Solved

duplicate entries in a drop down list and search results PHP

Posted on 2008-06-25
9
565 Views
Last Modified: 2013-12-12
Hi,
I have some code (attached) that displays a date extracted from a MySQL database (the format gets changed.)
Does someone know how to add some code to omit the duplicate entries?  That is, all entries that come up as January 2008 when the format is changed will only be displayed once?

When doing this, does it affect the searching, or will all matches be displayed?

I have added the section of the dropdown list code and all the Search code/

Thanks for your help
----------------------------
List.php
----------------------------
<select name="datelist"><option value="0" selected>Happening in...</option> 
<?php
$result = mysql_query("SELECT DATE_FORMAT(StartDate,'%Y-%m-%d') as valDate,  DATE_FORMAT(StartDate,'%M-%Y') as displayDate from event");
 
while ($row = mysql_fetch_assoc($result)) {
/*
    $date_pieces = explode("-",$row["StartDate"]);
        $tstamp = mktime(0,0,0,$date_pieces[1],$date_pieces[2],$date_pieces[0]);
        $displaydate = date("F-Y",$tstamp); 
*/
        ?>
 
    <option value="<?=$row['valDate']?>"><?=$row['displayDate']?></option>
    <?php
        } 
        ?>
	   </select>
<input type="submit" value="Submit">
 
--------------------------------------------
Search.php
--------------------------------------------
 
<?php
include_once("Settings.php"); //Path to Settings.php
 
$ErrorCount = 0;
 
if (isset($_SERVER['REQUEST_METHOD'])) { //The action of the form passes ?action=insert on the end of the URL. This line checks for that to see if the form has been submitted
   //Tells the page if the record should be inserted
 
   if (!empty($_REQUEST["typelist"])) { //If an input named EventType has been submitted
      $EventType = $_REQUEST["typelist"]; //Store its value in $EventType
   } else {
      $EventType = ""; //Declare $EventType as an empty string to prevent errors later
      echo "You must select an event type.";
      $ErrorCount++; //Increment the error count
 
   }
   
   if (!empty($_REQUEST["locationlist"])) { //If an input named Country has been submitted
      $Country = $_REQUEST["locationlist"]; //Store its value in $Country
   } else {
      $Country = ""; //Declare $Country as an empty string to prevent errors later
      echo "You must select a country.";
      $ErrorCount++; //Increment the error count
 
   }
   
   if (!empty($_REQUEST["datelist"])) { //If an input named StartDate has been submitted
      $StartDate = $_REQUEST["datelist"]; //Store its value in $StartDate
   } else {
      $StartDate = ""; //Declare $StartDate as an empty string to prevent errors later
      echo "You must select a start date.";
      $ErrorCount++; //Increment the error count
   }
   
}
 
//If the for was submitted with no errors, run the SQL query and display the results.
if ($ErrorCount == 0) {
 
?>
<table>
<tr>
<td>Event Type</td>
<td>City</td>
<td>Country</td>
<td>StartDate</td>
<td>EndDate</td>
<td>Title</td>
<td>Description</td>
<td>WebLink</td>
<td>BuyTicket</td>
<td>Images</td>
</tr>
<?
      $sql= "SELECT * FROM event WHERE EventType = '" . $EventType . "' AND Country = '" . $Country . "' AND DATE_FORMAT(StartDate,'%Y-%m-%d') = '" . $StartDate . "' ORDER BY StartDate ASC";
 
//echo "SQL Statement: " . $sql ; 
 
$result = mysql_query($sql) or die("Sql Query Failed: " . mysql_error());
 
while ($row = mysql_fetch_assoc($result)){
      //Ive echo'd all of the fields below without any HTML formating, so you can build your HTML around them.
?>
<tr>
<td><? echo $row["EventType"]; ?>
<td><? echo $row["City"]; ?>
<td><? echo $row["Country"]; ?>
<td><? echo $row["StartDate"]; ?>
<td><? echo $row["EndDate"]; ?>
<td><? echo $row["Title"]; ?>
<td><? echo $row["Description"]; ?>
<td><? echo $row["WebLink"]; ?>
<td><? echo $row["BuyTicket"]; ?>
<td><? echo $row["Images"]; ?>
</tr>
<?
     }
}
?>
</table>
<?
@mysql_close($connection);
?>

