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Sep 21st, 2003, 04:41 AM
#1
Thread Starter
Lively Member
parse error...
Well, i'm uploading the script to my server, and i'm getting this error when i try an access the script.
Parse error: parse error in /home/jclegg/public_html/st/addsoldier.php on line 13
my script is:
<?
//config//
$user = "usergoeshere";
$pass = "passgoeshere";
$db = "dbnamegoeshere";
//dont change anything under this//
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );
if ( $action == "" ) {
</form>
<form name="form1" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Real Name:
<input name="realname" type="text" id="realname">
</p>
</blockquote>
</form>
<form name="form2" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Americas Army name (if you play)
<input name="americasarmyname" type="text" id="americasarmyname">
</p>
</blockquote>
</form>
<form name="form3" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Military branch::
<input name="militarybranch" type="text" id="militarybranch">
</p>
</blockquote>
</form>
<form name="form4" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Rank:
<input name="rank" type="text" id="rank">
</p>
</blockquote>
</form>
<form name="form5" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">MOS (occupation civilian terms)
<input name="mos" type="text" id="mos">
</p>
</blockquote>
</form>
<form name="form6" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Based at (ship if Navy)
<input name="basedat" type="text" id="basedat">
</p>
</blockquote>
</form>
<blockquote>
<form name="form8" method="post" action="">
<div align="center">
<p>How long have you served? (or serving)
<input name="servedanserving" type="text" id="servedanserving">
</p>
<p> </p>
</div>
</form>
<form name="form7" method="post" action="add">
<div align="center">
<input type="submit" name="Submit" value="Submit">
</div>
</form>
<p align="center"> </p>
</blockquote>
}
if ( $action == "add" ) {
mysql_query("INSERT into tblmil ( fdrealn,fdaan,fdmilbranch,fdrank,fdmos,fdbased,fdserving) values ( '$Information has been added' );
echo "$Information has been added";
}
mysql_close( $link );
?>
PS: and of course i have the things filled up above.
To get somthing out of life, you always need to be persistant.
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Sep 21st, 2003, 10:13 AM
#2
Conquistador
<?
//config//
$user = "usergoeshere";
$pass = "passgoeshere";
$db = "dbnamegoeshere";
//dont change anything under this//
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );
if ( $action == "" ) {
?>
</form>
<form name="form1" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Real Name:
<input name="realname" type="text" id="realname">
</p>
</blockquote>
</form>
<form name="form2" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Americas Army name (if you play)
<input name="americasarmyname" type="text" id="americasarmyname">
</p>
</blockquote>
</form>
<form name="form3" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Military branch::
<input name="militarybranch" type="text" id="militarybranch">
</p>
</blockquote>
</form>
<form name="form4" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Rank:
<input name="rank" type="text" id="rank">
</p>
</blockquote>
</form>
<form name="form5" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">MOS (occupation civilian terms)
<input name="mos" type="text" id="mos">
</p>
</blockquote>
</form>
<form name="form6" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Based at (ship if Navy)
<input name="basedat" type="text" id="basedat">
</p>
</blockquote>
</form>
<blockquote>
<form name="form8" method="post" action="">
<div align="center">
<p>How long have you served? (or serving)
<input name="servedanserving" type="text" id="servedanserving">
</p>
<p> </p>
</div>
</form>
<form name="form7" method="post" action="add">
<div align="center">
<input type="submit" name="Submit" value="Submit">
</div>
</form>
<p align="center"> </p>
</blockquote>
<?
}
if ( $action == "add" ) {
mysql_query("INSERT into tblmil ( fdrealn,fdaan,fdmilbranch,fdrank,fdmos,fdbased,fdserving) values ( '$Information has been added' );
echo "$Information has been added";
}
mysql_close( $link );
?>
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Sep 21st, 2003, 11:07 AM
#3
Thread Starter
Lively Member
I keep getting parse errors on like their was one on line 82, well their was nothing on line 82.... then 81 nothin on 81 then, um...tried deletn couple things, that didn't work...
