使用php mysql和内部联接复制结果
[英]Duplication in results using php mysql and inner join
问题描述
我有三个下拉列表，其中包含从3个不同表中从数据库中检索到的值
省：
- Governorosrate_id
- Governorate_name
区：
- district_id
- 区名
村：
- ID
- village_name
Memebrs：
- 用户身份
- 用户名
- 省
- 区
- 当用户选择三个或全部一个下拉列表中的一个时，我要的是系统必须显示与所选之一相关的结果
但是问题是，当用户从省选择时，它工作正常，但是当选择省和地区时，它会复制与所选值相关的结果；如果用户选择省和地区，结果将变成三倍
任何人都可以帮助我吗？
我将显示三种类型的所有代码
如果有人抱怨，那就不要打败我
码：
由省
//**********search by locationn***************************************//
if(isset($_POST['listbyq']))
{
//********************by governorate**************************************//
if($_POST['listbyq']=="by_gov")
{
$bygov = $_POST['governorate'];
$sql = mysql_query("SELECT user_id,first_name, last_name, birth_date, registered_date,
s.specialization_name,
g.governorate_name,
d.district_name,
v.village_name
FROM members u
INNER JOIN specialization s
ON u.specialization = s.specialization_id
INNER JOIN governorate g
ON u.governorate = g.governorate_id
INNER JOIN districts d
ON u.district = d.district_id
INNER JOIN village v
ON u.village = v.id
WHERE governorate = '$bygov'")or die(mysql_error("Error: querying the governorate"));
$num_row = mysql_num_rows($sql);
if($num_row > 0 )
{
while($row = mysql_fetch_array($sql))
{
$row_id = $row['user_id'];
$row_first_name = $row['first_name'];
$row_last_name = $row['last_name'];
$row_birthdate = $row['birth_date'];
$row_registered_date = $row['registered_date'];
$row_spec = $row['specialization_name'];
$row_gov = $row['governorate_name'];
$row_dist = $row['district_name'];
$row_village = $row['village_name'];
////***********for the upload image*************************//
$check_pic="members/$row_id/image01.jpg";
$default_pic="members/0/image01.jpg";
if(file_exists($check_pic))
{
$user_pic="<img src=\"$check_pic\"width=\"120px\"/>";
}
else
{
$user_pic="<img src=\"$default_pic\"width=\"120px\"/>";
}
$outputlist.='
<table width="100%">
<tr>
<td width="23%" rowspan="5"><div style="height:120px;overflow:hidden;"><a href = "http://localhost/newadamKhoury/profile.php?user_id='.$row_id.'" target="_blank">'.$user_pic.'</a></div></td>
<td width="14%"><div align="right">Name:</div></td>
<td width="63%"><a href = "http://localhost/newadamKhoury/profile.php?user_id='.$row_id.'" target="_blank">'.$row_first_name.' '.$row_last_name.'</a></td>
</tr>
<tr>
<td><div align="right">Birth date:</div></td>
<td>'.$row_birthdate.'</td>
</tr>
<tr>
<td><div align="right">Registered:</div></td>
<td>'.$row_registered_date.'</td>
</tr>
<tr>
<td><div align="right">Job:</div></td>
<td>'.$row_spec.'</td>
</tr>
<tr>
<td><div align="right">Location:</div></td>
<td>'.$row_gov.'__'.$row_dist.'__'.$row_village.'</td>
</tr>
</table>
<hr />
';
}
}
}
else
{
$errorMSG = "No member within this selected governorate";
}
按地区
if($_POST['listbyqa']=="by_dist")
{
@ $bydist = $_POST['district'];
$sql = mysql_query("SELECT user_id,first_name, last_name, birth_date, registered_date,
s.specialization_name,
g.governorate_name,
d.district_name,
v.village_name
FROM members u
INNER JOIN specialization s
ON u.specialization = s.specialization_id
INNER JOIN governorate g
ON u.governorate = g.governorate_id
INNER JOIN districts d
ON u.district = d.district_id
INNER JOIN village v
ON u.village = v.id
WHERE district = '$bydist'")or die(mysql_error("Error: querying the district"));
$num_row = mysql_num_rows($sql);
if($num_row > 0 )
{
while($row = mysql_fetch_array($sql))
{
$row_id = $row['user_id'];
$row_first_name = $row['first_name'];
$row_last_name = $row['last_name'];
$row_birthdate = $row['birth_date'];
$row_registered_date = $row['registered_date'];
$row_spec = $row['specialization_name'];
$row_gov = $row['governorate_name'];
$row_dist = $row['district_name'];
$row_village = $row['village_name'];
////***********for the upload image*************************//
$check_pic="members/$row_id/image01.jpg";
$default_pic="members/0/image01.jpg";
