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PHP - Dropdown Box Query
Hi,
I'm having a tough time combining two separate pieces of code. To summarise, I have an entire folder hierachy which I want to access via a group of dropdown boxes, so there is a box for months, a box for the dates and a box for the year. The code I have found on the internet (and slightly modified with the dates etc.) is this: Code: [Select] <html> <head> <script language="javascript" type="text/javascript"> function jumpTo(url) { document.location.href=url; } </script> <script type="text/javascript"> var categories = []; categories["startList"] = ["January","February","March","April","May","June","July","August","September","October","November","December"] categories["January"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["February"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29"]; categories["March"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["April"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30"]; categories["May"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["June"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30"]; categories["July"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["August"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["September"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30"]; categories["October"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["November"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30"]; categories["December"] = ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"]; categories["1"] = ["2008","2009","2010","2011"]; categories["2"] = ["2008","2009","2010","2011"]; categories["3"] = ["2008","2009","2010","2011"]; categories["4"] = ["2008","2009","2010","2011"]; categories["5"] = ["2008","2009","2010","2011"]; categories["6"] = ["2008","2009","2010","2011"]; categories["7"] = ["2008","2009","2010","2011"]; categories["8"] = ["2008","2009","2010","2011"]; categories["9"] = ["2008","2009","2010","2011"]; categories["10"] = ["2008","2009","2010","2011"]; categories["11"] = ["2008","2009","2010","2011"]; categories["12"] = ["2008","2009","2010","2011"]; categories["13"] = ["2008","2009","2010","2011"]; categories["14"] = ["2008","2009","2010","2011"]; categories["15"] = ["2008","2009","2010","2011"]; categories["16"] = ["2008","2009","2010","2011"]; categories["17"] = ["2008","2009","2010","2011"]; categories["18"] = ["2008","2009","2010","2011"]; categories["19"] = ["2008","2009","2010","2011"]; categories["20"] = ["2008","2009","2010","2011"]; categories["21"] = ["2008","2009","2010","2011"]; categories["22"] = ["2008","2009","2010","2011"]; categories["23"] = ["2008","2009","2010","2011"]; categories["24"] = ["2008","2009","2010","2011"]; categories["25"] = ["2008","2009","2010","2011"]; categories["26"] = ["2008","2009","2010","2011"]; categories["27"] = ["2008","2009","2010","2011"]; categories["28"] = ["2008","2009","2010","2011"]; categories["29"] = ["2008","2009","2010","2011"]; categories["30"] = ["2008","2009","2010","2011"]; categories["31"] = ["2008","2009","2010","2011"]; var nLists = 3; // number of select lists in the set function fillSelect(currCat,currList){ var step = Number(currList.name.replace(/\D/g,"")); for (i=step; i<nLists+1; i++) { document.forms['tripleplay']['List'+i].length = 1; document.forms['tripleplay']['List'+i].selectedIndex = 0; } var nCat = categories[currCat]; for (each in nCat) { var nOption = document.createElement('option'); var nData = document.createTextNode(nCat[each]); nOption.setAttribute('value',nCat[each]); nOption.appendChild(nData); currList.appendChild(nOption); } } function getValue(L3, L2, L1) { alert("Your selection was:- \n" + L1 + "\n" + L2 + "\n" + L3); } function init() { fillSelect('startList',document.forms['tripleplay']['List1']) } navigator.appName == "Microsoft Internet Explorer" ? attachEvent('onload', init, false) : addEventListener('load', init, false); </script> </head> <body> <form name="tripleplay" action=""> <select name='List1' onchange="fillSelect(this.value,this.form['List2'])"> <option selected>Make a selection</option> </select> <select name='List2' onchange="fillSelect(this.value,this.form['List3'])"> <option selected>Make a selection</option> </select> <select name='List3' onchange="getValue(this.value, this.form['List2'].value, this.form['List1'].value)"> <option selected >Make a selection</option> </select> </form> </body> </html> Then I have some code that displays a folder's contents on screen: Code: [Select] <?php ($_POST['tripleplay']) { $link = "../marketing/2011\February\20"; if (file_exists($link)) { if ($handle = opendir($link)) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != "..") { if(!preg_match("/\.pdf$/", $file)) continue; { echo "<a href=\"".$link."/".$file."\">".$file."</a><br>"; } } } echo "<br>"; closedir($handle); } } break; ?