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PHP - Populating Dropdown Lists With Mysql , Then Submitting Form To New Table
hello. i have an issue where the data stored with an image is not saving to a mysql table.
the image data is ok, just not the selections from the dropdown lists. here is the code <?php include ('connect.php'); // Insert any new image into database if(isset($_POST['xsubmit']) && $_FILES['imagefile']['name'] != "") { $fileName = $_FILES['imagefile']['name']; $fileSize = $_FILES['imagefile']['size']; $fileType = $_FILES['imagefile']['type']; $content = addslashes (file_get_contents($_FILES['imagefile']['tmp_name'])); $jeweltype = $_POST['jeweltype']; $jewelsize = $_POST['jewelsize_in']; $jewelcolour = $_POST['jewelcolour_in']; $jewelmaterial = $_POST['jewelmaterial_in']; $jewelgender = $_POST['jewelgender_in']; if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } // Checking file size if ($fileSize < 150000) { mysql_query ("INSERT into jewel_images (name,size,type,content,jeweltype,jewelsize,jewelcolour,jewelmaterial,jewelgender) " . "values ('$fileName','$fileSize','$fileType','$content','$jeweltype','$jewelsize','$jewelcolour','$jewelmaterial','$jewelgender')"); } else { $err = "The Image file to too large!"; } } // Find out about latest image $gotten = mysql_query("select * from jewel_images order by row_id desc"); $row = mysql_fetch_assoc($gotten); $bytes = $row['content']; // If this is the image request, send out the image if ($_REQUEST['pic'] == 1) { header("Content-type: $row[type];"); print $bytes; } ?> <html> <head> <title>Upload an image to a database</title> </head> <body> <font color="#FF3333"><?php echo $err ?></font> <table> <form name="Upload" enctype="multipart/form-data" method="post"> <tr> <td>Upload <input type="file" name="imagefile"><br /> Jewelery Type: <select> <?php $sql="SELECT jeweltype FROM jeweltypes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jeweltype" ><?php echo $data['jeweltype'] ?></option> <?php } ?> </select> <br /> Jewelery Size: <select> <?php $sql="SELECT jewelsize FROM jewelsizes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelsize" ><?php echo $data['jewelsize'] ?></option> <?php } ?> </select> <br /> Jewelery Colour: <select> <?php $sql="SELECT jewelcolour FROM jewelcolours"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelcolour_in" ><?php echo $data['jewelcolour'] ?></option> <?php } ?> </select> <br /> Jewelery Material: <select> <?php $sql="SELECT jewelmaterial FROM jewelmaterials"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelmaterial_in" ><?php echo $data['jewelmaterial'] ?></option> <?php } ?> </select> <br /> Jewelery Gender: <select> <?php $sql="SELECT jewelgender FROM jewelgenders"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelgender_in" ><?php echo $data['jewelgender'] ?></option> <?php } ?> </select> <br /> <input type="submit" name="xsubmit" value="Upload"> </td> </tr> <tr> <td>Latest Image</td> </tr> <tr> <td><img src="?pic=1"></td> </tr> </form> </table> </body> </html> ============================================== here is the query =============================================== <html> <head><title>Your Page Title</title></head> <body> <?php $database="josh_jewel"; mysql_connect ("localhost", "xxxxxxxxx", "yyyyyyyyyyyy"); @mysql_select_db($database) or die( "Unable to select database"); $result = mysql_query( "SELECT jewelcolour FROM jewel_images" ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); print "There are $num_rows records.<P>"; print "<table width=400 border=1>\n"; while ($get_info = mysql_fetch_row($result)){ print "<tr>\n"; foreach ($get_info as $field) print "\t<td><font face=arial size=1/>$field</font></td>\n"; print "</tr>\n"; } print "</table>\n"; ?> </body> </html> Similar TutorialsHi, Can someone help me with the following problem... I am populating a html table with results from a mysql query. These results populate the 1st of four columns. The second column is a "RAG STATUS" dropdown menu - so GREEN/AMBER/RED. When this is selected I want it to change the colour of the corresponding row it is on. I have had a play around but can only get it to change the colour of the first row, no matter which dropdown menu I change. Code is below. If anything is not clear please let me know. Any help appreciated Code: [Select] <?php include ("commonTop.php"); include("dbvariables.php"); include("functions.php"); ?