if(file_exists($check_pic))
{
$user_pic="<img src=\"$check_pic\"width=\"120px\"/>";
}
else
{
$user_pic="<img src=\"$default_pic\"width=\"120px\"/>";
}
$outputlist.='
<table width="100%">
<tr>
<td width="23%" rowspan="5"><div style="height:120px;overflow:hidden;"><a href = "http://localhost/newadamKhoury/profile.php?user_id='.$row_id.'" target="_blank">'.$user_pic.'</a></div></td>
<td width="14%"><div align="right">Name:</div></td>
<td width="63%"><a href = "http://localhost/newadamKhoury/profile.php?user_id='.$row_id.'" target="_blank">'.$row_first_name.' '.$row_last_name.'</a></td>
</tr>
<tr>
<td><div align="right">Birth date:</div></td>
<td>'.$row_birthdate.'</td>
</tr>
<tr>
<td><div align="right">Registered:</div></td>
<td>'.$row_registered_date.'</td>
</tr>
<tr>
<td><div align="right">Job:</div></td>
<td>'.$row_spec.'</td>
</tr>
<tr>
<td><div align="right">Location:</div></td>
<td>'.$row_gov.'__'.$row_dist.'__'.$row_village.'</td>
</tr>
</table>
<hr />
';
}
}
}
else
{
$errorMSG = "No member within this selected District";
}
按村庄
if($_POST['listbyqb']=="by_city")
{
echo $_POST['listbyqb'];
@ $byvillage = $_POST['village'];
$sql = mysql_query("SELECT user_id,first_name, last_name, birth_date, registered_date,
s.specialization_name,
g.governorate_name,
d.district_name,
v.village_name
FROM members u
INNER JOIN specialization s
ON u.specialization = s.specialization_id
INNER JOIN governorate g
ON u.governorate = g.governorate_id
INNER JOIN districts d
ON u.district = d.district_id
INNER JOIN village v
ON u.village = v.id
WHERE village = '$byvillage'")or die(mysql_error("Error: querying the district"));
$num_row = mysql_num_rows($sql);
if($num_row > 0 )
{
while($row = mysql_fetch_array($sql))
{
$row_id = $row['user_id'];
$row_first_name = $row['first_name'];
$row_last_name = $row['last_name'];
$row_birthdate = $row['birth_date'];
$row_registered_date = $row['registered_date'];
$row_spec = $row['specialization_name'];
$row_gov = $row['governorate_name'];
$row_dist = $row['district_name'];
$row_village = $row['village_name'];
////***********for the upload image*************************//
$check_pic="members/$row_id/image01.jpg";
$default_pic="members/0/image01.jpg";
if(file_exists($check_pic))
{
$user_pic="<img src=\"$check_pic\"width=\"120px\"/>";
}
else
{
$user_pic="<img src=\"$default_pic\"width=\"120px\"/>";
}
$outputlist.='
<table width="100%">
<tr>
<td width="23%" rowspan="5"><div style="height:120px;overflow:hidden;"><a href = "http://localhost/newadamKhoury/profile.php?user_id='.$row_id.'" target="_blank">'.$user_pic.'</a></div></td>
<td width="14%"><div align="right">Name:</div></td>
<td width="63%"><a href = "http://localhost/newadamKhoury/profile.php?user_id='.$row_id.'" target="_blank">'.$row_first_name.' '.$row_last_name.'</a></td>
</tr>
<tr>
<td><div align="right">Birth date:</div></td>
<td>'.$row_birthdate.'</td>
</tr>
<tr>
<td><div align="right">Registered:</div></td>
<td>'.$row_registered_date.'</td>
</tr>
<tr>
<td><div align="right">Job:</div></td>
<td>'.$row_spec.'</td>
</tr>
<tr>
<td><div align="right">Location:</div></td>
<td>'.$row_gov.'__'.$row_dist.'__'.$row_village.'</td>
</tr>
</table>
<hr />
';
}
}
}
else
{
$errorMSG = "No member within this selected District";
}
}
1 个解决方案
解决方案1
0 2013-05-19 03:00:48
一些简单的事情值得尝试：
echo或var_dump $ _POST结果，以确保您正在获取认为自己正在获取的数据。 这可能是造成问题的原因。 您可能会获得不止一个
杂乱的代码很可能是导致此问题的原因。 您具有三个输出部分，因此有三个可能的输出，当满足适当的IF条件时，可以触发所有这些输出。 尝试在每个块中回显“ Here”，“ There”和“ Everywhere”或您选择的单词-您很快就会看到可疑代码在哪里！
更好的选择可能是将where子句更改为：
WHERE village='[villagename from post]' OR district='[districtname from post'] OR governorate='[govname from post]'
这样，您只需要一个中频和一个输出。 它的伸缩性也更好。
尝试运行每个带有手动插入的测试数据的sql（PHP myAdmin对此非常方便），然后查看问题是否出在sql或php级别。
@-尝试避免。 最好以不会产生错误的方式编写代码。 您是否将其放置在那里以获取空索引错误？ 如果是这样，这可能是您的问题。 检查您的HTML并查看上面的答案1。
使用php mysql和内部连接的syantax错误
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