PHP - Displaying Results From Mysql Database In Columns

I have a search set up to search a table for the text entered in a textbox, I have two columns in the table, one with the first name of people, and the second with their last names, I am wondering how I can search both, so for instance:

I type in the search field: Roger Smith

in the database it would look like:

First_name-----|-----Last_name
-------------------|-------------------
Roger------------|-------Smith

my current query is:

Code: [Select]
$query = mysql_query("SELECT * FROM users WHERE fname LIKE '%$find%' OR lname LIKE '%$find%'");
But if I type both parts of the name it doesn't return anything. works fine if I just search for "Roger" OR "Smith".

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. 

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 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!

Hi All

i am wanting a column list like this

http://extensions.joomla.org/extensions

basically the database is set up as
(id, category, parent)

I want the parent to group the category section and list like the the joomla example in three columns

search various threads throughout the internet but none seem to cover this entirely

can any one please help?????

Hi Everyone,

Well I have a bit of a problem... I have created a search page that I would like to return results from my MySQL database. However for some reason I can not get my page to display the results. Instead I am getting an error page. Below is the code I am using:

In addition to this, should I wish for a user to be able to edit the returned results by clicking a link, how would I do that?

<?php
session_start();
?>
<link rel="stylesheet" type="text/css" href="css/layout.css"/>
<html>

<?php
$record = $_POST['record'];
echo "<p>Search results for: $record<br><BR>";

$host = "localhost";
$login_name = "root";
$password = "P@ssword";

//Connecting to MYSQL
MySQL_connect("$host","$login_name","$password");

//Select the database we want to use
mysql_select_db("schedules_2010") or die("Could not find database");

$result = mysql_query("SELECT * 
FROM schedule_september_2010 
WHERE 
champ LIKE '%$record%' ") 
or die(mysql_error());  


// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
        // Print out the contents of each row
echo "<br><p>Your Schedule<BR></p><br>";
echo "<table border=1>\n";
echo "<tr>
<td bgcolor=#444444 align=center><p><b>Champ</p></td>
<td bgcolor=#444444 align=center><p><b>Date</p></td>
<td bgcolor=#444444 align=center><p><b>Start Time</p></td>
<td bgcolor=#444444 align=center><p><b>End Time</p></td>
<td bgcolor=#444444 align=center><p><b>Department</p></td>
<td bgcolor=#444444 align=center><p><b>First Break</p></td>
<td bgcolor=#444444 align=center><p><b>Second Break</p></td>
<td bgcolor=#444444 align=center><p><b>Login ID</p></td>
</tr>\n";
do {




printf("<tr>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
<td><p>%s</p></td>
</tr>\n"
, $row["champ"]
, $row["date_time"]
, $row["start_time"]
, $row["end_time"]
, $row["department"]
, $row["first_break"]
, $row["second_break"]
, $row["login_id"]
);

} while ($row = mysql_fetch_array($result));
echo "</table>\n";
} else {
echo "$champ No Records Found"; 
} 

mysql_free_result($result);
mysql_close($con);
?>
</html>
 

Say a user puts in a support request, and for every request it generates a unqiue string, and enters it into the database.

Ok, now say there is a text field, when the user enters their unique string and it finds a match, it displays the data along with it.

How can I accomplish this? Im kind of new to mysql, but I know basic SQL.

Would be great if somebody could point me in the right direction!

Thanks

Hi,

I have a website where users can log on and edit their profile pic, name, biography etc.
I was wondering about the correct way to:-
Add data to the database through forms (Register.php) Display the data on a page Using mysql escape sting, however, the way I am currently using will display a '\' before any ' symbol. So it's >> it\'s ...

Here is a snippet of the code I am using...

Code: [Select]
//insert data
$about1 = mysql_real_escape_string($_POST['about']);

//get
$query = mysql_query("SELECT * FROM `staff` WHERE username='$username'");
$row = mysql_fetch_array($query);
$about = $row['about'];
echo $about;

Not really sure how to get the images I have stored in MySQL into a html form. I can call-up the text fields from the database but it cannot seem to find the index for the images.

Here is my code:-

<?php
session_start();
mysql_connect("localhost","root","abc") or die ("Error! Cannot connect to database");
mysql_select_db("theimageworks") or die ("Cannot find database");
$query = "SELECT * FROM jobs";
$result = mysql_query($query) or die (mysql_error());
?>


<?php
//display data in html table
echo "<table>";
echo "<tr><td>Username</td><td align='center'>Message</td><td>Product Image</td></tr>";

while($row = mysql_fetch_array($result)) {
    echo "</td><td>";
    echo $row['username'];
    echo "</td><td>";
    echo $row['message'];
    echo "</td></tr>";   
    echo $row['image'];
     }
echo "</table>";
?>

The error message I get is "Notice: Undefined index: image in....."

Thanks in advance!

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=308347.0

need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. 



require("./include/mysqldb.php");

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);
$result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40");


print "<table width=\"293\" height=\"111\" border=\"0\">
  <tr>\n";

while($row = mysql_fetch_array($result))
  {
print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['star'] . "</td>\n";
print "  </tr>\n";
print "  <tr>\n";
print "<td>" . $row['username_search'] . "</td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['phone_search'] . "</td>\n";
print "  </tr>   \n";

  }
print "</table>";

mysql_close($con);
?>


 

Need some help I have 2 tables in a database and I need to search the first table and use the results from that search, to search another table, can this be done? and if it can how would you recommend that I go about it?

Thanks For Your Help Guys! 

Hello - hope you are all well.   I am creating a website for a family friend who has a shop. See preview of the site in this picture ....  Picture1.jpg   40.32KB   0 downloads   Currently the results are showing in one column. I want to show them in three columns in the format of:   1 2 3 4 5 6  7 8 9 etc   But formatted with the picture and then the title and the border.   The current code i have is....     HEAD   <?php $sql = "SELECT * FROM abs_productcat WHERE catid = $catid ORDER BY catid ASC";
$can = mysql_query($sql);
$catname = mysql_fetch_assoc($can); ?>       BODY    <?php while ($row_pro = mysql_fetch_assoc($rs_proddetails)) { ?><tr>
                            <td><table width="243" border="0" align="left" cellpadding="15" cellspacing="0" class="sectionborders">
                             
                              <tr>
                                <td height="120" align="center"><p><strong><a href="indproducts.php?ProductID=<?php echo $row_pro['ProductID']; ?>"><img src="products/product_<?php echo $row_pro['ProductID']; ?>" alt="<?php echo $row_pro['ProductName']; ?>" height="120" class="sectionborders" border="0" /></a></strong></p>
                                  <p><strong><a href="indproducts.php?ProductID=<?php echo $row_pro['ProductID']; ?>" class="products"><?php echo $row_pro['ProductName']; ?></a></strong></p></td>
                              </tr>
                              </table></td>
                          </tr>
                          <tr>
                            <td>&nbsp;</td>
                          </tr> <?php } ?>       THANK YOU FOR YOUR HELP!!!