PHP - How To Write A Multi Criteria Search Script

 

How would i begin to start php coding to work for this form shown in the image? could anyone point me to useful tutorials, i would appreciate it, thanx

Hello. I am working on a php script for searching a database table. I am really new to this, so I used the this tutorial http://www.phpfreaks.com/tutorial/simple-sql-search

I managed to get all the things working the way I wanted, except one important and crucial thing.
Let me explain.
My table consist of three columns, like this:

ID(bigint20)    title(text)        link (varchar255)
=============================
ID1                title1                link-1
ID2                title2                link-2
etc...

Like I said, I managed to make the script display results for a search query based on the title. Want I want it to do more, but I can't seem to find the right resource to learn how, is to place a "Download" button under each search result with its corresponding link from the table.

Here is the code I used.

<?php

$dbHost = 'localhost'; // localhost will be used in most cases
// set these to your mysql database username and password.
$dbUser = 'user'; 
$dbPass = 'pass';
$dbDatabase = 'db'; // the database you put the table into.
$con = mysql_connect($dbHost, $dbUser, $dbPass) or trigger_error("Failed to connect to MySQL Server. Error: " . mysql_error());

mysql_select_db($dbDatabase) or trigger_error("Failed to connect to database {$dbDatabase}. Error: " . mysql_error());
// Set up our error check and result check array
$error = array();
$results = array();
// First check if a form was submitted. 
// Since this is a search we will use $_GET
if (isset($_GET['search'])) {
   $searchTerms = trim($_GET['search']);
   $searchTerms = strip_tags($searchTerms); // remove any html/javascript.
   
   if (strlen($searchTerms) < 3) {
      $error[] = "Search terms must be longer than 3 characters.";
   }else {
      $searchTermDB = mysql_real_escape_string($searchTerms); // prevent sql injection.
   }
   
   // If there are no errors, lets get the search going.
   if (count($error) < 1) {
      $searchSQL = "SELECT title, link FROM db WHERE title LIKE '%{$searchTermDB}%'";

      $searchResult = mysql_query($searchSQL) or trigger_error("There was an error.<br/>" . mysql_error() . "<br />SQL Was: {$searchSQL}");
      
      if (mysql_num_rows($searchResult) < 1) {
         $error[] = "The search term provided {$searchTerms} yielded no results.";
      }else {
         $results = array(); // the result array
         $i = 1;
         while ($row = mysql_fetch_assoc($searchResult)) {
            $results[] = "{$row['title']}<br /> Download - this is the button I want to link to the title results - and maybe other links too - <br /> ";

    $i++;

       

   
         }
      
      }
   }
}

function removeEmpty($var) {
   return (!empty($var)); 


}
?>

      <?php echo (count($error) > 0)?"The following had errors:<br /><span id=\"error\">" . implode("<br />", $error) . "</span><br /><br />":""; ?>
      <form method="GET" action="search?" name="searchForm">
         Search for title: <input type="text" name="search" value="<?php echo isset($searchTerms)?htmlspecialchars($searchTerms):''; ?>" />
        
         <input type="submit" name="submit" value="Search" />
      </form>

    <?php echo (count($results) > 0)?"Rezultate lucrari de licenta sau disertatie pentru {$searchTerms} :<br /><br />" . implode("", $results):""; ?>




 $results = array(); // the result array
         $i = 1;
         while ($row = mysql_fetch_assoc($searchResult)) {
            $results[] = "{$row['title']}<br /> Download - this is the button I want to link to the title results - and maybe other links too - <br /> ";

    $i++;

 
I would like the results to be displayed like this

Results for SearchItem:

Result 1
Download | Other link
Result 2
Download | Other link
etc.... or something like this.

