PHP Search Form - Help

blepblep · April 12, 2012

Hi, I was wondering could anyone help me here please, I'm pretty stuck and not sure what's wrong.

I have also posted this topic on Daniweb if that is OK as I'm kind of stuck for time on this (If I'm not allowed post on multiple forums this can be deleted). I have a database that contains files and I want to be able to search those files by putting in a ID number into a textbox on my homepage of a website I'm working on. In my databse my doc_id is the primary key, and I want the user to be able to enter a document ID and have all the information returned to them.

Can anyone help me here as I don't understand why it isn't working? Here is the code as it is at the moment -

<?php
include 'connect_db.php';
include 'newheader.php';

function sanitize_data($data)
    {
        $data = array_map('trim',$data);
        $data = array_map('strip_tags',$data);
        $data = array_map('htmlspecialchars',$data);
        $data = array_map('mysql_real_escape_string',$data);
        return $data;
    }
    $post = sanitize_data($_POST);

if (isset($_POST['searchID']))
{
$find = $_POST['find'];

$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

$field = $_POST['field'];

$data = mysql_query("SELECT * FROM tc_tool.forms WHERE upper(".$field.") LIKE'%$find%'") or die(mysql_error());

while ($result = mysql_fetch_array( $data ))
{
	echo $result['doc_number'];
	echo " ";
	echo $result['doc_title'];
	echo " ";
	echo "<br>";
	echo "<br>";
}
?>

The error returned to me is this -

Incorrect parameter count in the call to native function 'upper'

Here is the code to my button too if it is any use -

<body onLoad = "documentNumber.focus()">

			<input type= "text" id= "documentNumber" name= "searchbyequipmenttype" class= "eSearchFormInput"/>
			<input type= "submit" name= "searchID" value="Search By Doc ID" /><br />

And I have no idea why it is, can anyone help if possible? I have tried it a few ways so if needed I'll post the other ways I've tried it. Thanks if anyone can help

Muddy_Funster · April 12, 2012

changing this:

$data = mysql_query("SELECT * FROM tc_tool.forms WHERE upper(".$field.") LIKE'%$find%'") or die(mysql_error());

to this:

$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
UPPER('$field') LIKE '%$find%'
)
SQL;
$result = mysql_query($sql) or die(mysql_error())."<br>----------------<br>$sql");

will help you diagnose the problem with the query.

It's looking like $field is empty. Also, this code doesn't look like what your description said you wanted your search to perform.

As for not being allowed to post on other forums....we're here to help, having that rule would be somewhat counter productive. It's all about getting the solution to a problem, whether it be here or anywhere else.

blepblep · April 12, 2012

Thanks for the reply Muddy. Alright that's cool about the multiple posting, some forums don't like it as far as I no.

I've changed it to this now -

<?php

include 'connect_db.php';
include 'newheader.php';

function sanitize_data($data)
    {
        $data = array_map('trim',$data);
        $data = array_map('strip_tags',$data);
        $data = array_map('htmlspecialchars',$data);
        $data = array_map('mysql_real_escape_string',$data);
        return $data;
    }
    $post = sanitize_data($_POST);

if (isset($_POST['searchID']))
{
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

$field = $_POST['field'];

$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
UPPER('$field') LIKE '%$find%'
)
SQL;
$result = mysql_query($sql) or die(mysql_error())."<br>----------------<br>$sql";

while ($result = mysql_fetch_array( $data ))
{
	echo $result['doc_id'];
	echo " ";
	echo $result['doc_title'];
	echo " ";
	echo "<br>";
	echo "<br>";
}

?>

<?php include "footer.php" ?>

And this is my error I get -

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 38

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 48

I have also tried it with changing

while ($result = mysql_fetch_array( $data ))

to this

while ($result = mysql_fetch_array( $sql ))

but I still the get the first error shown above..

Sorry if I didnt explain what I am looking for properly. What I am trying to do is on my front page, I have a text box (Which the code for is in my first post). I want to be able to enter in an id such as 300 for example, and then click search, that search should then return the form that is associated with the id - 300 - I understand how to do the displaying part of the form correctly but I cannot get it to search the database for it, my query is probably wrong but I don't no how to fix it.

blepblep · April 12, 2012

I can't edit my above post but is this any more helpful to what is going wrong?