To get somthing out of life, you always need to be persistant.
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Sep 21st, 2003, 11:13 AM
#4
Frenzied Member
that is becasue you forgot a quote on the end of this line
mysql_query("INSERT into tblmil ( fdrealn,fdaan,fdmilbranch,fdrank,fdmos,fdbased,fdserving) values ( '$Information has been added') ");
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Sep 21st, 2003, 11:22 AM
#5
Thread Starter
Lively Member
when i hit "Submit" it goes to a 404 page... what do I do? cuz now it shows all the stuff so i filled it out to test it, so (thank you that works now! ) but i hit submit an it just gives me a 404!
oh an when i hit submit it doens't add the data to the db
Last edited by johnsx2002; Sep 21st, 2003 at 11:25 AM.
To get somthing out of life, you always need to be persistant.
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Sep 21st, 2003, 11:38 AM
#6
Frenzied Member
I just really looked at the code and found you have way to many form tags. you only use 1
this is where your page is going
<form name="form7" method="post" action="add">
nowhere.
try this
PHP Code:
<?
//config//
$user = "usergoeshere";
$pass = "passgoeshere";
$db = "dbnamegoeshere";
//dont change anything under this//
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );
if ( $action == "" ) {
?>
<form name="form1" method="post" action="addsoldier.php?action=add">
<blockquote>
<p align="center">Real Name:
<input name="realname" type="text" id="realname">
</p>
</blockquote>
<blockquote>
<p align="center">Americas Army name (if you play)
<input name="americasarmyname" type="text" id="americasarmyname">
</p>
</blockquote>
<blockquote>
<p align="center">Military branch::
<input name="militarybranch" type="text" id="militarybranch">
</p>
</blockquote>
<blockquote>
<p align="center">Rank:
<input name="rank" type="text" id="rank">
</p>
</blockquote>
<blockquote>
<p align="center">MOS (occupation civilian terms)
<input name="mos" type="text" id="mos">
</p>
</blockquote>
<blockquote>
<p align="center">Based at (ship if Navy)
<input name="basedat" type="text" id="basedat">
</p>
</blockquote>
<blockquote>
<div align="center">
<p>How long have you served? (or serving)
<input name="servedanserving" type="text" id="servedanserving">
</p>
<p> </p>
</div>
<div align="center">
<input type="submit" name="Submit" value="Submit">
</div>
</form>
<p align="center"> </p>
</blockquote>
<?
}
if ( $action == "add" ) {
mysql_query("INSERT into tblmil ( fdrealn,fdaan,fdmilbranch,fdrank,fdmos,fdbased,fdserving) values ( '$Information has been added' ) ");
echo "$Information has been added";
}
mysql_close( $link );
?>
then to have it add you can't use
$Information has been added
as that is nothing and will error out. you need to fix your insert query to add the correct variables to the correct columns. and that all depeends on if you have register_globals off or on. but seeing as you are only using $action it might be on and that is a security issue.
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Sep 21st, 2003, 02:55 PM
#7
Thread Starter
Lively Member
I think its on, is that a problme? how can i fix it an how can i make it add i am a complete php nooobie! please help me brotha!
To get somthing out of life, you always need to be persistant.
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Sep 21st, 2003, 04:00 PM
#8
Frenzied Member
it won't hurt you to read a little on the mysql and php sites either. on this line do this
PHP Code:
if ( $action == "add" ) {
mysql_query("INSERT into tblmil ( fdrealn,fdaan,fdmilbranch,fdrank,fdmos,fdbased,fdserving) values ( '$realname','$americasarmyname','$militarybranch','$rank','$mos','$basedat','$servedanserving' ) ");
echo "$Information has been added";
}
try that.
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Sep 21st, 2003, 04:20 PM
#9
Thread Starter
Lively Member
Alright cool i'll check that out when i get home, i have a sql book an never payed attention to the sql / oracle class in HS :P too bad for me, well that'll probably do the trick, so thank ya!
To get somthing out of life, you always need to be persistant.
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