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[SOLVED] What wrong with this? User ID seems to wreck it...


forumnz

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Yup im sure and I realize that Im probably not meant to connect to the database 5 times =)

 

<?php require('require.php'); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Designer Vision - Adventure Plan</title>
<style type="text/css">
<!--
ul{
list-style-image:url(file:///C|/Documents and Settings/Owner/My Documents/Sam/samweb/../images/../images/arrow2.jpg);
}
.style4 {font-size: 18px; color: #3A3A3A; font-weight: bold; }
.style6 {font-size: 12px; color: #3A3A3A; text-decoration: none; }
.style11 {font-size: 22px; color: #5B5E71; font-weight: bold; }
.style13 {font-size: 14px; color: #3A3A3A; text-decoration: none; font-weight: bold; }
.style14 {color: #3A3A3A; text-decoration: none; font-size: 14px;}
.style17 {color: #333333}
.style19 {color: #000000}
.style20 {color: #5B5E71}
.style1 {	font-size: 16px;
color: #3A3A3A;
text-decoration:none;
}
.style21 {font-size: 16px; color: #3A3A3A; font-weight: bold; }
.style66 {font-size: 14px; color: #3A3A3A; text-decoration:underline; font-weight:bold; }
.style57 {font-size: 12px; color: #3A3A3A; text-decoration: underline; }
ul{
list-style-image:url(../images/arrow2.jpg);
}
.style69 {font-size: 14px; color: #FF0000; font-weight: bold; }
-->
</style>
</head>

<body background="bg.gif">
<table width="100%" border="0" align="center">
  <tr>
    <td height="24"> </td>
  </tr>
  <tr>
    <td><table width="760" height="885" border="0" align="center" cellpadding="0" cellspacing="0">
      <tr>
        <td width="760" height="240" background="../images/webtop.gif"><table width="100%" height="240
	" border="0" cellpadding="0" cellspacing="0">
            <tr>
              <td width="760" height="240"><table width="94%" height="207" border="0" align="center" cellpadding="0" cellspacing="0">
                  <tr>
                    <td height="53"> </td>
                  </tr>
                  <tr>
                    <td height="127"> </td>
                  </tr>
                  <tr>
                    <td height="20" class="style13"><div align="right" class="style14"><a href="index.php" class="style14">home</a>  <a href="services.php" class="style14">services</a>   <a href="portfolio.php" class="style14">portfolio</a>   <a href="plans.php" class="style14">plans</a>   <a href="contact.php" class="style14">contact</a>  </div></td>
                  </tr>
              </table></td>
            </tr>
        </table></td>
      </tr>
      <tr>
        <td height="605" background="../images/webrep.gif"><table width="96%" height="605" border="0">
          <tr>
            <td width="3%" height="400" rowspan="3"> </td>
            <td width="22%" rowspan="3" align="center" valign="top" class="style4"><?php include("../menu.php"); ?></td>
            <td width="4%" rowspan="3"> </td>
            <td height="544" valign="top" class="style13"><table width="100%" border="0">
              <tr>
                <td><div align="right">
                  <?php include("log.php")
			?>
                </div></td>
              </tr>
            </table>
            <p>Here is the progress of your site. </p>
              <p>Please note that this will be updated daily or whenever more work for your site is completed. </p>
              <table width="100%" border="0">
                <tr>
                  <td><?php

$database=mysql_connect("localhost","***","**");

mysql_select_db("test",$database);
//echo $userid;
$query="select * from members WHERE userid=$userid";


$result = mysql_query($query) or die(mysql_error());


while($img=mysql_fetch_assoc($result)){

$image_num=$img['stg1'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif' />";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif' />";

}else {

echo" sorry there is no images to show";

}
}
?></td>
                </tr>
                <tr>
                  <td>
			  /*<?php

$database=mysql_connect("localhost","***","**");

mysql_select_db("test",$database);
//echo $userid;
$query="select * from members WHERE userid=$userid";


$result = mysql_query($query) or die(mysql_error());


while($img=mysql_fetch_assoc($result)){

$image_num=$img['stg2'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif' />";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif' />";

}else {

echo" sorry there is no images to show";

}
}
?></td>
                </tr>
                <tr>
                  <td><?php

$database=mysql_connect("localhost","***","**");

mysql_select_db("test",$database);
//echo $userid;
$query="select * from members WHERE userid=$userid";


$result = mysql_query($query) or die(mysql_error());


while($img=mysql_fetch_assoc($result)){

$image_num=$img['stg3'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif' />";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif' />";

}else {

echo" sorry there is no images to show";

}
}
?></td>
                </tr>
                <tr>
                  <td><?php

$database=mysql_connect("localhost","***","**");

mysql_select_db("test",$database);
//echo $userid;
$query="select * from members WHERE userid=$userid";


$result = mysql_query($query) or die(mysql_error());


while($img=mysql_fetch_assoc($result)){

$image_num=$img['stg4'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif' />";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif' />";

}else {

echo" sorry there is no images to show";

}
}
?></td>
                </tr>
                <tr>
                  <td><?php

$database=mysql_connect("localhost","***","**");

mysql_select_db("test",$database);
//echo $userid;
$query="select * from members WHERE userid=$userid";


$result = mysql_query($query) or die(mysql_error());


while($img=mysql_fetch_assoc($result)){

$image_num=$img['stg5'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif' />";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif' />";

}else {

echo" sorry there is no images to show";

}
}
?>*/</td>
                </tr>
              </table>              
              <p> </p></td>
          </tr>
          
          <tr>
            <td valign="top"><div align="center"><a href="contact.php" class="style66">Contact Us</a></div></td>
          </tr>
          
          
          <tr>
            <td> </td>
          </tr>
        </table></td>
      </tr>
      <tr>
        <td width="760" height="40" background="../images/webbot.gif"> </td>
      </tr>
    </table></td>
  </tr>
</table>
</body>
</html>

As long as all your queries are from the same db you only need to connect and select db once.

 In addition you have this for every query:

$query="select * from members WHERE userid=$userid";

 

This is telling it to select nothing from the db.

Should be something like select img or something.

You need to tell the db what you want selected from it.

In your table there is a field that you want brought to the page.

What is it?

ya '$userid' would maybe change something but it could still work without them if its only numbers after all if it has a space after the number lets say  $userid= "    3";  that would give a error and you have to use quotes than but well I guess he solved it

Isnt is in this part?

 

$query="select * from members WHERE userid=$userid";

 

Anyway, I have simplified it. Here is it:

 

<?php require('require.php'); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>

<?php

$database=mysql_connect("localhost", "aaaa", "aaaaa");

mysql_select_db("test",$database);
echo $userid;
$query="select * from members WHERE userid=$userid";


$result = mysql_query($query) or die(mysql_error());


while($img=mysql_fetch_assoc($result)){

$image_num=$img['stg1'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif' />";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif' />";

}else {

echo"sorry there is no images to show";

}
}
?>

</body>
</html>

sspoke - yes i just added it and it still says

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

 

Greaser9780 - $_SESSION['userid']

It also needs to be identified in the php. The query uses userid from php to relate to your table. Usually it is defined as a post userid or a session userid.

Somewhere in the code before the query it should have

$userid= then after the equal sign you define what exactly it is.

 

What I am trying to say is, how does this php file identify with $userid?

POST from a form, a session, or does it need passed on from the previous web page?

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