Sorry Muddy I should have said, but where I have $field = $_POST['field'], should 'field' be related to a row in the database? If it should be, I don't have any row called field. But I do have doc_id, and I changed it to this -

$doc_id = $_POST['doc_id'];

$data = mysql_query("SELECT * FROM tc_tool.forms WHERE upper(".$doc_id.") LIKE'%$find%'") or die (mysql_error());

And when I run it I get this error -

Incorrect parameter count in the call to native function 'upper'

Any idea's?

Muddy_Funster · April 12, 2012

my bad on the code I gave you, I closed the or die() brackets too soon - change to this and let's see what comes back

	$result = mysql_query($sql) or die(mysql_error()."<br>----------------<br>$sql");

blepblep · April 12, 2012

I get this..

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 39

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 49

Sorry, but we can not find an entry to match your query

Would it have anything got to do with my button, as in is it taking in what's typed? Or any idea if it has to do with this -

I have also tried it with changing

Code: [select]

while ($result = mysql_fetch_array( $data ))

to this

Code: [select]

while ($result = mysql_fetch_array( $sql ))

but I still the get the first error shown above..

Muddy_Funster · April 12, 2012

you need to address the form element by it's name in the $_POST array - so try this as a test:

$id = $_POST['searchbyequipmenttype'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;
$result = mysql_query($sql) or die(mysql_error()."<br>-----------<br>$sql);
$row = mysql_fetch_assoc($result);
var_dump($row);

blepblep · April 12, 2012

Ok I done this -

if (isset($_POST['searchID']))
{
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

$doc_id = $_POST['doc_id'];

$id = $_POST['searchbyequipmenttype'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;
$result = mysql_query($sql) or die(mysql_error()."<br>-----------<br>$sql");
$row = mysql_fetch_assoc($result);
var_dump($row);

This is the output -

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 6

-----------

SELECT * FROM tc_tool.forms WHERE ( doc_id = )

Muddy, would I be getting an error because doc_id isn't a form element? It's just used in the database, nowhere else. The only fields that I am using are these below -

                       doc_title, 
			doc_number, 
			doc_type,
			revision, 
			cdm_link,
			main_req_id,
			mars_link, 
			checklist_link, 
			link_internal_1_3, 
			impacted_products,
			scope, 
			impacted_products_2,
			review_class, 
			full_or_simplified, 
			earliest_date, 
			latest_date, 
			previous_tc_reviews,
			proj_name, 
			author_name, 
			req_list, 
			optional_list,
			information_only,
			chairperson,
			reviewers_required,
			review_duration, 
			document_abstract

I don't no why this isn't working, it's really annoying but probably something stupid that I'm missing.

Muddy_Funster · April 12, 2012

problem is getting the information from the form to the script. Your form element name that we want to look at is "searchbyequipmenttype" so we use the line $id = $_POST['searchbyequipmenttype']; to get it from the POST array, only problem is, it's not got anything in it. So we need to now look at the form. where is your form? on the same page or on a different one and do you have any other code that the form uses that you didn't post??

blepblep · April 12, 2012

Ok, well I have this form, which is the search bar one -

<?php

include 'connect_db.php';
include 'newheader.php';

function sanitize_data($data)
    {
        $data = array_map('trim',$data);
        $data = array_map('strip_tags',$data);
        $data = array_map('htmlspecialchars',$data);
        $data = array_map('mysql_real_escape_string',$data);
        return $data;
    }
    $post = sanitize_data($_POST);

if (isset($_POST['searchID']))
{
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

$field = $_POST['field'];

$id = $_POST['searchbyequipmenttype'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;
$result = mysql_query($sql) or die(mysql_error()."<br>-----------<br>$sql");
$row = mysql_fetch_assoc($result);
var_dump($row);

while ($result = mysql_fetch_array( $data ))
{
	echo $result['doc_id'];
	echo " ";
	echo $result['doc_title'];
	echo " ";
	echo "<br>";
	echo "<br>";
}
}
?>