Open in new window

0
Comment
Question by:Amanda Watson
9 Comments
 
LVL 82

Expert Comment

by:hielo
ID: 21870902
Instead of:
$result = mysql_query("SELECT DATE_FORMAT(StartDate,'%Y-%m-%d') as valDate,  DATE_FORMAT(StartDate,'%M-%Y') as displayDate from event");

try:
$result = mysql_query("SELECT DISTINCT DATE_FORMAT(StartDate,'%Y-%m-%d') as valDate,  DATE_FORMAT(StartDate,'%M-%Y') as displayDate from event");
0
 
LVL 11

Author Comment

by:Amanda Watson
ID: 21870996
Hi,
Well if you look at
http://threerobots.com/List.php

You can see that it worked for the entries that were already in the table,  but then I added another entry and the date is being duplicated.

The new entry has got a different Country too, so I am not sure what is going on?

I have attached the tables I am testing.

Can you see what is wrong?
robot4.jpg
0
 
LVL 11

Author Comment

by:Amanda Watson
ID: 21871195
Hi, can you let me know if you are abadoning this question as the line of questioning is changing a bit and you were successful on the immediate question.

It has got a bit more complicated so I am happy to finish this question and re-ask another one?
0
Live: Real-Time Solutions, Start Here

Receive instant 1:1 support from technology experts, using our real-time conversation and whiteboard interface. Your first 5 minutes are always free.

 
LVL 11

Author Comment

by:Amanda Watson
ID: 21871464
I have changed the database now.
Its not working!

Any other ideas?
0
 
LVL 16

Expert Comment

by:CWS (haripriya)
ID: 21874397
Try this:

$result = mysql_query("SELECT DISTINCT DATE_FORMAT(StartDate,'%Y-%m-%d') as valDate,  DISTINCT DATE_FORMAT(StartDate,'%M-%Y') as displayDate from event");
0
 
LVL 16

Expert Comment

by:CWS (haripriya)
ID: 21874438
I saw the html code and it show like this:

    <option value="2008-01-03">January-2008</option>
     
    <option value="2008-07-07">July-2008</option>
     
    <option value="2008-01-06">January-2008</option>

You are having two dates in the month January-2008, so it will display 'January-2008' twice. What you have at present is correct only. If you want to display 'January-2008' only once, then you cannot have two dates "2008-01-03" and "2008-01-06"
0
 
LVL 11

Author Comment

by:Amanda Watson
ID: 21878597
So is ther a way to then get the date lists to retreive the earliest and latest dates from the database and fill in the gap. (Which will stop the duplicate dates).
0
 
LVL 5

Accepted Solution

by:
mms_master earned 500 total points
ID: 21879044
>>>So is ther a way to then get the date lists to retreive the earliest and latest dates from the database and fill in the gap. (Which will stop the duplicate dates).

I have allready done this for you in your other question. I will now change the image from BLOB to varchar and use a folder to store the images instead of the database. I will also fix the links for you. Should be complete tomorrow. I will upload it in your first question when its done.

mms_master
0
 
LVL 11

Author Comment

by:Amanda Watson
ID: 21887922
0

Featured Post

Live: Real-Time Solutions, Start Here

Receive instant 1:1 support from technology experts, using our real-time conversation and whiteboard interface. Your first 5 minutes are always free.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Why do people dis php? 5 49
mysql update statement 3 22
Download tables into separate sheets 3 27
Checkout Page Input Field not aligned 1 22
Generating table dynamically is the most common issue faced by php developers.... So it seems there is a need of an article that explains the basic concept of generating tables dynamically. It just requires a basic knowledge of html and little maths…
Password hashing is better than message digests or encryption, and you should be using it instead of message digests or encryption.  Find out why and how in this article, which supplements the original article on PHP Client Registration, Login, Logo…
Explain concepts important to validation of email addresses with regular expressions. Applies to most languages/tools that uses regular expressions. Consider email address RFCs: Look at HTML5 form input element (with type=email) regex pattern: T…
The viewer will learn how to count occurrences of each item in an array.

776 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question