So, how do I add the data from the link row into a text(Dowload), within an <a href> tag (well, at least I guess it would go this way) ?
My first tries (fueled by my lack of knowledge) where things like

$results[] = "{$row['title']}<br /> <a href="{$row['link']}">Download</a> <br /> ";

but I keep getting lots of errors, and then I don't know much about arrays and stuff (except basic notions);
So there it is. I am really stuck and can't seem to find any workaround for this.
Any suggestions? (examples, documentation, anything would do, really)
Thanks,
Radu

hey guys im trying to make a file upload script so that when the user uploads a pic it gets shown in a div...when you add another it is the added by the side of the 1st pic...but some reason when selecting a image nothing shows...   what am i doing wrong?   here is my code below if anyone can tell me what it is wrong im doing...thank you

<script src="//ajax.googleapis.com/ajax/libs/jqueryui/1.11.2/jquery-ui.min.js"></script>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
<script>
$( document ).ready(function() {
	
	var files       = $('.input-file').prop("files");
	var file_count  = files.length;
	
	var names = $.map(files, function(val) { 
		return val.name; 
	});
	
	$('#input-file-button').click(function() {
		$('.input-file').click(); 
	});
	
	$('.input-file').change(function (){
		var file_name = $(this).val();
		var file_path = URL.createObjectURL(event.target.files[0]);

		$('#uploaded-images').append('<div><img src="'+ file_path +'" /></div>' + file_name);	
	 });
		
	if (file_count < 1)
	{
		$('#uploaded-images').text("No images are selected.");
	}
});


</script>
<style>
.input-file
{
	display: none;
}
</style>
</head>
<body>
<div id="uploaded-images"></div>
<br />
<input type="file" name="files[]" class="input-file"/><span id='input-file-button'>Upload Photo</span>
</body>
</html>

I am trying to make a database application. I encounterd a problem that. suppose multiple users tryto access data it may cause problem. to clearly understand please see the attachment. How can I resolve it?

Hi to all !   I would like to ask that :-   1. what is the best way to / or how to best test a multi-user login script. 2. The best way to test a multi user website that saves data from the users into a database.      Are there any tools that can hep me in testing my work locally on a localhost before I move them out to a website. ?   I have heard that testing routines can be written for such purposes to automate the testing. If so where should I begin to look for them? I have no idea at all about writing test routines / scripts. A tutorial , if any exists, would be a good place to start.   Thanks very much.         

Hello:
I have this snippet of code that will write the current year onto a page if it's embedded on the page itself:
Code: [Select]
<?php echo date("Y") ?>

Works fine.

But, when I include it in an included file and pull it onto the page with a function, it does not work.

Like this:
MyNav.php
Code: [Select]
<?php

function spFooter() {
$spFooter =
"
   <p>
   
   &copy; <?php echo date(\"Y\") ?>
 
   </p>

";

return $spFooter;
}
?>

MyPage.php
Code: [Select]
<?php
include('include/myNav.php');
?>
<html>
...

<?php echo spFooter(); ?>

...
</html>

What is the proper way to do this, and also what are the basic rules for properly nesting PHP scripts like this?

Thanks for the help.

Hello Friends,

Im trying to connect Perforce from Linux. I want to get(sync) file from perforce, then need to edit and submit it back to perforce....

Do you guys have any sample codings to do so??

Waiting for ur reply,

Thanks
Rekha

Something I have never done, but want to learn the best practice for doing so, is a listener script.

Example: one of our websites is in bespoke PHP and has two links that shows if there are any noticeboard 'notifications' or any new 'messages'.

Ideally I'd like to show 1, 2, 3... etc as a count of how many are remaining for Messages.  So if a new one comes in while they are on a page, that number literally changes while they are on the page.

I assume this is with Ajax, and not straight forward.  I don't know any DOM code or Javascript.  But would appreciate some help please.