> <script type="text/javascript"> function changeBGCol(status) { document.getElementById("colour").bgColor=status; } </script> <?php $Checkout_ID = 2; $result = mysql_query( "SELECT Task FROM Task T, Checkout C Where T.Checkout_ID = C.Checkout_ID and C.Checkout_ID= $Checkout_ID" ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); echo "This will be sent to the following recepients $num_rows records.<P>"; echo "<table width=70% border=3>\n"; echo "<tr><th>Check</th><th>STATUS</th><th>JIRA</th><th>Comments</th></tr>"; while ($get_info = mysql_fetch_row($result)){ echo "<tr id=colour>\n"; foreach ($get_info as $field) echo "\t<td><font face=arial size=1/>$field</font></td>\n"; echo "\t<td><select name='Status'> <option value='None'>-- Choose --</option> <option onclick='changeBGCol(this.value)' value='green' >GREEN</option> <option onclick='changeBGCol(this.value)' value='orange'>AMBER</option> <option onclick='changeBGCol(this.value)' value='red'> RED</option> </select></td>"; echo "\t<td><input type='text' name='JIRA'></td>"; echo "\t<td><input type='text' name='Comments'></td>"; echo "</tr>\n"; } print "</table>\n"; ?> [code] </code> Hello to all. I'm new here and i am very glad to be a part of this community. I am new in PHP/mysql but i try to learn as much as i can in php & mysql. And here is the problem: Into a php or html page i want to display 2 dropdown list to act as a filter for displaying a table. I managed to display a table based on option select from first dropdown but i don'y know how to make or activate the second dropdown list to act as a filter, and when i press "result" to display on the same page the results. Thanks a lot. Dear All, I have attached here with a schema, of MySql DB format i am refering to for the topic i need help on. I want to know what is the best way to achieve this problem / task. I have a profile table, and a category table, and a table to map profiles to category. Which allows me to define/add into the category_profile_mapping table, profile ids against a category id. Each category can have anywhere between 2 to 1,00,000 profiles mapped to it. I have another table called sms_out. hence on the site, when i select a category and submit the form value with message, i want to know what is the best way to populate sms_out with MSISDN from the table "Profile" and the message from the form. The list of MSISDN will be based on the category id selected, hence the profile id's will be available in the category_profile_mapping table. I am bad at explaining the problem, however incase you folks need more info do let me know. i shall be glad to provide the same to .. all of you guys who are willing to help me PS: I am asking this question, cuz i want to knw the best solution to achieve this, when there is entries in millions to be done at one form submit. [attachment deleted by admin] Ok, so i got this website project dropped in my lap, and I'm stuck not really having done much php coding ever. I have a database which is linked already, and it was working when pulling data from a specific table. The problem arose when I was asked to make it work with other tables. I have a table with the table names in it, and the drop down field to select the table populates. The next field is a multiple select field that is supposed to populate report dates from the previously selected table, and then finally the last multiple select field is supposed to populate with names that match the dates selected from the previous field. The last one I can live without, as the existing code takes into account if the name selected doesn't exist in the table with the selected date, but at the moment the date field just comes up blank. Any help would be greatly appreciated. I'm trying to retrieve data from my DB and have it populate the dropdown values inside a form: echo "<option value='0.00' " . ($array['roastturkey'] == '0.00' ? 'selected="selected"' : '') . ">0</option>";echo "<option value='1.00' " . ($array['roastturkey'] == '1.00' ? 