variable not passing

rschneid

Hello All,

I have created a script to edit rows in a table. When the editable part comes up (ser_change.php) it brings up 4 input boxes , containing the original data inputed.. two of the input boxes are drop down menus displaying automatically the intial value of inputed.. dropping the menus down we can see the entire list of choices that they can change it to. The problem i am having is when i am finshed editing .. i click amend and the variables from both the drop down boxes are not parsing to the next script (ser_changeit.php) to do further querying for the final input( ID of selected drop down box will be inputed.. not the name itself)

AS you can see i am trying to grab the selected values by using the ISSET on the ser_changeit.php however, always comes up blank while the other variables are parsing no problems(name, ip)
I believe my problem lies in the way i am displaying the value previously chosen as default and all possible values in the drop menu. but I don't know another way to do it.. Included here are both the parts of the script in question.. anyhelp would be appreicated..

Ser_change.php**********************************

echo "Office <BR>";
$sql4="SELECT * FROM sites LEFT JOIN servers ON servers.site_ID = sites.site_ID where servers.server_ID = '$row_id' ";
$mysql_result4 = mysql_query($sql4,$connection);
while ($row4=mysql_fetch_array($mysql_result4))

{
$site_name=$row4["site"];

}

echo "<SELECT NAME=\site_name\">";
echo "<OPTION SELECTED>$site_name";
$mysql_result3 = mysql_query($sql3,$connection);
while ($row3=mysql_fetch_array($mysql_result3))

{
$site=$row3["site"];

echo "<OPTION >$site";
}
echo "</SELECT>";
echo "<BR><BR>";

}

echo "Server Type <BR>";
$sql5="SELECT * FROM server_types LEFT JOIN servers ON servers.server_type = server_types.type_ID where servers.server_ID = '$row_id' ";
$mysql_result5 = mysql_query($sql5,$connection);
while ($row5=mysql_fetch_array($mysql_result5))

{
$type_nameALL=$row5["type_name"];

}

echo "<SELECT NAME=\type_name\">";
echo "<OPTION SELECTED>$type_nameALL";
$mysql_result2=mysql_query($sql2,$connection);
$num_rows2=mysql_num_rows($mysql_result2);

while ($row2=mysql_fetch_array($mysql_result2))

{
$type_name=$row2["type_name"];

echo "<OPTION>$type_name";
}
echo "</SELECT>";

    echo "<BR><BR>";

echo "<INPUT TYPE= SUBMIT VALUE=\"Amend Record\">";

mysql_close($connection);
?>
</FORM>
</BODY>
</HTML>

ser_changeit.php*******************************

<BODY>
<?php

_it.php

include file

include ("magops_info.php");

if we are sending back a protect form contents

if (isset($_GET['row_id']))

$row_id = $_GET['row_id'] ;

if (isset($_GET['type_ID']))

$type_ID = $_GET['type_ID'] ;

if (isset($_GET['site_name']))

$site_name = $_GET['site_name'] ;

if (isset($_GET['name']))

$name = $_GET['name'] ;

if (isset($_GET['ip']))

$ip = $_GET['ip'] ;

$sql5="SELECT name FROM servers Where name LIKE '$name'";
$mysql_result5=mysql_query($sql5,$connection);
$num_rows5=mysql_num_rows($mysql_result5);

$sql="SELECT site_ID, site FROM sites Where site='$site_name'";
$sql2="SELECT type_ID, type_name FROM server_types Where type_name ='$type_name'";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);

sarahk

you have some errors in your code, see comments below
Ser_change.php**********************************



echo "Office <BR>"; 
$sql4="SELECT * FROM sites LEFT JOIN servers ON servers.site_ID = sites.site_ID where servers.server_ID = '$row_id' "; 
$mysql_result4 = mysql_query($sql4,$connection); 
$mysql_result3 = mysql_query($sql3,$connection); 
//only want one row so we can use if not while
if ($row4=mysql_fetch_array($mysql_result4)) 
{ 
  $site_name=$row4['site']; 
} 
else $site_name = 'X';
//mismatched quotes around site_name
echo "<SELECT NAME='site_name'>"; 

//echo "<OPTION SELECTED>$site_name"; 
while ($row3=mysql_fetch_array($mysql_result3)) 
{ 
  $site=$row3["site"]; 
  // $selected will be an empty string if the row site is not
  // the current site.
  $selected = ($site == $site_name)? ' selected': '';
  echo "<OPTION {$selected}>{$site}"; 
} 
echo "</SELECT>"; 
echo "<BR><BR>"; 

echo "Server Type <BR>"; 
$sql5="SELECT * FROM server_types LEFT JOIN servers ON servers.server_type = server_types.type_ID where servers.server_ID = '$row_id' "; 
$mysql_result5 = mysql_query($sql5,$connection); 

//only want one row so we can use if not while
if ($row5=mysql_fetch_array($mysql_result5)) 
{ 
  $type_nameALL=$row5["type_name"]; 
} 

//mismatched quotes around type_name
echo "<SELECT NAME='type_name'>"; 
//echo "<OPTION SELECTED>$type_nameALL"; 
$mysql_result2=mysql_query($sql2,$connection); 
$num_rows2=mysql_num_rows($mysql_result2); 

while ($row2=mysql_fetch_array($mysql_result2)) 
{ 
  $type_name=$row2["type_name"]; 
  $selected = ($type_name == $type_nameALL)? ' selected' : '';
  echo "<OPTION {$selected}>{$type_name}"; 
} 
echo "</SELECT>"; 
echo "<BR><BR>"; 

echo '<INPUT TYPE= SUBMIT VALUE="Amend Record">'; 
mysql_close($connection); 
?> 
</FORM> 
</BODY> 
</HTML>

tekky

you may also want to consider using <form method=get action=ser_changeit.php>

post is the prefered method... maybe I missed it but i didnt see it in the code above....

sarahk

the form was submitting ok so I assumed that had been left off the snippet.

rschneid

Hi Guys, thanks so much for the replies!! Really appreciate it. I made the changes sarah specified however.. the variables are still not parsing.. I assumed that the new variable would be $selected (change to selectedSite and selectedType ) so I made the changes on the ser_changeit.php to relfect this but still no dice.. i did notice however that the url is grabbing the correct selected field as shown here....(what are those variables then?_

http://localhost/magops/test/ser_changeit.php?name=emeagc184muc&ip=123.1234&%5C%27site%27%22=paris&%5C%27type_name%27%22=Domain+Controller

attached in zip file are both scripts.. thanks again for the help