> Basically I need for the results of the dropdown boxes to help select the folder. So for the above example, if the dropdown says February 20, 2011, it would display that folders contents. I'm sorry that this is such a mess, but I would be grateful for any help you can give :shy: Similar TutorialsHey, I have the following coding: Quote <? $dbuser="*******"; $dbpass="*******"; $dbname="virtuda_db"; //the name of the database $chandle = mysql_connect("localhost", $dbuser, $dbpass) or die("Connection Failure to Database"); mysql_select_db($dbname, $chandle) or die ($dbname . " Database not found. " . $dbuser); $mainsection="license"; $query1="select name from license"; $result = mysql_db_query($dbname, $query1) or die("Failed Query of " . $query1); //do the query while($thisrow=mysql_fetch_row($result)) { $i=0; while ($i < mysql_num_fields($result)) { $field_name=mysql_fetch_field($result, $i); echo $thisrow[$i] . " "; //Display all the fields on one line $i++; } echo "<br>"; //put a break after each database entry } ?> How would I set up this so that instead of just "listing" them out on new lines, it would list the results into a drop down list? Thanks! I am querying a database and trying to display the names of categories in a drop-down menu. My problem is that the names are not visible. I have the following code: $result = mysql_query('SELECT name FROM category') or die(mysql_error()); echo('Choose a category'); echo('<select name="category">'); echo('<option value="general" >general</option>'); while($row = mysql_fetch_array($result)){ echo('<option style="color:black;" value = "'.$row['name'].'">'.$row['name'].'</option>'); } echo('</select><br /><br />'); The query is good and executing the query retrieves the expected number of results. My dropdown box is the proper length and width (showing that it is trying to print the names); however, nothing is displayed. Any ideas? I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$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
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) Hello all,
Based on the suggestion of you wonderful folks here, I went away for a few days (to learn about PDO and Prepared Statements) in order to replace the MySQLi commands in my code. That's gone pretty well thus far...with me having learnt and successfully replaced most of my "bad" code with elegant, SQL-Injection-proof code (or so I hope).
The one-and-only problem I'm having (for now at least) is that I'm having trouble understanding how to execute an UPDATE query within the resultset of a SELECT query (using PDO and prepared statements, of course).
Let me explain (my scenario), and since a picture speaks a thousand words I've also inlcuded a screenshot to show you guys my setup:
In my table I have two columns (which are essentially flags i.e. Y/N), one for "items alreay purchased" and the other for "items to be purchased later". The first flag, if/when set ON (Y) will highlight row(s) in red...and the second flag will highlight row(s) in blue (when set ON).
I initially had four buttons, two each for setting the flags/columns to "Y", and another two to reverse the columns/flags to "N". That was when I had my delete functionality as a separate operation on a separate tab/list item, and that was fine.
Now that I've realized I can include both operations (update and delete) on just the one tab, I've also figured it would be better to pare down those four buttons (into just two), and set them up as a toggle feature i.e. if the value is currently "Y" then the button will set it to "N", and vice versa.
So, looking at my attached picture, if a person selects (using the checkboxes) the first four rows and clicks the first button (labeled "Toggle selected items as Purchased/Not Purchased") then the following must happen:
1. The purchased_flag for rows # 2 and 4 must be switched OFF (set to N)...so they will no longer be highlighted in red.
2. The purchased_flag for row # 3 must be switched ON (set to Y)...so that row will now be highlighted in red.
3. Nothing must be done to rows # 1 and 5 since: a) row 5 was not selected/checked to begin with, and b) row # 1 has its purchase_later_flag set ON (to Y), so it must be skipped over.
Looking at my code below, I'm guessing (and here's where I need the help) that there's something wrong in the code within the section that says "/*** loop through the results/collection of checked items ***/". I've probably made it more complex than it should be, and that's due to the fact that I have no idea what I'm doing (or rather, how I should be doing it), and this has driven me insane for the last 2 days...which prompted me to "throw in the towel" and seek the help of you very helpful and intellegent folks. BTW, I am a newbie at this, so if I could be provided the exact code, that would be most wonderful, and much highly appreciated.
Thanks to you folks, I'm feeling real good (with a great sense of achievement) after having come here and got the great advice to learn PDO and prepared statements.
Just this one nasty little hurdle is stopping me from getting to "end-of-job" on my very first WebApp. BTW, sorry about the long post...this is the best/only way I could clearly explaing my situation.
Cheers guys!