Then once searchID which is the name of the button is pressed, it goes to this -

<?php

include "newheader.php";
include "connect_db.php";
include "lib/toolfunctions.php";

?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
    <meta http-equiv="Content-Language" content="en" />
</head>

<body>
		<p>
		<form action = "searchByDocID.php" method = "post">
		<!-- body onLoad focuses textbox on load
			-- Only works in Chrome at the moment -->
			<body onLoad = "documentNumber.focus()">

			<input type= "text" id= "documentNumber" name= "searchbyequipmenttype"/>
			<input type= "submit" name= "searchID" value="Search By Doc ID" /><br />

			<font size = '2'>EG: 123</font>
		</p>
	</form>		

<div id="toolFooterWrapper">
	<div id="toolFooter">

		TC Tool<br/>Version 1.00

	</div>
</div>
</body>
</html>

They are the only two forms that are being used at the moment. So once the user types a number into the search bar, I want it to go to the searchByDocID.php file, and in the searchByDocID.php file I have the query that should return the file.

So everything I have in the above is how it runs at the moment, and when I run it I get this error -

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 6

-----------

SELECT * FROM tc_tool.forms WHERE ( doc_id = )

Could you possibly explain this to me please -

 name= "searchbyequipmenttype"

Is that relevant or can I delete it? I'm getting confused myself from looking at it all day!!

Muddy_Funster · April 12, 2012

ok, assuming this to be your searchByDocID.php :

<?php

include 'connect_db.php';
include 'newheader.php';

function sanitize_data($data)
    {
        $data = array_map('trim',$data);
        $data = array_map('strip_tags',$data);
        $data = array_map('htmlspecialchars',$data);
        $data = array_map('mysql_real_escape_string',$data);
        return $data;
    }
    $post = sanitize_data($_POST);

if (isset($_POST['searchID']))
//...

if it is I would like you to add a line at the top of the page:

<?php
die(print_r($_POST));

Let me know what you get on the page from that.

blepblep · April 12, 2012

Yep thats searchByDocID.php. I done that and got this when it ran -

Array ( [searchbyequipmenttype] => [searchID] => Search By Doc ID ) 1

What is it doing with searchbyequipmenttype there?!

Thanks for all the help so far by the way.

Muddy_Funster · April 12, 2012

I know this sounds crazy, but you are filling in a value before you click the button to submit the form? right?

blepblep · April 12, 2012

Yeah I am I have only one entry in the DB which is number 340, and when I type 340 in and click search I get this -

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 36

Sorry, I only realised when I leave the search box empty I get this -

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 6

-----------

SELECT * FROM tc_tool.forms WHERE ( doc_id = )

AND, when I leave in

var_dump($row);

and enter 340 into the search bar, I get this result -

array(27) { ["doc_id"]=> string(3) "340" ["doc_title"]=> string(11) "david flynn" ["doc_number"]=> string(4) "0012" ["doc_type"]=> string(2) "IS" ["revision"]=> string(14) "www.google.com" ["cdm_link"]=> string(4) "blab" ["main_req_id"]=> string(0) "" ["mars_link"]=> string(0) "" ["checklist_link"]=> string(0) "" ["link_internal_1_3"]=> string(0) "" ["impacted_products"]=> string(0) "" ["scope"]=> string(0) "" ["impacted_products_2"]=> string(0) "" ["review_class"]=> string(0) "" ["full_or_simplified"]=> string(0) "" ["earliest_date"]=> string(0) "" ["latest_date"]=> string(0) "" ["previous_tc_reviews"]=> string(0) "" ["proj_name"]=> string(13) "OSS-RC Energi" ["author_name"]=> string(0) "" ["chairperson"]=> string(0) "" ["req_list"]=> string(0) "" ["optional_list"]=> string(0) "" ["information_only"]=> string(0) "" ["reviewers_required"]=> string(0) "" ["review_duration"]=> string(0) "" ["document_abstract"]=> string(0) "" }

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 37

Would that have something got to do with this taking in $data when I don't have anything else named $data?

while ($result = mysql_fetch_array( $data ))

blepblep · April 13, 2012

Anyone have any idea's?

