2011-05-19 44 views
0

我有一个表单将提交将数据插入数据库(公司详细信息)的信息。提交工作正常,但我也发现进入包含表单的页面会在数据库中插入一个空行。任何想法,为什么这可能是?代码如下:在PHP - 数据库在载入页面时产生空白行

if(!empty($_POST)){ 
... 
} 

包装所有的底码,因此这种方式只有在有后它会做插件:

<?php 
echo "<form action='addnew.php' method='post'>"; 
echo "<input type='text' name='categoryAdd' value='Travel'/><br/>"; 
echo "<input type='text' name='EstablishmentNameAdd' value='EstablishmentName'/><br/>"; 
echo "<input type='text' name='Address1Add' value='Address1'/><br/>"; 
echo "<input type='text' name='Address2Add' value='Address2'/><br/>"; 
echo "<input type='text' name='Address3Add' value='Address3'/><br/>"; 
echo "<input type='text' name='Address4Add' value='Address4'/><br/>"; 
echo "<input type='text' name='PostcodeAdd' value='Postcode'/><br/>"; 
echo "<input type='text' name='NearestStationAdd' value='NearestStation'/><br/>"; 
echo "<input type='text' name='TelAdd' value='Tel'/><br/>"; 
echo "<input type='text' name='FaxAdd' value='Fax'/><br/>"; 
echo "<input type='text' name='EmailAdd' value='Email'/><br/>"; 
echo "<input type='text' name='WebsiteAdd' value='Website'/><br/>"; 
echo "<input type='text' name='DescriptionAdd' value='Description'/><br/>"; 
echo "<input type='submit' value='test'/>"; 
echo "</form>"; 

$EstablishmentNameAdd = mysql_real_escape_string($_POST['EstablishmentNameAdd']); 
$CategoryAdd = $_POST['categoryAdd']; 
$Address1Add = mysql_real_escape_string($_POST['Address1Add']); 
$Address2Add = mysql_real_escape_string($_POST['Address2Add']); 
$Address3Add = mysql_real_escape_string($_POST['Address3Add']); 
$Address4Add = mysql_real_escape_string($_POST['Address4Add']); 
$PostcodeAdd = $_POST['PostcodeAdd']; 
$NearestStationAdd = mysql_real_escape_string($_POST['NearestStationAdd']); 
$TelAdd = $_POST['TelAdd']; 
$FaxAdd = $_POST['FaxAdd']; 
$EmailAdd = mysql_real_escape_string($_POST['EmailAdd']); 
$WebsiteAdd = mysql_real_escape_string($_POST['WebsiteAdd']); 
$DescriptionAdd = mysql_real_escape_string($_POST['DescriptionAdd']); 

$result1 = mysql_query("INSERT into establishment_id (EstablishmentName) values ('$EstablishmentNameAdd')"); 
$result2 = mysql_query("SELECT * from establishment_id where EstablishmentName = '$EstablishmentNameAdd'"); 
while($row = mysql_fetch_array($result2)) 
    { 
$ID = $row['EstablishmentID']; 
$NAME = $row['EstablishmentName']; 

}; 
$result3 = mysql_query("INSERT into establishmentdetails (EstablishmentID, EstablishmentName, category, Address1, Address2, Address3, Address4, Postcode, NearestStation, Tel, Fax, 
Email, Website, Description) values('$ID', '$EstablishmentNameAdd', '$CategoryAdd', '$Address1Add', '$Address2Add', '$Address3Add', '$Address4Add', '$PostcodeAdd', '$NearestStationAdd', '$TelAdd', '$FaxAdd', '$EmailAdd', 
'$WebsiteAdd', '$DescriptionAdd')"); 

?> 

回答

1

所有的回声后添加此。

+0

这是一个好办法做到这一点。 'if($ _SERVER ['REQUEST_METHOD'] =='POST')'保证工作。如果提交内容合法为空,则您的失败。 – 2011-05-19 20:03:37

+1

@MarcB,是的,但是OP不希望插入一个空的帖子,那就是问题所在! – Neal 2011-05-19 20:04:30

+0

实际上,他正在GET上插入,这是整个问题。简单地说“不要空POST”不会停止空白字段。他将不得不添加更多的服务器端验证。 – 2011-05-19 20:06:43

0

发生这种情况是因为您正在回显表单,然后立即将其插入到您的数据库中,但尚不存在。

你可以试试这个:

if(count($_POST) > 0 && in_array('EstablishmentNameAdd',$_POST)) { 
$EstablishmentNameAdd = mysql_real_escape_string($_POST['EstablishmentNameAdd']); 
$CategoryAdd = $_POST['categoryAdd']; 
$Address1Add = mysql_real_escape_string($_POST['Address1Add']); 
$Address2Add = mysql_real_escape_string($_POST['Address2Add']); 
$Address3Add = mysql_real_escape_string($_POST['Address3Add']); 
$Address4Add = mysql_real_escape_string($_POST['Address4Add']); 
$PostcodeAdd = $_POST['PostcodeAdd']; 
$NearestStationAdd = mysql_real_escape_string($_POST['NearestStationAdd']); 
$TelAdd = $_POST['TelAdd']; 
$FaxAdd = $_POST['FaxAdd']; 
$EmailAdd = mysql_real_escape_string($_POST['EmailAdd']); 
$WebsiteAdd = mysql_real_escape_string($_POST['WebsiteAdd']); 
$DescriptionAdd = mysql_real_escape_string($_POST['DescriptionAdd']); 

$result1 = mysql_query("INSERT into establishment_id (EstablishmentName) values ('$EstablishmentNameAdd')"); 
$result2 = mysql_query("SELECT * from establishment_id where EstablishmentName = '$EstablishmentNameAdd'"); 
while($row = mysql_fetch_array($result2)) 
    { 
$ID = $row['EstablishmentID']; 
$NAME = $row['EstablishmentName']; 

}; 
$result3 = mysql_query("INSERT into establishmentdetails (EstablishmentID, EstablishmentName, category, Address1, Address2, Address3, Address4, Postcode, NearestStation, Tel, Fax, 
Email, Website, Description) values('$ID', '$EstablishmentNameAdd', '$CategoryAdd', '$Address1Add', '$Address2Add', '$Address3Add', '$Address4Add', '$PostcodeAdd', '$NearestStationAdd', '$TelAdd', '$FaxAdd', '$EmailAdd', 
'$WebsiteAdd', '$DescriptionAdd')"); 
} 
+0

:P对不起,这是一个错字..我拉了!谢谢 – 2011-05-19 20:06:29

+0

你为什么选择'EstablishmentNameAdd'作为该帖子的警告? – Neal 2011-05-19 20:07:00

+0

只是一个例子..我选择了其他人,因为它似乎最有可能是必要的。例如,你可以有一个企业,但没有一个网站。或者,也许你正在让你的机构成立..所以你还没有电话号码。但是,所有的机构都会有一个名称。在任何时候,与该机构有关的每个人都会知道该名称。 – 2011-05-19 20:11:11