case "update-delete":
if(isset($_POST['highlight-purchased']))
{
// ****** Setup customized query to obtain only items that are checked ******
$sql = "SELECT * FROM shoplist WHERE";
for($i=0; $i < count($_POST['checkboxes']); $i++)
{
$sql=$sql . " idnumber=" . $_POST['checkboxes'][$i] . " or";
}
$sql= rtrim($sql, "or");
$statement = $conn->prepare($sql);
$statement->execute();
// *** fetch results for all checked items (1st query) *** //
$result = $statement->fetchAll();
$statement->closeCursor();
// Setup query that will change the purchased flag to "N", if it's currently set to "Y"
$sqlSetToN = "UPDATE shoplist SET purchased = 'N' WHERE purchased = 'Y'";
// Setup query that will change the purchased flag to "Y", if it's currently set to "N", "", or NULL
$sqlSetToY = "UPDATE shoplist SET purchased = 'Y' WHERE purchased = 'N' OR purchased = '' OR purchased IS NULL";
$statementSetToN = $conn->prepare($sqlSetToN);
$statementSetToY = $conn->prepare($sqlSetToY);
/*** loop through the results/collection of checked items ***/
foreach($result as $row)
{
if ($row["purchased"] != "Y")
{
// *** fetch one row at a time pertaining to the 2nd query *** //
$resultSetToY = $statementSetToY->fetch();
foreach($resultSetToY as $row)
{
$statementSetToY->execute();
}
}
else
{
// *** fetch one row at a time pertaining to the 2nd query *** //
$resultSetToN = $statementSetToN->fetch();
foreach($resultSetToN as $row)
{
$statementSetToN->execute();
}
}
}
break;
}
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Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] 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 '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 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 $cat is being set, but the first dropdown no longer has the state selected and the second dropdown is populated with All cities and not just one ones from the selected state. Any ides why this is working fine on one machine and not the other? Selected code from reports.php
<?php
?>
/////// for second drop down list we will check if State is selected else we will display all the cities/////
echo "<form method=post action='brewerylistbycity.php'>";
}
////////// Starting of second drop downlist /////////
//// End Form /////
If you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> Say I have this query: site.com?var=1 ..I have a form with 'var2' field which submits via get. Is there a way to produce: site.com?var=1&var2=formdata I was hoping there would be a quick way to affix, but can't find any info. Also, the query could sometimes be: site.com?var2=formdata&var=1 I would have to produce: site.com?var2=updatedformdata&var=1 Is my only option to further parse the query? I was just wondering if it's possible to run a query on data that has been returned from a previous query? For example, if I do Code: [Select] $sql = 'My query'; $rs = mysql_query($sql, $mysql_conn); Is it then possible to run a second query on this data such as Code: [Select] $sql = 'My query'; $secondrs = mysql_query($sql, $rs, $mysql_conn); Thanks for any help I'm trying to update every record where one field in a row is less than the other. The code gets each row i'm looking for and sets up the query right, I hope I combined the entire query into one string each query seperated by a ; so it's like UPDATE `table` SET field2= '1' WHERE field1= '1';UPDATE `table` SET field2= '1' WHERE field1= '2';UPDATE `table` SET field2= '1' WHERE field1= '3';UPDATE `table` SET field2= '1' WHERE field1= '4';UPDATE `table` SET field2= '1' WHERE field1= '5'; this executes properly if i run the query in phpMyAdmin, however when I run the query in PHP, it does nothing... Any advice? here's the code: Code: [Select] $companyName = 'big company'; $address1 = 'big bay #8'; $address2 = 'some big warehouse'; $city = 'big city'; $province = 'AB'; $postalCode = 'T1T0N0'; $phone = '0123456789'; $email2 = 'bigKahuna@bigKahuna.edu'; $query = "INSERT INTO clients ( companyName, address1, address2, city, province, postalCode, phone, email) VALUES ( ". $companyName.",".$address1.",".$address2.",".$city.",".$postalCode.",".$phone.",".$email2.")"; $result = mysql_query($query, $connexion); if ($result) { // Success! echo "Fabulous! check the DB, we did it! :D<br>"; ?> <pre> <?php print_r($result); ?> </pre> <?php } else { // Fail! echo"CRAAAAAPP! something went wrong. FIX IT! :P<br>"; echo mysql_error(); } if (isset($connexion)) { mysql_close($connexion); } i copied it over from an old *working* file to illustrate how a simple INSERT works. this is the error i get: Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'company,big bay #8,some big warehouse,big city,T1T0N0,0123456789,bigKahuna@bigKa' at line 4 looks completely valid to me. all the database table elements are set to VARCHAR(80), so it can't be a space/type issue... halp! WR! What would be the correct way to close a mysql query? At current the second query below returns results from the 1st query AND the 2nd query The 3rd query returns results from the 1st, 2nd and 3rd query. etc etc. At the moment I get somthing returned along the lines of... QUERY 1 RESULTS Accommodation 1 Accommodation 2 Accommodation 3 QUERY 2 RESULTS Restaurant 1 Restaurant 2 Restaurant 3 Accommodation 1 Accommodation 2 Accommodation 3 QUERY 3 RESULTS Takeaways 1 Takeaways 2 Takeaways 3 Restaurant 1 Restaurant 2 Restaurant 3 Accommodation 1 Accommodation 2 Accommodation 3 Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <?php include($_SERVER['DOCUMENT_ROOT'].'/include/db.php'); ?> <title>Untitled Document</title> <style type="text/css"> <!