Simon

SET UP: Windows vista
# XAMPP 1.7.3,
# Apache 2.2.14 (IPv6 enabled) + OpenSSL 0.9.8l
# MySQL 5.1.41 + PBXT engine
# PHP 5.3.1
# phpMyAdmin

So I'm getting this error message: 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 8. I can't see where the '", )' code is in that script. I've looked and contrl + find it but no joy. I've looked through the previous pages and mysql database to ensure the variables and the tables match up. Any help would be appreciated. Maybe some good mysql debug advice. Is there a good debug open source program out there? Any help would be appreciated.

<?php   

//let's start our session, so we have access to stored data
session_start();   

include 'db.inc.php';

$db = mysql_connect('localhost', 'root', '') or 
    die ('Unable to connect. Check your connection parameters.');
mysql_select_db('ourgallery', $db) or die(mysql_error($db));
   
//let's create the query
$query = sprintf("INSERT INTO subscriptions (
name, email_address, membership_type,
terms_and_conditions, name_on_card,
credit_card_number,
credit_card_expiration_data)
VALUES ('%s','%s','%s','%s','%s','%s','%s')",
$_SESSION['email_address'],
$_SESSION['membership_type'],
$_SESSION['terms_and_conditions'],
$_POST['name_on_card'],
$_POST['credit_card_number'],
$_POST['credit_card_expiration_data']);
//let's run the query
mysql_query($insert_query);
?>

I have approximately 2,300 videos embedded on my site, and I am constantly adding more. The videos are hosted by YouTube. When a video is deleted from YouTube, it becomes unavailable on my site.

Each video's ID is in my MySQL database. A video ID looks like this:

ZFOuxAx-dkc

I need a script that will do the following: select the IDs from my database check if the videos are still available on YouTube create a list of the videos that are unavailable. If possible, it would be great if the script could also tell me if any of the videos have been blocked in certain countries.

The script has to be fully commented so that I can understand it.

I want to run the script manually every few days. I don't want to have to change my database tables.

I've been told that you can use YouTube's API or YouTube's oEmbed interface. I've also been told that, because there are so many videos, you will have to use cURL multi handles.

If interested in this project, please PM me your price.   This is an urgent project. I have placed ads elsewhere.

This is posting to any row in mssql that has a matching 'start_date'.  I would like it to post rows where start_date = '$start_date' and it also matches the 'meter_id' .  It should perform this check of both parameters before posting to mssql. 

<?php
$session_id = $_SESSION['id'];
$comm_id = $_GET['meter_id'];
$start_date = $_GET['start_date'];
$payment_date = $_POST['payment_date'];
$amount_paid = $_POST['amount_paid'];
$check_number = $_POST['check_number'];

$sql = "UPDATE [radiogates].[dbo].[ops_invoice_history] SET payment_date = '$payment_date', amount_paid = '$amount_paid', check_number = '$check_number' where start_date = '$start_date'";

$query = sqlsrv_query($conn, $sql);

Thank you for any help.

I need a php/mysql star rating which has 3 criteria to rate and at the bottom a link when clicked saves the ratings. Any idea of where to get that?


Something like this

Rate criteria 1

Rate criteria 2

Rate criteria 3


Save ratings

Hi,   I can I include a date range criteria to query with in the following code? The date field in the table (t_persons) is IncidentDate.

$criteria = array('FamilyName', 'FirstName', 'OtherNames', 'NRCNo', 'PassportNo', 'Gender', 'IncidenceCountryID', 'Status', 'OffenceKeyword', 'AgencyID', 'CountryID', 'IncidenceCountryID' );
    $likes = "";
    $url_criteria = '';
    foreach ( $criteria AS $criterion ) {
            if ( ! empty($_POST[$criterion]) ) {
                    $value = ($_POST[$criterion]);
                    $likes .= " AND `$criterion` LIKE '%$value%'";
                    $url_criteria .= '&'.$criterion.'='.htmlentities($_POST[$criterion]);
            } elseif ( ! empty($_GET[$criterion]) ) {
                    $value = mysql_real_escape_string($_GET[$criterion]);
                    $likes .= " AND `$criterion` LIKE '%$value%'";
                    $url_criteria .= '&'.$criterion.'='.htmlentities($_GET[$criterion]);
            } //var_dump($likes);
    }
    $sql = "SELECT * FROM t_persons WHERE PersonID>0" . $likes . " ORDER BY PersonID DESC";

Kind regards.  

