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PHP - Dropdown Search
Hey All.
I'm trying to modify a search I have setup. What I want to do is take out the text field and replace it with a dropdown menu. It starts with an html form Old working code: Code: [Select] <form action="result.php" method="post"> <div align="center">Search: <input type="text" name="search" /> <input type="submit" /> </div> </form> New non working code: Code: [Select] <form action="result.php" method="POST"> <select name="dropdown"> <option value="option1">option1</option> <option value="option2">option2</option> <option value="option3">option3</option> </select><input type="submit" name="search" /> </form> That should open up result.php which searches two fields in a mysql database and prints of the results. result.php Code: [Select] $var = $_POST["search"]; $data = mysql_query("select * FROM database1 WHERE field1 LIKE '$var' || field2 LIKE '$var'") or die(mysql_error()); echo $data; ?> If result.php not receiving the info in the dropdown or am I handling it wrong in the php? Thanks so much. Similar TutorialsThis topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=353763.0 Hello I am having a trouble with creating page that will search records from database using dropdown menu, when I select a category, for example By: Gender and after clicking the Submit button, there is no records displays but when I select By: ALL, the records displays. What's wrong with my code? I'm sorry I'm fairly new to PHP. Here is my code: Code: [Select] <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <title>Search</title> </head> <body> <form action="" method="POST" name="records"> <label>SEARCH: </label> <select name="select"> <option value="all" selected="selected">ALL</option> <option value="year">By Year</option> <option value="gender">By Gender</option> <option value="course">By Course</option> </select> <input name="submit" value="GO" type="submit" /> <br /> </form> </body> </html> <?php include('collegeinfo_connect.php'); mysql_connect("$server", "$user", "$pass")or die("cannot connect"); mysql_select_db("$db")or die("cannot select DB"); if (isset($_POST['submit'])) { $selection = $_POST['select']; if($selection == "all") $query = "SELECT * FROM collegeinfo_tbl"; else if($selection == "year") $query = "SELECT * FROM collegeinfo_tbl WHERE Year='year'"; else if($selection == "gender") $query = "SELECT * FROM collegeinfo_tbl WHERE Gender='gender'"; else if($selection == "course") $query = "SELECT * FROM collegeinfo_tbl WHERE Course='course'"; } if (isset($query)) { $result = mysql_query($query) or die(mysql_error()); echo "<br /> <br />"; echo "<table border='1' cellpadding='10'>"; echo "<tr> <th>ID Number:</th> <th>First Name:</th> <th>Last Name:</th> <th>Gender:</th> <th>Year:</th> <th>Course:</th> </tr>"; while ($row = mysql_fetch_array($result)) { echo "<tr>"; echo '<td>' . $row['ID'] . '</td>'; echo '<td>' . $row['FirstName'] . '</td>'; echo '<td>' . $row['LastName'] . '</td>'; echo '<td>' . $row['Gender'] . '</td>'; echo '<td>' . $row['Year'] . '</td>'; echo '<td>' . $row['Course'] . '</td>'; echo "</tr>"; } echo "</table>"; } ?> hi, im new to php and need some help with my project. i wanted to create a search form with dropdown list provided that the values in the dropdown list are the field names of the database that the form that call the search form. Ex. i have a A.php form that uses Table A from database X. then A.php press a button inside the form and then a search form will show. when i search for something in that table the dropdownlist will be populated by the field names of Table A. this same goes to B.php that uses Table B of database X.. and so on.. can anyone help me with this?? thanks!! Hi, I have a search form where users can search by age and country. Users can also save their search so they can go back and do it again but the problem is when they reload their saved search, I need it to select the country that has been saved in the drop down. So if someone searched the United Kingdom, I need it to show United Kingdom in the select drop down instead of (Select Country). Is there any easy and quick way around this? Many Thanks I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID. Index.php