Muddy_Funster · April 13, 2012

need to change these lines:

	$result = mysql_query($sql) or die(mysql_error())."<br>----------------<br>$sql";

while ($result = mysql_fetch_array( $data ))

to

	$data = mysql_query($sql) or die(mysql_error())."<br>----------------<br>$sql";

while ($result = mysql_fetch_array( $data ))

blepblep · April 13, 2012

Done that there and this is the result -

Array ( [searchbyequipmenttype] => 340 [searchID] => Search By Doc ID ) 1

I don't no why but I have a feeling it has something to do with my button code -

<form action = "searchByDocID.php" method = "post">
<!-- body onLoad focuses textbox on load
    -- Only works in Chrome at the moment -->
<body onLoad = "documentNumber.focus()">
<input type= "text" id= "documentNumber" name= "searchbyequipmenttype"/>
<input type= "submit" name= "searchID" value="Search By Doc ID" /><br />

Which goes to this (searchByDocID.php) -

if (isset($_POST['searchID']))
{
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

$field = $_POST['field'];

$id = $_POST['searchID'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;

$data = mysql_query($sql) or die(mysql_error())."<br>-----------<br>$sql";

while ($result = mysql_fetch_array( $data ))
        {
         .........
         }

Muddy_Funster · April 13, 2012

take out the line we added at the top :

 die(print_r($_POST));

blepblep · April 13, 2012

Done that and get this -

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'By Doc ID )' at line 5

When I change

<input type= "submit" name= "searchID" value="Search By Doc ID" /><br />

to

<input type= "submit" name= "searchID" value="SearchByDocID" /><br />

I get this though -

Unknown column 'SearchByDocID' in 'where clause'

I don't understand why I'm getting that though?!

Muddy_Funster · April 13, 2012

OK, post up the full code for both pages as it is now so we're both working from the same page

blepblep · April 13, 2012

Ok, thanks for all the help so far too.

Here is my index page -

<?php

//----------------
// Include files 
// - Connect to DB
//----------------

include "newheader.php";
include "connect_db.php";
include "lib/toolfunctions.php";

?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
    <meta http-equiv="Content-Language" content="en" />
</head>

<body>
		<p>
		<form action = "searchByDocID.php" method = "post">
		<!-- body onLoad focuses textbox on load
			-- Only works in Chrome at the moment -->
			<body onLoad = "documentNumber.focus()">

			<input type= "text" id= "documentNumber" name= "searchbyequipmenttype"/>
			<input type= "submit" name= "searchID" value="Search By Doc ID" /><br />

			<font size = '2'>EG: 123</font>
		</p>
	</form>		

<div id="toolFooterWrapper">
	<div id="toolFooter">

		TC Tool<br/>Version 1.00

	</div>
</div>
</body>
</html>

Here is searchByDocID.php -

<?php

include 'connect_db.php';
include 'newheader.php';

function sanitize_data($data)
    {
        $data = array_map('trim',$data);
        $data = array_map('strip_tags',$data);
        $data = array_map('htmlspecialchars',$data);
        $data = array_map('mysql_real_escape_string',$data);
        return $data;
    }
    $post = sanitize_data($_POST);

if (isset($_POST['searchID']))
{
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

$field = $_POST['field'];

$id = $_POST['searchID'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;

$data = mysql_query($sql) or die(mysql_error())."<br>-----------<br>$sql";

while ($result = mysql_fetch_array( $data ))
{
	echo $row  ['doc_id'] . 			" <br /> " 
		. $row ['doc_title'] . 			" <br /> "
		. $row ['doc_number'] .			" <br /> "
		. $row ['doc_type'] . 			" <br /> " 
		. $row ['revision'] . 			" <br /> " 
		. $row ['cdm_link'] . 			" <br /> "
		. $row ['mars_link'] .			" <br /> "
		. $row ['checklist_link'] .		" <br /> "
		. $row ['link_internal_1_3']. 	" <br /> "
		. $row ['impacted_products'].	" <br /> "
		. $row ['scope'].				" <br /> "
		. $row ['impacted_products_2'] ." <br /> "
		. $row ['review_class'].		" <br /> "
		. $row ['full_simplified'] .	" <br /> "
		. $row ['pref_earliest'] . 		" <br /> "
		. $row ['pref_latest'] . 		" <br /> "
		. $row ['prev_reviews'] . 		" <br /> "
		. $row ['proj_name'] .			" <br /> "
		. $row ['auth_name'] . 			" <br /> "
		. $row ['req_list'] .			" <br /> "
		. $row ['optional_list'] .		" <br /> "
		. $row ['information_only'] . 	" <br /> "
		. $row ['chairperson'] .		" <br /> "
		. $row ['req_reviewers'] .		" <br /> "
		. $row ['review_duration'] .	" <br /> "
		. $row ['document_abstract'] . 	" <br /> "; 
}
}
?>
<?php include "footer.php" ?>

