Please can someone help?

robbaust · November 18, 2010

Hi new to PHPfreaks and I'm sorry if this as been asked before but I'm having a problem

basically my question is how would i get ajax live search to return results from a mysql database I have an example here www.spinfish.co.uk/test.html if you type in say salmon it shows the title but if you click on any of the titles shown it just doesn't do anything, what I want is for a user to be able to click the title then get the description from the database so say a user clicks the first title it comes up with a description of that river

this is my code that I am using and the html file

<?php

mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

$partialFishing = $_GET['partialfish'];

$fishing = mysql_query("SELECT title FROM fishing WHERE title LIKE '%$partialFishing%'");
while ($fish = mysql_fetch_assoc($fishing)){
     echo "<div>"."<a href='title'>".$fish['title']."</a>"."</div>";
}
?>

<html>
<head>
	<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js"></script>

	<script type="text/javascript">
		function getfishing(value) {
			$.get("getfishing.php",{partialfish:value},function(data){
				$("#results").html(data);
		});
	}
</script>
</head>
<body>
	<input name="search" onKeyUp="getfishing(this.value)" id="search" size="35" />
        <input type="submit" name="submit" id="submit" value="Submit" />
  </form>
	<br>
	<div id="results"></div>
</body> 
</html>

Thank you so much in advance for any help you can give

Rob

seanlim · November 19, 2010

Your ajax seems to be working fine. The problem lies here:

echo "<div>"."<a href='title'>".$fish['title']."</a>"."</div>";

href="title" will redirect users to a non-existent file named "title". you will want to do one of the following:

1. intercept the click on the link and show the description by more AJAX. i.e. <a href="..." onclick="getDescription('titlehere')

2. create another page so to display description with a parameter passed i.e. <a href="description.php?title=titlehere"

Note: it would be preferable to retrieve and ID column together with the title so that the correct description can be retrieved

robbaust · November 19, 2010

Hi Seanlim

Thank you for your reply it's really appreciated

Sorry to ask this but where whould i put it as i forgot to say i'm new to php as well but slowly getting the hand of it

Thank you again for the help

Rob

seanlim · November 19, 2010

you've got quite a long way to go... i can't really help you much further without more details

What are the columns in your table? Are the Descriptions stored in the same table? Do you have an ID column? How do you want the results displayed? On the same page as the results shown (AJAX) or on a separate page (PHP)?

robbaust · November 19, 2010

Hi seanlim

the columns in the table are id, title, description, keywords, url and images but i don't need the keywords and yes the descriptions are in the same table

i would like the results to display on the same page if possible but if not a separate page is fine

[attachment deleted by admin]

seanlim · November 19, 2010

$fishing = mysql_query("SELECT id, title FROM fishing WHERE title LIKE '%$partialFishing%'");
while ($fish = mysql_fetch_assoc($fishing)){
     echo "<div>"."<a href='description.php?id=".$fish['id']."'>".$fish['title']."</a>"."</div>";
}

description.php

<?php
if(!isset($_POST['id']))
exit("No ID set");

// add mysql connection stuff here


$fishing = mysql_query("SELECT description FROM fishing WHERE id='".mysql_real_escape_string(intval($_POST['id']))."'");

if(mysql_num_rows($fishing)==0)
exit("Record doesn't exist");

$fish = mysql_fetch_assoc($fishing);
echo $fish['description'];
?>

that's the jist of it. if you want it on the same page, use ajax to display the same information output-ed by description.php

robbaust · November 19, 2010

thank you seanlim

where do i out this first part

$fishing = mysql_query("SELECT id, title FROM fishing WHERE title LIKE '%$partialFishing%'");
while ($fish = mysql_fetch_assoc($fishing)){

echo "<div>"."<a href='description.php?id=".$fish['id']."'>".$fish['title']."</a>"."</div>";

}

seanlim · November 19, 2010

replace the last 4 lines in your original PHP file with that 4 lines of code

robbaust · November 19, 2010

Thanks

done that but now just getting "no ID set"

really sorry about all this

seanlim · November 19, 2010

what is the URL shown when you get the message "No ID set"?

work on the copy online if possible, easier to debug remotely

robbaust · November 19, 2010

http://localhost/description.php?id=

seanlim · November 19, 2010

upload whatever you have done.

do all your records have a valid id?

in the lines that you replaced:

while ($fish = mysql_fetch_assoc($fishing)){
     // add this next line
     var_dump($fish);

     echo "<div>"."<a href='description.php?id=".$fish['id']."'>".$fish['title']."</a>"."</div>";
}

then try going to http://localhost/getfishing.php?partialfish=SALMON%20River%20Dee,%20Ballogie%20Estates,%20Aberdeenshire, view source and paste everything here

robbaust · November 19, 2010

this is what the sorce code is saying

array(1) {

["title"]=>

string(50) "SALMON River Dee, Ballogie Estates, Aberdeenshire "

}

<div><a href='description.php?id='>SALMON River Dee, Ballogie Estates, Aberdeenshire </a></div>

and yes every record as an id

seanlim · November 19, 2010

geez... you didn't change the code right?

in the four lines of code i posted, the MySQL query said "SELECT id, title FROM....". did you copy those over?

if you're not sure, copy those four lines of code into your PHP file (getfishing.php) again!

