PHP - Displaying Mysql Results Table Side By Side

Hi

I have got results being displayed after clicking the search button in a form on my home page but it brings up all the results which is ok but how do I get onlt the results a user searches for for example a location or property type etc as its for a property website

The coding is below for the results page

Also sorry how do I add a background image to the php page, I tried using css but wouldn't work

Code: [Select]
<style type="text/css">
body {background-image:url('images/greybgone.png');}
</style>

<?php
 
mysql_connect ("2up2downhomes.com.mysql", "2up2downhomes_c","mD8GsJKQ")  or die (mysql_error());
mysql_select_db ("2up2downhomes_c");

echo $_POST['term'];

$sql = mysql_query("select * from properties where typeProperty like '%$term%' or location like '%$term%'");

while ($row = mysql_fetch_array($sql)){
    echo 'Type of Property: '.$row['typeProperty'];
    echo '<br/> Number of Bedrooms: '.$row['bedrooms'];
    echo '<br/> Number of Bathrooms: '.$row['bathrooms'];
echo '<br/> Garden: '.$row['garden'];
echo '<br/> Description: '.$row['description'];
echo '<br/> Price: '.$row['price'];
echo '<br/> Location: '.$row['location'];
echo '<br/> Image: '.$row['image'];
    echo '<br/><br/>';
    }
 
?>

Hello everyone,

So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help.

A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right.

Some of this probably doesnt even make sense but I'll show you the kind of things I've tried...


            <select name="gender" size="1" id="gender">
              <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option>
              <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option>
            </select>



OR


     <select name="gender" id="gender">
              <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option>
              <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option>
            </select>


OR



            <select name="gender" size="1" id="gender">
              <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option>
              <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option>
            </select>


OR

            <select name="gender" id="gender">
              <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option>
              <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option>
            </select>


Honestly man, I've got no idea.


The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen.  And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol.

Any help would be tremendous and greatly appreciated.

Thanks very much, Learner

P.S  Man!

Hey guys,
Having a slight problem with part of the code in my index.php file

Code: [Select]
mysql_select_db('db_name', $con);

$result = mysql_query("SELECT * FROM spy ORDER BY id desc limit 25");
$resulto = mysql_query("SELECT * FROM spy ORDER BY id desc");
$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))
  {
?>
<div class="contentDiv">Someone is looking at <?=$row[title];?> Stats for  "<a href="/<?=$row[type];?>/<?=$row[code];?>/<?=$row[city];?>"><?=$row[code];?> <?=$row[city];?></a>"</div>
<?}?>
</div>
  <div id="login"></div>
          <? include("footer.php"); ?>
</div>
</body>
</html>
I'm getting the following error when viewing the file

Code: [Select]
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 70

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 71

where lines 70 and 71 are

$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))

Any ideas on how to fix this?

Hello,
 
I have a quick question about methods for retrieving records from a mysql table and displaying
them as a links

For example,  imagine I have three tables called countries, cities and city_info.
I'd like to be able to select a country and have a list of that country's city names returned as links. 
I'd then like to be able to click on the link for London, say, and that would trigger a mysql query to retrieve
the entry in city_info about London.

Are there any functions that allow this?

If anyone could point me in the right direction for further research I'd be grateful.

Thanks. 

Hi,
This has been baffling me for a couple hours now and i cant seem to figure it out.
I have some code which creates an array and gets info from a mysql database and then displays in a list. This works great but after adding more and more rows to my database the list is now becoming quite large and doesnt look great on my site.
Is it possible to split the list into multiple columns of about 25 and if possible once 3 or 4 columns have been created start another column underneath.
To help explain i would be looking at a layout as follows:
Code: [Select]
line 1      line 1      line 1
line 2      line 2      line 2
...      ...      ...
line 25      line 25      line 25

line 1      line 1      line 1
line 2      line 2      line 2
...      ...      ...
line 25      line 25      line 25Im guessing there should be some sort of if statement to check how many items are being displayed and to create a new column if necessary. Is this correct?

Thanks,
Alex

Hi guys,

I am wanting to do the same as Blogger.com (aka Blogspot) and Tumblr. And allow people to set their own domain names for "pages" or blog sites.

So if I was to tell them to set their own CNAME (on domain) to domains.mydomain.com

What would be the PHP part to make this work? Or how would I make this work if its not PHP?

Like Blogger and Tumblr it would not be a redirect (to view) (unless I am wrong about this).

Any and all help is much appreciated.

(PS - Am running Apache/PHP and MySQL on dedicated server).

I was wondering how does one go about showing results from SELECT query in columns in a html table.

I have a list of products in a table, and would like to show them on the page in 4 columns. I have done many searches on google to try and find the sulution, but the majority of what im finding instead is about displaying a table from phpmyadmin as a table in html.

If its a large operation to do this, I would be very happy if someone could poiint me in the direction of a tutorial maybe.

Here is the code I have so far to display the products, but for some reason, it only show 1 row instead of all the rows from my table.