<!doctype html>
<?PHP
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass", "Database" =>
"database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
?>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<section id="formaT2" class="formaT2 formContent">
<div class="row">
<div class="col-md-2 col-3 row-color remove-mob"></div>
<div class="col-md-5 col-9 bg-img" style="padding-left: 0;
padding-right: 0;">
<h1>Form</h1>
<div class="rest-text">
<div class="contactFrm">
<p class="statusMsg <?php echo
!empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
<form action="connection.php" method="post">
<div>machinery</div>
<select id="machinery">
<option value="0">--Please Select Machinery--</option>
<?php
// Fetch Department
$sql = "SELECT Machinery FROM T013";
$machanery_data = sqlsrv_query($conn2,$sql);
while($row = sqlsrv_fetch_array($machanery_data) ){
$id = $row['Id'];
$machinery = $row['Machinery'];
// Option
echo "<option value='".$id."' >".$machinery."</option>";
}
?>
</select>
<div class="clear"></div>
<div>Sub Machinery</div>
<select id="sub_machinery">
<option value="0">- Select -</option>
</select>
<input type="submit" name="submit"
id="submit" class="strelka-send" value="Insert">
<div class="clear"> </div>
</form>
</div>
</div>
</div>
</div>
</section>
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#machinery").change(function(){
var machinery_id = $(this).val();
$.ajax({
url:'ajaxfile.php',
type: 'post',
data: {machinery:machinery_id},
dataType: 'json',
success:function(response){
var len = response.length;
$("#sub_machinery").empty();
for( var i = 0; i<len; i++){
var machinery_id = response[i]['machinery_id'];
var machinery = response[i]['machinery'];
$("#sub_machinery").append("<option
value='"+machinery_id+"'>"+machinery+"</option>");
}
}
});
});
});
</script>
</body>
</html>
Ajaxfile.php
<?php
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass",
"Database" => "database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
$machineryID = $_POST['machinery']; // department id
$sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
$result = sqlsrv_query($conn2,$sql);
$machinery_arr = array();
while( $row = sqlsrv_fetch_array($result) ){
$machinery_id = $row['ID'];
$machinery = $row['MachineID'];
$machinery_arr[] = array("ID" => $machinery_id, "MachineID" =>
$machinery);
}
// encoding array to json format
echo json_encode($machinery_arr);
?>
Edited May 6, 2019 by davidd
The result pages is supposed to have pagination like google help me please
Hi freaks, I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database. The code I'm using to populate dropdown: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if(!$con) { die ('Can not connect to : '.mysql_error()); } mysql_select_db("ims",$con); $result=mysql_query("select cat_id,cat_name from category"); echo "<select name=cat>"; while($nt=mysql_fetch_array($result)) { echo "<option value=$nt[cat_id]> $nt[cat_name] </option>"; } echo "</select>"; mysql_close($con); ?> Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]"); Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow. Help on this will be highly appreiable . (Tell me if I'm not clear with my question) I require a page to be added to my website. This page will facilitate a refined search via Google, Bing and Yahoo search engine simultaneously , show the top 5 of each engine. Is this possible using php ? Thank's Hello, Does anyone know a tutorial I can follow to create my own search engine over the items I have in my SQL database? I find a lot of tutorials but they are all 'one word searches' which means if you type two words you will get all results that contain either word (too many hits). If the search engine tutorial displays result with AJAX even better. Thanks for help. df I am developing a intranet forum in Php and MySQL and I am using ajax to display searched results on the same page but right now I am using query LIKE text% to search in database which is slower. but I want to make it fast search engin which can parse *,+ and show result. Since I am using ajax i am not able to use free search engin,so if possible pls provide a complete solution I have code to search a database of members, I can search by lastname but having a little problem. I want to be able to search by lastname starting with the first letter. (ie: a or b or c or d and so on). As it is right now I can only search by first letter of last name if I add % to the search (ie: a% b% c%) will give me all members with the last names starting with the approiate letter designation. I'm not sure how to handle this, any help would be appreciated. Thanks. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title></title> </head> <body> <center><table cellspacing="10" cellpadding="10" width="750" border="0"> <h4>Search</h4> <form name="search" method="post" action="<?php $PHP_SELF?>"> Seach for: <input type="text" name="find" /> in <Select NAME="field"> <Option VALUE="lname">Last Name</option> <input type="hidden" name="searching" value="yes" /> <input type="submit" name="search" value="Search" /> </form> <?php // check to see if anything is posted if (isset($_POST['find'])) {$find = $_POST['find'];} if (isset($_POST['searching'])) {$searching = $_POST['searching'];} if (isset($_POST['field'])) {$field = $_POST['field'];} //This is only displayed if they have submitted the form if (isset($searching) && $searching=="yes") { echo "<h4>Results</h4><p>"; // If they did not enter a search term we give them an error if (empty($find)) { echo "<p>You forgot to enter a search term"; exit; } // Otherwise we connect to our Database mysql_connect("localhost", "root", "1910") or die(mysql_error()); mysql_select_db("cmc_member") or die(mysql_error()); // We preform a bit of filtering $find = strtoupper($find); $find = strip_tags($find); $find = trim ($find); // Now we search for our search term, in the field the user specified $results = mysql_query("SELECT * FROM members WHERE upper($field) LIKE'$find'"); // And we display the results if($results && mysql_num_rows($results) > 0) { $i = 0; $max_columns = 3; while($row = mysql_fetch_array($results)) { // make the variables easy to deal with extract($row); // open row if counter is zero if($i == 0) echo "<tr>"; // make sure we have a valid product ALIGN='CENTER' if($fname != "" && $fname != null) echo "<td ALIGN='CENTER'><FONT COLOR='red'><b>$fname $lname</b></FONT><br> $address <br> $city $state $zip <br>$phone<br><a href=\"mailto: $email\">$email</a><br></td>"; // increment counter - if counter = max columns, reset counter and close row if(++$i == $max_columns) { echo "</tr>"; $i=0; } // end if } // end while } // end if results // clean up table - makes your code valid! if($i < $max_columns) { for($j=$i; $j<$max_columns;$j++) echo "<td> </td>"; } //This counts the number or results - and if there wasn't any it gives them a little message explaining that $anymatches = mysql_num_rows($results); if ($anymatches == 0) { echo "Sorry, but we can not find an entry to match your query<br><br>"; } //And we remind them what they searched for echo "<b>Searched For:</b> " .$find; } ?> </tr> </table></center> </body> </html> Friends, I want to extract the Search Keyword from the URL, a visitor came from. I am using a PHP CMS and want to show the Keyword on my Blog. So if they search "abcd" from google, i want to extract "abcd" and echo on my blog. Here is the coding i could got hold of, but its not working, not echoing anything <?php function pk_stt2_function_get_delimiter($ref) { $search_engines = array('google.com' => 'q', 'go.google.com' => 'q', 'images.google.com' => 'q', 'video.google.com' => 'q', 'news.google.com' => 'q', 'blogsearch.google.com' => 'q', 'maps.google.com' => 'q', 'local.google.com' => 'q', 'search.yahoo.com' => 'p', 'search.msn.com' => 'q', 'bing.com' => 'q', 'msxml.excite.com' => 'qkw', 'search.lycos.com' => 'query', 'alltheweb.com' => 'q', 'search.aol.com' => 'query', 'search.iwon.com' => 'searchfor', 'ask.com' => 'q', 'ask.co.uk' => 'ask', 'search.cometsystems.com' => 'qry', 'hotbot.com' => 'query', 'overture.com' => 'Keywords', 'metacrawler.com' => 'qkw', 'search.netscape.com' => 'query', 'looksmart.com' => 'key', 'dpxml.webcrawler.com' => 'qkw', 'search.earthlink.net' => 'q', 'search.viewpoint.com' => 'k', 'mamma.com' => 'query'); $delim = false; if (isset($search_engines[$ref])) { $delim = $search_engines[$ref]; } else { if (strpos('ref:'.