'selected="selected"' : '') . ">1</option>";echo "<option value='2.00'> " . ($array['roastturkey'] == '2.00' ? 'selected="selected"' : '') . ">2</option>";echo "<option value='3.00'> " . ($array['roastturkey'] == '3.00' ? 'selected="selected"' : '') . ">3</option>";echo "<option value='4.00'> " . ($array['roastturkey'] == '4.00' ? 'selected="selected"' : '') . ">4</option>";echo "<option value='5.00'> " . ($array['roastturkey'] == '5.00' ? 'selected="selected"' : '') . ">5</option>";echo "<option value='6.00'> " . ($array['roastturkey'] == '6.00' ? 'selected="selected"' : '') . ">6</option>";echo "<option value='7.00'> " . ($array['roastturkey'] == '7.00' ? 'selected="selected"' : '') . ">7</option>";echo "<option value='8.00'> " . ($array['roastturkey'] == '8.00' ? 'selected="selected"' : '') . ">8</option>";echo "<option value='9.00'> " . ($array['roastturkey'] == '9.00' ? 'selected="selected"' : '') . ">9</option>";echo "<option value='10.00'> " . ($array['roastturkey'] == '10.00' ? 'selected="selected I'm using this code, but cannot get the data that's inside the DB to populate the dropdown value. I want it to get the info from the database so that the option that is saved in the database will be new default when the page is loaded. If I don't change it to the previous info it will update the database with the default option rather then the actual option. Could anyone point me in the correct direction of how to go about excluding results from a drown down menu, Basically i'm calling everyone who's a contact to the user in the code below, id like to remove contacts who already have a bank account with the user, So basically its something along the lines of if the player_id is already in the table bank_accounts as PlayersID, then its excluded, Hope this makes sense and any help would be great, Thanks Code: [Select] <select name="contact1" class="maintablepstats" id="contact1"> <option value=""></option> <?php $sql = "SELECT player_id, contact_id, name, is_active FROM contacts as c JOIN players as p ON c.contact_id = p.id WHERE c.player_id = $playerID AND is_active = 1 ORDER BY name ASC"; $que = mysql_query($sql) or die(mysql_error()); while($list = mysql_fetch_array($que)) { ?> <option value="<?php echo $list['contact_id'] ?>"><?php echo $list['name'] ?></option> <?php } ?> </select> Hello, I'm trying to populate a dropdown box in a RSForm using this code:
//<code>
$db = JFactory::getDbo();
$db->setQuery("SELECT Bruel_ID FROM mpctz_rsform_bruels");
return $db->loadObjectList();
//</code>
However, it displays nothing in the box and some code outside of it (see attached file). Can anyone help?Thanks, Dani Attached Files APNAE.jpg 16.92KB 0 downloads I am trying to figure out how to display member records after selecting it from the box. I've got the dropdown box working which retrieves the members name but cannot figure out how to populate that members details on the same page? I'm using php/mysql and although I want to display the records I also want a user to update that record aswell, creating it for administrators to update accounts? Any help would be appreciated I've researched that ajax or javascript might be helpful but not totally familiar with them. Hi, I don know how to explain this in words too well, so i have create 2 images to show what I need. What I need to do is use whats in the first image (table.jpg)(the database) to create whats in the second image (dropdown.jpg)(the drop-down menu) The results of the form will be entered into the database. category_id will be the value thats inserted. Many kind regards Eoin [attachment deleted by admin] Hello there! I've been banging my head on this for a while and I just can't seem to get it to work properly. I have a dropdown menu which selects information from table1 using a select statement (this table is called 'lid'). It selects the firstname, lastname and member id from this table and shows it in the dropdown menu. I'm glad I got that part working but the hard thing is inserting the data that the user selects into another table. So when you select the id member from this dropdown menu it only inserts a blank row into table2 (which is called 'teamlid'). Can you guys help me? How can I insert the id member into my table2? What am I doing wrong here? Thanks a million! This is my first post so if I'm doing anything wrong, let me know and I'll fix it asap! My code:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Boast & Drive</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<style type="text/css">