Code: [Select]
<?php

$dbhost = "localhost";
$dbuser = "user";
$dbpass = "pass";
$dbname = "dbname";

mysql_connect ($dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

$result = mysql_query("SELECT * FROM mcproducts");
while($row = mysql_fetch_array($result))
 {
  $products_local_id = $row['products_local_id'];
  $productname = $row['product_name'];
  $thumburl = $row['image_from_url'];
  $productlink = $row['product_local_url'];
  $thumbnail = $row['product_image_small'];
  $currencysymbol = $row['product_currency'];
  $price = $row['product_price'];
  $flagicon = $row['product_country_from'];

 }

?>
<html>
<head>
<link href="style/stylesheet.css" rel="stylesheet" type="text/css" />
</head>
<body>

<div class="displaybox">
                <div class="productimage">
                <a href="<?php echo $productlink; ?><?php echo $products_local_id; ?>"><img src="<?php echo $thumburl; ?><?php echo $thumbnail ?>" width="150" height="150"></a>
                </div>
                <div class="productdescription">
                  <div class="pro_name">
      <a href="<?php echo $productlink ?><?php echo $products_local_id; ?>"><?php echo $productname; ?></a>
                  </div>
                  <div class="pro_description">
                  </div>
                  <?php
                  if ($flagicon=="Ireland") {
  $flagicon = "<img src=\"flags/ireland.jpg\">";
  }
  
  elseif ($flagicon=="UK") {
  $flagicon = "<img src=\"flags/uk.jpg\">";
  }  else
  echo "";
  ?>
                  <div class="pro_description"><?php echo $flagicon; ?><?php echo $currencysymbol ?> <?php echo $price ?></div>
                </div>
              </div>
             
</body>
</html>


Many thanks,

DB

Hey Everyone,
I'm creating a site that will show images uploaded for certain days working on a job site. Kind of a day-to-day photo journal for the customer. On the site, the user gets here, sees 3 large images, and a series of thumbnails if more than 3 images exist for that day (works fine).
However, underneath that I want to display a 3-4 column setup of "Archived Dates" that provide a link to the images of the other dates. I have this working correctly, but the results are displayed as follows:

Date 1: Date 2: Date 3:

etc....

I want them to display like this;
Day 1 Day 4
Day 2 Day 5
Day 3 Day 6

and so on..... in a 3 column format. Here is the code I have right now just looping through to display these link results, not the rest of the page. I am trying to do it tableless right now, but if that isn't the right way to go, please let me know.
Thanks to anyone in advance,
Nick

 $SQLRowe = "SELECT DISTINCT RoweImgDate from tblRowe WHERE RoweImgDate !='" . $_GET['date'] . "' Order by RoweImgDate DESC Limit 0, 30";
        //echo $SQLRowe;
        $rsSQLRowe = mysql_query($SQLRowe);

<span class="rowe">Archived Photos:</span><br/>
                    <div id="archive">
                    <?php
                    
                        while($row = mysql_fetch_array($rsSQLRowe)){

                            //echo "<a href='index.php?id="' . $row[RoweImgID] . '"' class='link'>$row[RoweImgDate]</a></br>";
                                    echo "<div id='archivedates'>";
                                    echo "<a href='index.php?date=" . $row[RoweImgDate] . "' class='link'>$row[RoweImgDate]</a>";
                                    echo "</div>";
                                    //echo "<img src='images/$row[RoweImage]'/><br/>";
                                    //echo "<span class='FeatDesc'><p>$row[RoweImgDesc]</p></span><br/>";
                                }
                    ?>
                    </div>

Greetings,   My current code logs into a database, opens a table named randomproverb, randomly selects 1 proverb phrase, and then SHOULD display the proverb in the footer of my web page. As of right now, the best I can do is get it to display "Array", but not the text proverb... this code below actually causes my whole footer to not even show up. Please help!

<?php

include("inc_connect.php"); //Connects to the database, does work properly, already tested

$Proverb = "randomproverb";
$SQLproverb = "SELECT * FROM $Proverb ORDER BY RAND() LIMIT 1";
$QueryResult = @mysql_query($SQLproverb, $DBConnect);
	while (($Row = mysql_fetch_assoc($QueryResult)) !== FALSE) {
	echo "<p style = 'text-align:center'>" . {$Row[proverb]} . "</p>\n";
	}
	$SQLString = "UPDATE randomproverb SET display_count = display_count + 1 WHERE proverb = $QueryResult[]";
	$QueryResult = @mysql_query($SQLstring, $DBConnect);
...
?>

Hello guys. I was thinking on posting this under the mysql thread. but i think this can be solved by php.. anyway, i want to dump mysql data on the client side whenever a user logs out. how can i do this? thanks in advance.

I'm in the habit of verifying input client side. For example, there is a field to enter a number, and I ensure that it is within a particular range using Javascript.

My backend PHP code does not do such checking... Is this bad? Should I be doing checking using my backend code and send a response back to the client to display a message? Admittedly, I avoid it, since it's a little extra work -- well more extra than just javascripts, IF-THEN-ALERT type statement.

Hi guys,
I'm using this EOD thingi in order to print a table and get it PDF. I am using TCPDF and following their examples to get the PDF output.
I need to print a php variable's value within the EOD.
Can you please help me out.
Below is the piece of code I'm running.

Code: [Select]
$tbl = <<<EOD
<table cellspacing="0" cellpadding="1" border="1">
    <tr>
        <td rowspan="4"><img src="images/m24.jpg"></td>
        <td align="right" width="150">From the desk of</td>
        <td>: $contat</td>
    </tr>
    <tr>
        <td align="right">E-Mail</td>
<td>: $email</td>
    </tr>
    <tr>
       <td align="right">Telephone</td>
   <td>: $telephone</td>
    </tr>
      <tr>
       <td align="right">Mobile</td>
   <td>: $mobile</td>
    </tr>
</table>
EOD;

the $variables are simply ignored. or even if I echo, there is no PHP value.
Can you guys please shed some light on this issue.
Pretty pleeeaaassseee

I'm trying to learn how to validate a form using the php server side method. Everything I search is very different and old. Anyone have knowledge of a good tutorial or source that I could use? None of my books have anything about it either!

Thanks!