$ref,'google')) $delim = "q"; elseif (strpos('ref:'.$ref,'search.atomz.')) $delim = "sp-q"; elseif (strpos('ref:'.$ref,'search.msn.')) $delim = "q"; elseif (strpos('ref:'.$ref,'search.yahoo.')) $delim = "p"; elseif (preg_match('/home\.bellsouth\.net\/s\/s\.dll/i', $ref)) $delim = "bellsouth"; } return $delim; } /** * retrieve the search terms from search engine query * */ function pk_stt2_function_get_terms($d) { $terms = null; $query_array = array(); $query_terms = null; $query = explode($d.'=', $_SERVER['HTTP_REFERER']); $query = explode('&', $query[1]); $query = urldecode($query[0]); $query = str_replace("'", '', $query); $query = str_replace('"', '', $query); $query_array = preg_split('/[\s,\+\.]+/',$query); $query_terms = implode(' ', $query_array); $terms = htmlspecialchars(urldecode(trim($query_terms))); return $terms; } /** * get the referer * */ function pk_stt2_function_get_referer() { if (!isset($_SERVER['HTTP_REFERER']) || ($_SERVER['HTTP_REFERER'] == '')) return false; $referer_info = parse_url($_SERVER['HTTP_REFERER']); $referer = $referer_info['host']; if(substr($referer, 0, 4) == 'www.') $referer = substr($referer, 4); return $referer; } $referer = pk_stt2_function_get_referer(); if (!$referer) return false; $delimiter = pk_stt2_function_get_delimiter($referer); if( $delimiter ){ $term = pk_stt2_function_get_terms($delimiter); } echo $term; ?> May someone help? Natasha T Hi all How do I modify the below code to search multiple tables in mySQL database? $query = "select * from store_items where description like \"%$trimmed%\" or title like \"%$trimmed%\" or dimensions like \"%$trimmed%\" order by id ASC"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); It is for a search function and I need it to search another table called 'about_text' and 'faq_text' Thanks Pete 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 Hello, I'm developing a website that asks the user to submit a keyword for a search. The results should display matches for this keyword, but also show matches for related keywords (in order of relevenace!). I'm planning on building up a library of which search terms users use in the same sessions (e.g. if someone searches for "it jobs" and "php jobs", I'll know the terms are correlated), and I'll also measure the click-through rates of the items on the results list. I've been spending all weekend trying to map out the design for this but it's proving incredibly complicated and I think the solution is likely to be on the internet somewhere already?! Please could someone point me in the right direction if you've come accross this problem before? Thanks a million, Stu Hello All, I am not a developer but I can modify code to work for me. The following code works on my test machine (Windows 10, IIS, PHP 7.4) but doesn't work on my website (cPanel, Some version of Linux, PHP 7.4). The two dropdowns are for State and City. You are supposed to be able to select the state and then select a city from that state then bring up a report for craft breweries in the city. When selecting State from the first dropdown, the page refreshes, the URL is correct with the reports.php?cat=<STATE> so $cat is being set, but the first dropdown no longer has the state selected and the second dropdown is populated with All cities and not just one ones from the selected state. Any ides why this is working fine on one machine and not the other? Selected code from reports.php
<?php
?>
/////// for second drop down list we will check if State is selected else we will display all the cities/////
echo "<form method=post action='brewerylistbycity.php'>";
}
////////// Starting of second drop downlist /////////
//// End Form /////
Need some help I have 2 tables in a database and I need to search the first table and use the results from that search, to search another table, can this be done? and if it can how would you recommend that I go about it? Thanks For Your Help Guys! I want to know how to display results from mysql database by filling in a form. but i have found a tutorial which shows typing in a text box which displays results. if i follow this tutorial will it help me to understand and create php coding to display results for my form? |
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