.wrapper{
width: 650px;
margin: 0 auto;
}
.page-header h2{
margin-top: 0;
}
table tr td:last-child a{
margin-right: 15px;
}
</style>
</head>
<body>
<div class="container-fluid">
<div class="row">
<div class="col-md-12">
<div class="page-header clearfix">
<h2 class="pull-left">Teamleden</h2>
<div class="btn-toolbar">
<a href="read.php" class="btn btn-primary btn-lg pull-right">Terug</a>
</div>
</div>
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
error_reporting(E_ALL);
//Verbinding maken met de database
require_once "login.php";
$sql = "SELECT tl.teamnaam,
tl.tl_ID,
tl.lidnummer,
l.voornaam,
l.achternaam
FROM teamlid tl
JOIN lid l ON tl.lidnummer = l.lidnummer
ORDER BY tl.teamnaam;";
if($result = mysqli_query($conn, $sql)) {
if(mysqli_num_rows($result) > 0) {
echo "<table class='table table-bordered table-striped'>";
echo "<thead>";
echo "<tr>";
echo "<th>Teamnaam</th>";
echo "<th>Tl_ID</th>";
echo "<th>Lidnummer</th>";
echo "<th>Voornaam</th>";
echo "<th>Achternaam</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['teamnaam'] . "</td>";
echo "<td>" . $row['tl_ID'] . "</td>";
echo "<td>" . $row['lidnummer'] . "</td>";
echo "<td>" . $row['voornaam'] . "</td>";
echo "<td>" . $row['achternaam'] . "</td>";
echo "<td>";
echo "<a href='update.php?id=". $row['lidnummer'] ."' title='Gegevens wijzigen' data- toggle='tooltip'><span class='glyphicon glyphicon-pencil'></span></a>";
echo "<a href='delete.php?id=". $row['lidnummer'] ."' title='Lid verwijderen' data- toggle='tooltip'><span class='glyphicon glyphicon-trash'></span></a>";
echo "</td>";
echo "</tr>";
}
echo "</tbody>";
echo "</table>";
mysqli_free_result($result);
} else{
echo "<p class='lead'><em>Er zijn geen gegevens om weer te geven.</em></p>";
}
} else{
echo "De volgende fout is gevonden: " . mysqli_error($conn);
}
?>
<form name="dropdown" method="post">
<div class="page-header clearfix">
<h2 class="pull-left">Teamlid toevoegen</h2>
</div>
<p>Selecteer hieronder met behulp van het dropdown menu een lid welke je aan bovenstaand team wilt toevoegen</p>
<div class="container-fluid">
<div class="row">
<?php
// Variabelen aanmaken en tonen met lege waardes
$teamnaam = $lidnummer = '';
// Code voor dropdown. Selecteert voornaam, achternaam en lidnummer van tabel lid)
$sql = "SELECT voornaam, achternaam, lidnummer FROM lid ORDER BY achternaam";
$result = mysqli_query($conn, $sql);
echo "<select id='teamLid' name='teamLid'>";
echo "<option>--Selecteer Lid--</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['lid'] . "'>" . $row['voornaam'] . " " . $row['achternaam'] . " " . $row['lidnummer'] . "</option>";
}
echo "</select>";
if
(isset($_POST["id"]) && !empty($_POST["id"])) {
$id = $_POST["teamLid"];
$stmt = $conn->prepare("INSERT INTO teamlid (teamnaam, lidnummer) VALUES (?,?)");
$stmt->bind_param('si', $param_teamnaam, $param_lidnummer);
$param_teamnaam = $teamnaam;
$param_lidnummer = $lidnummer;
$stmt->execute();
}
// Verbinding sluiten
mysqli_close($conn);
?>
<div>
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input type="submit" name="submit" class="btn btn-primary" value="Toevoegen">
</div>
</div>
</div>
</form>
</div>
</div>
</body>
</html>
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 /////
Hello,
I have been trying to figure this for a while now and reading the tutorials are not helping, I think I'm a little over my head on this one and was hoping someone could help me out with this issue.
I am making a User Edit page and would like to have the access level part of the form show the users access current access level thats set in the database when the page loads, and if it needs to be changed you can press the dropdown box and select a new access level. I can't figure out how to show the current access level as default and populate the drop down box with the other access levels in my table.