That's the two of them..

Muddy_Funster · April 13, 2012

ok on searchByDocID.php you need to change this line back :

$id = $_POST['searchID'];

to read

$id = $_POST['searchbyequipmenttype'];

and as long as the field in the form is filled in you should get a result.

blepblep · April 13, 2012

Alright I done that, and get a blank page. No errors though or anything so I'm wondering is it my query?

Is it right in passing in $data to the $result as seen here?

$id = $_POST['searchbyequipmenttype'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;

$data = mysql_query($sql) or die(mysql_error())."<br>-----------<br>$sql";

while ($result = mysql_fetch_array( $data ))

Because when I pass in $sql to where $data is so it looks like this -

$data = mysql_query($sql) or die(mysql_error())."<br>-----------<br>$sql";

while ($result = mysql_fetch_array( $sql ))

I get this -

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /opt/htdocs/webpages/TC_Tool/Tool/searchByDocID.php on line 36

Muddy_Funster · April 13, 2012

ok, replace your searchByDocID.php code (all of it) with this:

<?php

include 'connect_db.php';
include 'newheader.php';

function sanitize_data($data)
    {
        $data = array_map('trim',$data);
        $data = array_map('strip_tags',$data);
        $data = array_map('htmlspecialchars',$data);
        $data = array_map('mysql_real_escape_string',$data);
        return $data;
    }
    $post = sanitize_data($_POST);

if (isset($post['searchID']) && $post['searchbyequipmenttype' != '')
{
$id = $post['searchbyequipmenttype'];
$sql = <<<SQL
SELECT *
FROM tc_tool.forms
WHERE
(
doc_id =$id
)
SQL;

$data = mysql_query($sql) or die(mysql_error()."<br>-----------<br>$sql");

while ($result = mysql_fetch_array( $data ))
{
	echo $row  ['doc_id'] . 			" <br /> " 
		. $row ['doc_title'] . 			" <br /> "
		. $row ['doc_number'] .			" <br /> "
		. $row ['doc_type'] . 			" <br /> " 
		. $row ['revision'] . 			" <br /> " 
		. $row ['cdm_link'] . 			" <br /> "
		. $row ['mars_link'] .			" <br /> "
		. $row ['checklist_link'] .		" <br /> "
		. $row ['link_internal_1_3']. 	" <br /> "
		. $row ['impacted_products'].	" <br /> "
		. $row ['scope'].				" <br /> "
		. $row ['impacted_products_2'] ." <br /> "
		. $row ['review_class'].		" <br /> "
		. $row ['full_simplified'] .	" <br /> "
		. $row ['pref_earliest'] . 		" <br /> "
		. $row ['pref_latest'] . 		" <br /> "
		. $row ['prev_reviews'] . 		" <br /> "
		. $row ['proj_name'] .			" <br /> "
		. $row ['auth_name'] . 			" <br /> "
		. $row ['req_list'] .			" <br /> "
		. $row ['optional_list'] .		" <br /> "
		. $row ['information_only'] . 	" <br /> "
		. $row ['chairperson'] .		" <br /> "
		. $row ['req_reviewers'] .		" <br /> "
		. $row ['review_duration'] .	" <br /> "
		. $row ['document_abstract'] . 	" <br /> "; 
}
}
else{
  echo "there was a problem with the form, please try again and make sure you have filled in a document number";
  }
include "footer.php" ?>

blepblep · April 13, 2012

I get this now -

There was a problem with the form, please try again and make sure you have filled in a document number

Seems as if its just skipping the query..

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