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blepblep

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    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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..

     

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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?!

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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 ))
        {
         .........
         }

     

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    Yeah I am :D 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 ))

     

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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.

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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!!

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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.

     

     

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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..

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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?

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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.

    PHP Search Form - Help

    in PHP Coding Help

    Posted

    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 :)

     

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