My Tables look like this
Users table (users):
---------------------------------------------------------------------------------------------------
| id | username | password | flag | realfirst | reallast | dept |
---------------------------------------------------------------------------------------------------
1 loderd 9 test guy Service
Auth Table (auth):
--------------------------------------------
| id | auth_level | descrip |
--------------------------------------------
1 1 Service Tech
2 2 Office Staff
3 9 Super Admin
My SQL Query looks like this
$users = $db->fetch_all_array("SELECT users.id, users.username, users.password, users.realfirst, users.reallast, users.dept, users.flag, auth.auth_level, auth.descrip, auth.id
FROM users LEFT JOIN auth
ON users.flag = auth.auth_level
WHERE users.id = '".$_GET['id']."'");
I can't seem to figure out how I can do this for the Access Level dropdown box.
<tr>
<td width="19%" align="right" valign="top">Access Level :</td>
<td width="1%" align="left"> </td>
<td width="80%" align="left" valign="top">
<?php
echo "<select name='flag' id='flag'>";
foreach ($users as $row){
if($row[auth_level]==$row[auth_level]){
echo "<option value=$row[auth_level] selected>$row[auth_level] - $row[descrip]</option>";
}else{
echo "<option value=$row[auth_level]>$row[auth_level] - $row[descrip]</option>";
}
}
echo "</select>"; ?>
</td>
</tr>
Any help would be greatly appreciatedHi everyone, I've read through the FAQs for Dynamic Dropdown Menus on this site, as well as others, and I can't figure out why my code won't work. I have two dropdown boxes that need to be populated with data from a mysql table; one menu for route types, and one for route numbers. When I choose the route from the 'Route' menu, the 'Number' menu automatically populates with all of the possible numbers, rather than only those that correspond with the route type. I can tell that the problem has something to do with the value of 'Route' not being recognized, but I don't know why. I'm a beginner when it comes to PHP, so any suggestions or help would be much appreciated! Thanks! The code is as follows: Code: [Select] <html> <body> <basefont face='calibri' color='#7E2217'> <?php // set variables $mileTable = $_GET['mileTable']; $routeType = isset($_POST['Roadtype'])? $_POST['Roadtype']: 0; include 'opendbMile.php'; include 'error.php'; // Connect to the MySQL DBMS if (!($connection = @ mysql_connect($hostName, $username, $password))) die("Could not connect"); if (!mysql_select_db($databaseName, $connection)) showerror( ); // Start a query... $query = "SELECT ID, Roadtype FROM Alabama GROUP BY Roadtype"; // execute the SQL statement $result = mysql_query($query, $connection) or die(mysql_error()); echo '<form name="mileform" method="post" action="MileQuery.php">'; echo '<p>Route: <select name="routeType" id="routeType" onchange="this.form.submit();"> <option value="0"'.($routeType == 0? ' SELECTED': '').'>Route</option>'; while($row = mysql_fetch_array($result)){ echo ' <option value="'.$row[0].'"'.(($routeType == $row[0])? ' SELECTED': '').'>'.$row[1].'</option>'; } echo ' </select> </p>'; // create the SQL statement $query2 = "SELECT ID, Roadnumber FROM Alabama GROUP BY Roadnumber"; if($mnucategory != 0){ // Filter road numbers $query2 .= " WHERE Roadtype='".$routeType."'"; } // execute the SQL statement $result2 = mysql_query($query2, $connection) or die(mysql_error()); echo '<p>Number: <select name="routeNumber" id="routeNumber" onchange="this.form.submit();"> <option value="0"'.($routeNumber == 0? ' SELECTED': '').'>Number</option>'; while($row2 = mysql_fetch_array($result2)){ echo ' <option value="', $row2[0].'"'.(($routeNumber == $row2[0])? ' SELECTED': '').'