?contactstatus=Cold">Cold</a></li>
</td>
</tr>
</table>
</div>

$status = ''; 
$todotype = '';  

if(isset($_GET['contactstatus']) 
&& in_array($_GET['contactstatus'], array('Hot', 'Warm', 'Cold')))
{     
$status = $_GET['contactstatus'];     
 $query = "SELECT * FROM contacts,contacttodo,contactnotes WHERE contacts.ID = contacttodo.contacts_id = contactnotes.contacts_id AND contacts.contactstatus = '".$status."' ORDER BY contacts.firstname ASC"; 
}  

if(isset($_GET['type']) 
&& in_array($_GET['type'], array('Appointment', 'Email', 'Call')))
{     
$todotype = $_GET['type'];     
$query = "SELECT * FROM contacts,contacttodo,contactnotes WHERE contacts.ID = contacttodo.contacts_id = contactnotes.contacts_id AND contacttodo.type = '".$todotype."' ORDER BY contacts.firstname ASC"; 
} 
$result=mysql_query($query);

while ($row = mysql_fetch_array($result)) {

<div id="contact-results">
<table width="100%" cellspacing="0" cellpadding="0" class="contact-results"> 
<tr>
<td align="left" width="15%"><a href="/dbs/editcontact.php?ID=<? echo $row['ID']; ?>"><strong><? echo $row['firstname']; ?> <? echo $row['lastname']; ?></strong></a></td>
<td align="left" width="5%"><? echo $row['contactstatus']; ?></td>
<td align="left" width="15%"><? echo $row['contacttype']; ?></td>
</tr>
</table>
</div>
<?
}
mysql_close();
?>

默认情况下，如何显示所有“热”'暖''冷'潜在客户？

Answer 1

您没有连接联系人和contacttodo表的连接条件，因此您将获得一个交叉产品。你需要这样的东西：

SELECT * 
    FROM contacts c
        INNER JOIN contacttodo ctd
            ON c.contact_id = ctd.contact_id /* Obviously, I guessed on the column names */
    WHERE c.contactstatus = '".$status."' 
        OR ctd.type = 'Appointment'

Answer 2

试

if(isset($_GET['contactstatus']) && in_array($_GET['contactstatus'], array('Hot', 'Warm', 'Cold')))
{
  $status = contacts.contactstatus = '".$_GET['contactstatus']."' ; 
} 
else 
{       
 $status = '0'; 
}

$query = "SELECT * FROM contacts,contacttodo WHERE contacttodo.id =  contacts.id AND (  ".$status." OR contacttodo.type = 'Appointment')";

需要针对您的结构更新contacttodo.id = contacts.id

Answer 3

在您对表格定义的反馈后，我正在更新我的答案。

$query = "SELECT DISTINCT t2.type FROM contacts t1, contacttodo t2 WHERE t1.id=t2.contacts_id AND (t1.contactstatus IN ('Hot', 'Warm', 'Cold') OR t2.type = 'Appointment')";

Answer 4

if(isset($_GET['contactstatus']) && in_array($_GET['contactstatus'], array('Hot', 'Warm', 'Cold'))){
    $status = "'" . $_GET['contactstatus'] . "'";
} else {
    $status = 'contact.contactstatus';
    // since contact.contactstatus ALWAYS equals contact.contactstatus
    // I wonder about the wisdom of doing it this way, but this is roughly how I do it anyway
}

$query = "SELECT *
    FROM contacts
        INNER JOIN contacttodo ON contacts.contact_id=contacttodo.contact_id
    WHERE contacts.contactstatus = " . $status . " OR contacttodo.type = 'Appointment'";

另外，从Joe的关于连接的回答中稍微补充一点，因为如果你把条件设置为true，那么在WHERE子句中加入连接条件会真的搞砸了结果。

从mysql中提取数据

4 个答案: