[RESOLVED] Creating an efficient keyword search script.

swatisonee

Hi, I looked up the following post

http://www.phpbuilder.com/board/showthread.php?s=&threadid=10208147

but am not sure how i can adapt it to my requirement. I need to update a customer table. So if a user types in abc company in the form field, a list of all firms with the words abc company in them should come up.

I dont think this script i have is particularly efficient and i would like to convert both the scripts into a single one like the one i have seen above . Can anyone guide me on how i could accomplish this ? Thanks.

1st file

<? include("protect.php"); ?>


<form method="post" action="update_customer.php">

Company:<input type="text" size="12" name="customer">
<input type="submit" value="Search">

</form>

2nd file

<? include("protect.php"); ?>


<?php
$customer = $_POST["customer"];
?>
<b>These are the customers in your database in the selected alphabet. Select one to update.</b></font><p>
<table border="1" style="border-collapse: collapse" bordercolor="#111111" width="100%">
<tr bgcolor="#DEEDFE">

<form method="post" action="update_customer_chosen.php">
<td><b><font size=2 face=Tahoma color=blue>Select</b></td>
<td><b><font size=2 face=Tahoma color=blue>Company</b></td>
<td><b><font size=2 face=Tahoma color=blue>City</b></td>
<td><b><font size=2 face=Tahoma color=blue>State</b></td>
<td><b><font size=2 face=Tahoma color=blue>First Name</b></td>
<td><b><font size=2 face=Tahoma color=blue>Last Name</b></td>
<tr>

<?php

mysql_connect("localhost", $dbname, $dbpasswd )
    	or die ("Unable to connect to server.");

mysql_select_db("$database")
    	or die ("Unable to select database.");



$resulta = mysql_query("SELECT * FROM `Customers`WHERE  `Company` LIKE '%$customer%' ORDER BY `Company` asc ") or die (mysql_error ());
echo $sqla;

do {

printf("<tr><td><input type=\"checkbox\" name=\"choice\" value=%d><td>
<font size=2 face=Tahoma color=blue>%s<td>
<font size=2 face=Tahoma color=blue>%s<td>
<font size=2 face=Tahoma color=blue>%s<td>
<font size=2 face=Tahoma color=blue>%s<td>
<font size=2 face=Tahoma color=blue>%s<td>
 </tr>",
 $myrowa["Customerid"],
 $myrowa["Company"],
 $myrowa["Business City"],
 $myrowa["Business State"],
 $myrowa["First Name"],
 $myrowa["Last Name"]);

  } while ($myrowa = mysql_fetch_array($resulta));

?>

</table>


<p><input type="submit" value="Select">
</form>

bpat1434

<?php
include('protect.php');

if(isset($_POST) && !empty($_POST['customer']))
{

?>
<b>These are the customers in your database in the selected alphabet. Select one to update.</b></font><p>
<table border="1" style="border-collapse: collapse" bordercolor="#111111" width="100%">
<tr bgcolor="#DEEDFE">

<form method="post" action="update_customer_chosen.php">
<td><b><font size=2 face=Tahoma color=blue>Select</b></td>
<td><b><font size=2 face=Tahoma color=blue>Company</b></td>
<td><b><font size=2 face=Tahoma color=blue>City</b></td>
<td><b><font size=2 face=Tahoma color=blue>State</b></td>
<td><b><font size=2 face=Tahoma color=blue>First Name</b></td>
<td><b><font size=2 face=Tahoma color=blue>Last Name</b></td>
<tr>

<?php

mysql_connect("localhost", $dbname, $dbpasswd )
        or die ("Unable to connect to server.");

mysql_select_db("$database")
        or die ("Unable to select database.");



$resulta = mysql_query("SELECT * FROM `Customers`WHERE  `Company` LIKE '%$customer%' ORDER BY `Company` asc ") or die (mysql_error ());
echo $sqla;

while($row = mysql_fetch_array($resulta))
{
printf("<tr><td><input type=\"checkbox\" name=\"choice\" value=%d><td>
    <font size=2 face=Tahoma color=blue>%s<td>
    <font size=2 face=Tahoma color=blue>%s<td>
    <font size=2 face=Tahoma color=blue>%s<td>
    <font size=2 face=Tahoma color=blue>%s<td>
    <font size=2 face=Tahoma color=blue>%s<td>
</tr>",
$myrowa["Customerid"],
$myrowa["Company"],
$myrowa["Business City"],
$myrowa["Business State"],
$myrowa["First Name"],
$myrowa["Last Name"]);
}

?>

</table>
<p><input type="submit" value="Select">
</form>
<?php
}
else
{
  showForm();
}


function showForm()
{
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

Company:<input type="text" size="12" name="customer">
<input type="submit" value="Search">

</form>
<?php
}
?>

That's in one file...

swatisonee

Oh great ! Thanks !

swatisonee

bpat1434,
I've a small issue. The script works like a dream except that if i alter

then the uid value is not getting passed into the next script. I even made it a hidden but it doesnt get passed once php self gets executed although i see it in the address bar before the above script is run.

any suggestions please ?
thanks.

bpat1434

Where do you set UID? And how are you accessing it?

swatisonee

Hi, In an earlier page , i have this link

<?php
          echo "
          <a href=\"update_customer.php?uid=$uid\">Update Customer</a>
          ";
          ?>

and when this link is clicked the page that you showed me pops up with the uid in the addressbar .

I also modified update_customer.php as under :

<?php 
include('protect.php'); 

$uid=$_POST["uid"];

if(isset($_POST) && !empty($_POST['customer'])) 
{ 


?>

But once update_customer.php is processed, i dont find the uid coming up in subsequent pages nor does it get passed as a hidden variable.

Thanks.

maxtech

if u r passing the $uid thru the url then go for $_GET['uid'] in the next page........so that u can get the value of $uid.

swatisonee

Assuming you meant replacing $POST with $GET, I changed this

<?
$uid = $_GET["uid"];
echo $uid;

?>

The uid gets echoed out but doesnt get passed into

<form method="post" action="update_customer_chosen.php?uid=<? echo $uid ?>"> 

And I even tried changing 


<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

to


<form method="post" action="update_customer.php?uid=<? echo $uid ?>">

but the value is just not going thru...

Any clues ?

maxtech

change the following :
<form method="post" action="update_customer_chosen.php?uid=<? echo $uid ?>">

to :

swatisonee

no luck there either, i'm afraid.

the odd thing is that i can see the uid value in the addressbar when update_customer.php is run so why its not getting passed is a mystery.

maxtech

:queasy:

can u b plz proper in what u r posting.............plz post what u exactly want from us

swatisonee

ok heres what i have done thus far :

<?php 
include('protect.php'); 

$uid = $_GET["uid"]; 
echo $uid;  // this echoes uid correctly 

$uid = $_POST["uid"]; 
echo $uid;  // this does not echo uid at all  

if(isset($_POST) && !empty($_POST['customer'])) 
{ 

?> 
<b>These are the customers in your database in the selected alphabet. Select one to update.</b></font><p> 
<table border="1" style="border-collapse: collapse" bordercolor="#111111" width="100%"> 
<tr bgcolor="#DEEDFE"> 

<input type="hidden" name="uid" value="<? echo $uid ?>" >
<form method="post" action="update_customer_chosen.php?uid=<? echo $uid ?>">

//<form method="get" action="update_customer_chosen.php?uid=<? echo $uid ?>"> : doesnt make a difference as uid still doesnt get passed

<td><b><font size=2 face=Tahoma color=blue>Select</b></td> 
<td><b><font size=2 face=Tahoma color=blue>Company</b></td> 
<td><b><font size=2 face=Tahoma color=blue>City</b></td> 
<td><b><font size=2 face=Tahoma color=blue>State</b></td> 
<td><b><font size=2 face=Tahoma color=blue>First Name</b></td> 
<td><b><font size=2 face=Tahoma color=blue>Last Name</b></td> 
<tr> 

<?php 

mysql_connect("localhost", $dbname, $dbpasswd ) 
        or die ("Unable to connect to server."); 

mysql_select_db("$database") 
        or die ("Unable to select database."); 



$resulta = mysql_query("SELECT * FROM `Customers`WHERE  `Company` LIKE '%$customer%' ORDER BY `Company` asc ") or die (mysql_error ()); 
echo $sqla; 

while($row = mysql_fetch_array($resulta)) 
{ 
printf("<tr><td><input type=\"checkbox\" name=\"choice\" value=%d><td> 
    <font size=2 face=Tahoma color=blue>%s<td> 
    <font size=2 face=Tahoma color=blue>%s<td> 
    <font size=2 face=Tahoma color=blue>%s<td> 
    <font size=2 face=Tahoma color=blue>%s<td> 
    <font size=2 face=Tahoma color=blue>%s<td> 
</tr>", 
$myrowa["Customerid"], 
$myrowa["Company"], 
$myrowa["Business City"], 
$myrowa["Business State"], 
$myrowa["First Name"], 
$myrowa["Last Name"]); 
} 

?> 

</table> 
<p><input type="submit" value="Select"> 
</form> 
<?php 
} 
else 
{ 
  showForm(); 
} 


function showForm() 
{ 
?> 

<input type="hidden" name="uid" value="<? echo $uid ?>" >

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 

//<form method="post" action="update_customer_chosen.php?uid=<? echo $uid ?>"> // tried this too but it does not pass the uid 


Company:<input type="text" size="12" name="customer"> 
<input type="submit" value="Search"> 

</form> 
<?php 
} 
?>

maxtech

put that hidden field inside the form tag..........................try it out

bpat1434

Right here is your problem:

$uid = $_GET["uid"];
echo $uid;  // this echoes uid correctly

$uid = $_POST["uid"];
echo $uid;  // this does not echo uid at all

Remove the second one (the one with $_POST) and you'll be fine.

You first set $uid to the uid in the address bar (what you want), then you set it to nothing, because $POST doesn't contain it. Remove the line $uid = $POST['uid'] and it should work.

swatisonee

I did remove the $_POST but it still doesnt get passed to the form..should i pass the hidden within the top form tags or the php self form tags?

Also, Maxtech has been recommending i use get instead of post for form action but i always use post in all scripts so wont that be a problem ?

Thanks !

bpat1434

heh... I found your problem.... look in the function showForm/b.

<input type="hidden" name="uid" value="<? echo $uid ?>" >

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

If it's not within the <form> tags, it's not going to be passed to the next page. Since your action doesn't give the uid in the url, and your hidden field is outside the form, it won't get passed. Fix this, and you should be on yoru way!!!

swatisonee

You're a genius Bpat1434 ! Thank you, thank you, thank you ! and thanks too Maxtech for your suggestions !