robbaust · November 19, 2010

sorry yes i copied the code as you pointed out

this is all i'm getting

array(2) {

["id"]=>

string(1) "1"

["title"]=>

string(50) "SALMON River Dee, Ballogie Estates, Aberdeenshire "

}

<div><a href='description.php?id=1'>SALMON River Dee, Ballogie Estates, Aberdeenshire </a></div>

really sorry for putting on you like this

robbaust · November 19, 2010

Seanlim

these are the files i have

getfishing.php

<?php

mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

$partialFishing = $_GET['partialfish'];

$fishing = mysql_query("SELECT id, title FROM fishing WHERE title LIKE '%$partialFishing%'");
while ($fish = mysql_fetch_assoc($fishing)){
// add this next line
     var_dump($fish);
     echo "<div>"."<a href='description.php?id=".$fish['id']."'>".$fish['title']."</a>"."</div>";
}
?>

the description.php you gave me

and the index.html

<html>
<head>
	<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.js"></script>

	<script type="text/javascript">
		function getfishing(value) {
			$.get("getfishing.php",{partialfish:value},function(data){
				$("#results").html(data);
		});
	}
</script>
</head>
<body>
	<input name="search" onKeyUp="getfishing(this.value)" id="search" size="35" />
        <input type="submit" name="submit" id="submit" value="Submit" />
  </form>
	<br>
	<div id="results"></div>
</body> 
</html>

seanlim · November 19, 2010

okay, now remove

     // add this next line
     var_dump($fish);

it should be working, no? any error messages?

robbaust · November 19, 2010

i have uploaded the files to my website

www.spinfish.co.uk/test.html

seanlim · November 19, 2010

my bad, replace description.php with this:

<?php
if(!isset($_GET['id']))
exit("No ID set");

// add mysql connection stuff here


$fishing = mysql_query("SELECT description FROM fishing WHERE id='".mysql_real_escape_string(intval($_GET['id']))."'");

if(mysql_num_rows($fishing)==0)
exit("Record doesn't exist");

$fish = mysql_fetch_assoc($fishing);
echo $fish['description'];
?>

robbaust · November 19, 2010

WORKING

Thank you so much Seanlim

I really really appreciate what you have done

Thank You

Robert

seanlim · November 19, 2010

no problem, glad i could help

and remember to mark this topic as solved

robbaust · November 20, 2010

hi seanlim

sorry i have kind of lost the plot today

as stated in my PM to you how would i get images to show the images are coming from a folder on my server but can't seem to get them to show

if you look at my site www.spinfish.co.uk/test.html it just shows the file location or an broken image icon (firefox)

description.php

<?php
if(!isset($_GET['id']))
exit("No ID set");

mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

$fishing = mysql_query("SELECT title, description, images FROM fishing WHERE id='".mysql_real_escape_string(intval($_GET['id']))."'");
$images = mysql_query("SELECT images FROM fishing WHERE id='".mysql_real_escape_string(intval($_GET['id']))."'");

if(mysql_num_rows($fishing)==0)
exit("Record doesn't exist");

$fish = mysql_fetch_assoc($fishing);
echo $fish['title'] . "<BR/>\n";
echo $fish['description'] . "<BR/>\n" . "<BR/>\n";
echo $fish['images'];
echo '<img src=/images/';

?>

getfishing.php

<?php

mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

$partialFishing = $_GET['partialfish'];

$fishing = mysql_query("SELECT id, title, images FROM fishing WHERE title LIKE '%$partialFishing%'");
while ($fish = mysql_fetch_assoc($fishing)){

     echo "<div>"."<a href='description.php?id=".$fish['id']."'>".$fish['title']."</a>"."</div>";

}
?>

index.html

<html>
   <head>
      <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.js"></script>

      <script type="text/javascript">
         function getfishing(value) {
            $.get("getfishing.php",{partialfish:value},function(data){
               $("#results").html(data);
         });
      }
   </script>
</head>
<body>
      <input name="search" onKeyUp="getfishing(this.value)" id="search" size="35" />
        <input type="submit" name="submit" id="submit" value="Submit" />
  </form>
      <br>
      <div id="results"></div>
   </body>
</html>

and this is my database

[attachment deleted by admin]

seanlim · November 20, 2010

The only problem is with description.php now. When you click the link, it takes you to description.php?id=[id-here], which is perfectly good.

However, the code I gave only shows the raw, unformatted database data. The page should be a proper HTML page to display correctly i.e. proper <html> & <body> tags which you are missing. Furthermore, you can do a simple View Source to see that your image tag is not showing properly:

http://www.spinfish.co.uk/images/066.jpg<img src=/images/

Your <img> tag must have a "src" attribute pointing to a proper file for it to be displayed. The last line of description.php:

echo '<img src="/images/'.$fish['images'].'">';
?>

robbaust · November 20, 2010

thanks seanlim I think i have it working

www.spinfish.co.uk/test.html

robbaust · November 21, 2010

Hi sorry my i ask two final questions please

if a person as typed something in the search box then backspaces how would i get my script to stop showing all the results until they start typing in again and do you know when you type something in a box and the popup box thing shows that as something you typed in before how can you stop that

I have uploaded a picture to show what i mean

thanks for your time again

[attachment deleted by admin]

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