I am working on a project that involves providing details for county government, and if applicable, cities independent of county oversight.  Virginia is a state with a lot of independent cities, so I wanted to echo them in a separate list underneath the county governments.   This code works fine when displaying Virginia information.  However, I would like to hide the Independent Cities heading in every other state.  For some reason, this code still echoes the Independent Cities heading in states without independant cities in the database.   Any thoughts or wisdom?   (sorry for pasting the whole bulky code)

$query = "SELECT * FROM state WHERE st_id=$statecode";
            $result = mysql_query($query) or die("Query $query failed : " . mysql_error());
            $row = mysql_fetch_assoc($result);
            $state_name=$row["st_name"];
            $state_note=$row["state_notes"];

            echo "<h2>Counties of " . $state_name; "</h2>\n";
            
            $query = "SELECT * FROM counties WHERE st_id=$statecode AND visible=1 ORDER BY county_name ASC";
            $result = mysql_query($query) or die("Query $query failed : " . mysql_error());
        
            echo "<table width=100%><tr><td valign=top width=33%>\n";

            $count = 1;
            $col = 0;
            $rowcount = (int)(mysql_num_rows($result) / 3) + 1;
            $remainder = mysql_num_rows($result) - ($rowcount - 1) * 3;
            /* ********************************************************* */
            while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
                echo "<a href=\"services.php?county=".$row["county_id"]."\">".$row["county_name"] . "</style></a><br />\n";
                $count = $count + 1;
                if ((($remainder > 0) && ($count == ($rowcount + 1))) || (($remainder <= 0) && ($count == $rowcount))) {
                    $col = $col + 1;
                    if ($col!=3) {
                        echo "</td><td valign=top width=33%>\n";
                        $count = 1;
                        if ($remainder > 0)
                            $remainder = $remainder - 1;
                    }
                }
            }
            /* ********************************************************* */
            echo "</td></tr></table><br>\n";
            mysql_free_result($result);
            /* ********************************************************* */
            
            $query2 = "SELECT * FROM cities WHERE st_id=$statecode AND visible=1 ORDER BY city_name ASC";
            $result2= mysql_query($query2) or die("Query $query failed : " . mysql_error());
            $city_id = $row['city_id'];
            $city_name = $row['city_name'];
            
            if ($city_id!="") {
                echo "<h3>Independent cities of " . $state_name; "</h3>\n";
                echo "<table width=100%><tr><td valign=top width=33%>\n";
                $count2 = 1;
                $col2 = 0;
                $rowcount2 = (int)(mysql_num_rows($result2) / 3) + 1;
                $remainder2 = mysql_num_rows($result2) - ($rowcount2 - 1) * 3;
                /* ********************************************************* */
                while ($row = mysql_fetch_array($result2, MYSQL_ASSOC)) {
                    echo "<a href=\"services.php?county=".$row["city_id"]."\">".$row["city_name"] . "</style></a><br />\n";
                    $count2 = $count2 + 1;
                    if ((($remainder2 > 0) && ($count2 == ($rowcount2 + 1))) || (($remainder2 <= 0) && ($count2 == $rowcount2))) {
                        $col2 = $col2 + 1;
                            if ($col2!=3) {
                            echo "</td><td valign=top width=33%>\n";
                            $count2 = 1;
                            if ($remainder2 > 0)
                                $remainder2 = $remainder2 - 1;
                    }
                }
            
                
            }
            /* ********************************************************* */
            echo "</td></tr></table><br>\n";
            mysql_free_result($result2);
            /* ********************************************************* */
            
            
            
            }
            
        ?>

Edited by myotch, 06 September 2014 - 09:19 PM.