-- --> </style> <link href="a.css" rel="stylesheet" type="text/css" /> </head><body> <div id="listhold"> <!------------------------------------------------------------------------------------------------------------------------------------------------------> <div class="list"><a href="Placestostay.html">Places To Stay</a><br /> <?php $title ="TITLE GOES HERE"; $query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Accommodation' AND confirmed ='Yes' ORDER BY name"); echo mysql_error(); while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0]; $i = -1; foreach($nt as $value) {$i++; $FileName = str_replace(' ','_',$nt[$i]) . ".php"; $FileUsed = str_replace('_',' ',$nt[$i]); echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>" . $nt[$i] . "</a>" . "<br/>"; $FileHandle = fopen($FileName, 'w') or die("cant open file"); $pageContents = file_get_contents("header.php"); fwrite($FileHandle,"$pageContents");} fclose($FileHandle); ?> </div> <!------------------------------------------------------------------------------------------------------------------------------------------------------> <div class="list"><a href="Eatingout.html">Eating Out</a><br /> <?php $title ="TITLE GOES HERE"; $query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Restaurant' AND confirmed ='Yes' ORDER BY name"); echo mysql_error(); while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0]; $i = -1; foreach($nt as $value) {$i++; $FileName = str_replace(' ','_',$nt[$i]) . ".php"; $FileUsed = str_replace('_',' ',$nt[$i]); echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>" . $nt[$i] . "</a>" . "<br/>"; $FileHandle = fopen($FileName, 'w') or die("cant open file"); $pageContents = file_get_contents("header.php"); fwrite($FileHandle,"$pageContents");} fclose($FileHandle); ?> </div> <!------------------------------------------------------------------------------------------------------------------------------------------------------> <div class="list"><a href="Eatingin.html">Eating In</a><br /> <?php $title ="TITLE GOES HERE"; $query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Takeaways' AND confirmed ='Yes' ORDER BY name"); echo mysql_error(); while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0]; $i = -1; foreach($nt as $value) {$i++; $FileName = str_replace(' ','_',$nt[$i]) . ".php"; $FileUsed = str_replace('_',' ',$nt[$i]); echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>" . $nt[$i] . "</a>" . "<br/>"; $FileHandle = fopen($FileName, 'w') or die("cant open file"); $pageContents = file_get_contents("header.php"); fwrite($FileHandle,"$pageContents");} fclose($FileHandle); ?> </div> <!------------------------------------------------------------------------------SKILLED TRADES BELOW---------------------------------------------------> <div class="list"><a href="Skilledtrades.html">Skilled Trades</a><br/> <?php $title ="TITLE GOES HERE"; $query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Skilled Trades' AND confirmed ='Yes' ORDER BY name"); echo mysql_error(); while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0]; $i = -1; foreach($nt as $value) {$i++; $FileName = str_replace(' ','_',$nt[$i]) . ".php"; $FileUsed = str_replace('_',' ',$nt[$i]); echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>" . $nt[$i] . "</a>" . "<br/>"; $FileHandle = fopen($FileName, 'w') or die("cant open file"); $pageContents = file_get_contents("header.php"); fwrite($FileHandle,"$pageContents");} fclose($FileHandle); ?> </div> I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? I am trying to create a query which reads and uses a previous query which could go on for upto four queries. For example: Query: $carcolour(red), Query: $carmodel(ford), Query: $enginesize(1600), Query: $carlocation(New York) This displays all red cars, which are Ford, which 1600CC, which are located in New York. or Query: $enginesize(1600), Query: $carcolour(red), Query: $carmodel(ford), Query: $carlocation(New York) This displays all 1600CC cars, which red car, which are Ford, which are located in New York. (Same result as above) I have found this guide but Im not sure it what I am looking for. I have also come across the Join function. However this seems to be based on joining two seperate queries. http://www.suite101.com/content/how-tor-run-multiple-mysql-queries-with-php-a105672 Can anyone advise on the best way to create a set of queries which reads and uses the results of the previous query? Good Day Guys
I have a bit of a urgent problem.
Here is my Query:
$query = "SELECT distinct Img.propertyId as PropertyId, Title, ImageUrl, Location, Bedrooms, Bathrooms, Parking, Price FROM PROPERTIES as Prop LEFT JOIN IMAGES as Img ON Img.PropertyId = Prop.PropertyId WHERE 1=1 AND Price >=1000 AND Price <=5000 "; I currently have a web page which the user selects the value from the drop menu. I then want that value to go into another form in the same page inside a hidden field. Which will then be submitted to the processing php. The second option is the user selects the drop menu then the processing php would have to grab the value without a submit button because I am using a save button on the same page in the other form. Any help Hi, I'm wondering how i would go about making a drop down menu in php to insert the option into users > Maint of my database, if anyone has any tutorials or examples it would be great, Thanks a lot. Hi, I have a dropdown box, and I want it so that if dropdownbox.value='selected:' then show error message. The checkbox is dynamically populated. Pleaee Advise, - stuart |
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