[RESOLVED] Why can't I read the Id field from my database!

[cdeszaq]

I have a databse set up, and I can read all of the other fields, but I cannot get any information out of the "Id" field.

Here is the code:

<?php

/*
********************************************************************************
Please set the MySQL connection variables below, according to your configuration
********************************************************************************
*/
$dbhost="######"; // the url of your MySQL Server
$dbuser="######"; // the username of your MySQL Server
$dbpass="######"; // the password of your MySQL Server
$dbname="news"; // the name of your MySQL Database
$tblName="articles"; //the name of your MyQSQL Database Table
$tblID="Id"; // the name of your MySQL Database Table field
$tblDate="Date"; // the name of your MySQL Database Table Date field
$tblSnipit="Snipit"; // the name of your MySQL Date Table Snipit fild
$tblText="Text"; // the name of your MySQL Info Table Text field
/*
********************************************************************************
Please use the following SQL Statement if the Date Field has a type such as 'datetime'
********************************************************************************
*/
$SQL="SELECT UNIX_TIMESTAMP(".$tblDate.") AS ".$tblDate.", ".$tblID.", ".$tblSnipit.", ".$tblText." FROM ".$tblName." ORDER BY ".$tblDate." DESC";
/*
********************************************************************************
Connect to Database and send the query
********************************************************************************
*/
$conID=@mysql_connect($dbhost,$dbuser,$dbpass);
@mysql_select_db($dbname, $conID) or die( "Unable to select database");

$sqlReturn=@mysql_query($SQL) or die( "Unable to read database");

?>

<table class="body_container">
 <tr>
  <td>
   <table align="left" bordercolor="black" border="1">
    <tr>
     <td>
      <table align="left">
       <tr>
        <td>
         <font size="14">News</font>
         <hr width="%100"></hr>
        </td>
       </tr>

<?php
/*
********************************************************************************
Gets all dates from your database and set the calendar events html classes (for the layout)
********************************************************************************
*/

$num=mysql_numrows($sqlReturn);
$i=$num;

while ($data=@mysql_fetch_array($sqlReturn, MYSQL_BOTH)){
$mysqlID=$i;//$date[$tblID] or die("Failed to read ID");
$mysqlDay=date("j",$data[$tblDate]) or die("Failed to read Date"); // makes a day out of the database date
$mysqlMonth=date("n",$data[$tblDate]) or die("Failed to read Date"); // makes a month out of the database date
$mysqlYear=date("Y",$data[$tblDate]) or die("Failed to read IDDate"); // makes a year out of the database date
$mysqlSnipit=$data[$tblSnipit] or die("Failed to read Snipit");
//$mysqlArticle=$data[$tblText];
$mysqlLink="http://www.mailletime.com/news/bob.php?id=".$mysqlID or die("Failed to set Link"); // gets the article link

$mysqlMonth=getMonthName($mysqlMonth) or die("Failed to select month"); //makes textual month from numbers

echo  "<tr><td><a href=\"".$mysqlLink."\" class=\"link_plain\"><font size=\"5\">".$mysqlMonth." ".$mysqlDay.", ".$mysqlYear."</font></a></br>".$mysqlSnipit."...<a href=\"".$mysqlLink."\" class=\"link_more\">  (More)</a></br></td></tr><tr><td><strong></strong></td></tr>";

$i--;

};
mysql_close();

?>
       <tr>
	    <td>
	     <hr width="%100"></hr>
	     Last Updated: June 18, 2005
	    </td>
       </tr>
      </table>
     </td>
    </tr>
   </table>
  </td>
 </tr>
</table>

Please help,
Thank You

[toplay]

Change this:

$mysqlID=$i;

to this:

$mysqlID = $data[$tblID];

or

$mysqlID= $data['Id'];

Please don't double post.

.

[cdeszaq]

Sorry about the double post, I did not know.

Even after I make the chage you suggested, It still fails to read the Id field. The $i is a workaround, untill I can get the code to read the Id field.

Thank you

[toplay]

Well, then your ID column in your table is not called "Id". Double check what it is. Maybe it's called "id" or "ID".

FYI - The "or die()" only works with a TRUE and FALSE return value. So remove them all from the assignment statements. Example:

Change this:

$mysqlLink="http://www.mailletime.com/news/bob.php?id=".$mysqlID or die("Failed to set Link"); // gets the article link

To this:

$mysqlLink = "http://www.mailletime.com/news/bob.php?id=$mysqlID";

Remove all the '@' signs while testing this because they suppress errors. You want to see any errors/warnings/notices.

🙂

[cdeszaq]

Thank you for helping me, but unfortunatly, it still is not working.

This line: $mysqlID=$date[$tblID];

is where the problem is, unless it is earlier. The $mysqlID variable is not getting set.

Thank You

[toplay]

cdeszaq wrote:

Thank you for helping me, but unfortunatly, it still is not working.

This line: $mysqlID=$date[$tblID];

is where the problem is, unless it is earlier. The $mysqlID variable is not getting set.

Thank You

I never said $date[$tblID].

It's $data as in:

$mysqlID = $data[$tblID];

.

[cdeszaq]

Oh my god. Such a simple typo! It was there all along, I just did not see it. THank you so much for your help, even though it was just a typo.