>'. $row2[1], '</option>'; } echo ' </select> </p>'; // Close the DBMS connection mysql_close($connection); ?> </form> </body> </html> <?php $con=mysql_connect("localhost","root",""); if(!$con) { die('could not connect' .mysql_error()); } mysql_select_db("hrc_fault",$con); $query = "SELECT * " . "FROM fault_book"; $result = mysql_query($query, $con) or die(mysql_error()); $num_movies = mysql_num_rows($result); $registered_comp=<<<EOD <h2><center>Registered Fault HRC</center></h2> <table width="70%" border="1" cellpadding="2" cellspacing="2" align="center"> <tr> <th>Job Cd No</th> <th>Date</th> <th>Section</th> <th>Item Description</th> <th>Item Sl.No</th> <th>Fault</th> </tr> EOD; $fault_details = ''; while ($row = mysql_fetch_array($result)) { $jc_no = $row['jc_no']; $date = $row['date']; $section = $row['section']; $itm_des = $row['itm_des']; $itm_slno = $row['itm_slno']; $fault_brf = $row['fault_brf']; $fault_details .=<<<RAM <tr> <td>$jc_no</td> <td>$date</td> <td>$section</td> <td>$itm_des</td> <td>$itm_slno</td> <td>$fault_brf</td> </tr> RAM; } $movie_footer ="</table>"; $movie =<<<MOVIE $jc_no $date $section $itm_des $itm_slno $fault_brf MOVIE; echo "There are $num_movies complains in our database"; echo $movie; ?> I m using above code but it only display the last record please debug the code I have read around and can't seem to find the right coding for what I need on this forum and some other other forums. I have a contact form (as listed below) and I need 2 locations (Print Name and Title) fields to auto-populate on a separate form (can be a doc, pdf, etc. any form of document which is easiest) and this form can be totally back end and the individual using the form never is going to see the form. It's going on a contract form, that we would like to auto-populate. Also is there a simple attachment code so individuals can attach documents to the code? <p style: align="center"><form action="mailtest.php" method="POST"> <?php $ipi = getenv("REMOTE_ADDR"); $httprefi = getenv ("HTTP_REFERER"); $httpagenti = getenv ("HTTP_USER_AGENT"); ?> <input type="hidden" name="ip" value="<?php echo $ipi ?>" /> <input type="hidden" name="httpref" value="<?php echo $httprefi ?>" /> <input type="hidden" name="httpagent" value="<?php echo $httpagenti ?>" /> <div align="center"> <p class="style1">Name</p> <input type="text" name="name"> <p class="style1">Address</p> <input type="text" name="address"> <p class="style1">Email</p> <input type="text" name="email"> <p class="style1">Phone</p> <input type="text" name="phone"> <p class="style1">Debtor</p> <input type="text" name="debtor"> <p class="style1">Debtor Address</p> <input type="text" name="debtora"> <br /> <br /> <a href="authoforms.php" target="_blank" style="color:#ffcb00" vlink="#ffcb00">Click here to view Assignment Agreement and Contract Agreement</a> <p class="style1"><input type='checkbox' name='chk' value='I Have read and Agree to the terms.'> I have read and agree to the Assignment and Contract Agreement <br></p> <p class="style1">Print Name</p> <input type="text" name="pname"> <p class="style1">Title</p> <input type="text" name="title"> <p class="style1">I hear by agree that the information I have provided is true, accurate and the information I am submitting is <br /> not fraudulent. Please click the agree button that you adhere to Commercial Recovery Authority Inc.'s terms:</p> <select name="agree" size="1"> <option value="Agree">Agree</option> <option value="Disagree">Disagree</option> </select> <br /> <br /> <p class="style1">Employee ID:</p> <input type="text" name="employee"> <br /> <input type="submit" value="Send"><input type="reset" value="Clear"> </div> </form> </p> The mailtest php is this ?php $ip = $_POST['ip']; $httpref = $_POST['httpref']; $httpagent = $_POST['httpagent']; $name = $_POST['name']; $address = $_POST['address']; $email = $_POST['email']; $phone = $_POST['phone']; $debtor = $_POST['debtor']; $debtora = $_POST['debtora']; $value = $_POST['chk']; $pname = $_POST['pname']; $title = $_POST['title']; $agree = $_POST['agree']; $employee = $_POST['employee']; $formcontent=" From: $name \n Address: $address \n Email: $email \n Phone: $phone \n Debtor: $debtor \n Debtor's Address: $debtora \n 'Client' has read Assignment and Contract Agreement: $value \n Print Name: $pname \n Title: $title \n I hear by agree that the information I have provided is true, accurate and the information I am submitting is not fraudulent. Please click the agree button that you adhere to Commercial Recovery Authority Inc.'s terms: $agree \n \n Employee ID: $employee \n IP: $ip"; $recipient = "mail@crapower.com"; $subject = "Online Authorization Form 33.3%"; $mailheader = "From: $email \r\n"; mail($recipient, $subject, $formcontent, $mailheader) or die("Error!"); echo "Thank You!" . " -" . "<a href='index.php' style='text-decoration:none;color:#ffcb00;'> Return Home</a>"; $ip = $_POST['visitoraddress'] ?> Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike. In each table their are 2 fields, one INDEX_ID and MODELS. Under that for rows I have every model bike named. A quick question before I get onto what I want to do: Can data be added underneath the rows? For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit. To make it easier to understand I'll post the SQL code for the brand Honda: CREATE TABLE `Honda` ( `INDEX_ID` int(3) NOT NULL auto_increment, `MODELS` varchar(20) collate latin1_general_ci NOT NULL, PRIMARY KEY (`INDEX_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ; -- -- Dumping data for table `Honda` -- INSERT INTO `Honda` VALUES(1, 'CR85'); INSERT INTO `Honda` VALUES(2, 'CR125'); INSERT INTO `Honda` VALUES(3, 'CR250'); INSERT INTO `Honda` VALUES(4, 'CRF100'); INSERT INTO `Honda` VALUES(5, 'CRF150'); INSERT INTO `Honda` VALUES(6, 'CRF230'); INSERT INTO `Honda` VALUES(7, 'CRF250X'); INSERT INTO `Honda` VALUES(8, 'CRF250R'); INSERT INTO `Honda` VALUES(9, 'CRF450X'); INSERT INTO `Honda` VALUES(10, 'CRF450R'); INSERT INTO `Honda` VALUES(11, 'CRF50'); INSERT INTO `Honda` VALUES(12, 'CRF70'); INSERT INTO `Honda` VALUES(13, 'CRF80'); INSERT INTO `Honda` VALUES(14, 'XR650'); INSERT INTO `Honda` VALUES(15, 'CR500'); INSERT INTO `Honda` VALUES(16, 'XR100'); INSERT INTO `Honda` VALUES(17, 'XR200'); INSERT INTO `Honda` VALUES(18, 'XR250'); INSERT INTO `Honda` VALUES(19, 'XR400'); INSERT INTO `Honda` VALUES(20, 'XR50'); INSERT INTO `Honda` VALUES(21, 'XR70'); INSERT INTO `Honda` VALUES(22, 'XR80'); Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?: <? $connection = mysql_connect("localhost","user","pass"); $fields = mysql_list_fields("dbname", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; ?> Thanks. How would I force a certain table format and then populate it with the SQL results in their respective columns and rows? i.e. Code: [Select] +------------------+----------+----------+----------+ | | Col 1 | Col 2 | Col 3 | +-------------------+----------+----------+----------+ | row 1 lab data | 16777216 | | 573 | +-------------------+----------+----------+----------+ | row 2 lab data | 23454235 | 87247247 | 65743 | +-------------------+----------+----------+----------+ | row 3 lab data | 16777216 | 47832364 | | +-------------------+----------+----------+----------+ Currently I'm just using a while loop to populate it horizontally, but that doesn't work for row 1. Code: [Select] <?PHP while($row = mysql_fetch_array($result) { ?> blah blah blah html here <?PHP echo $row[0]; ?> more html <?PHP echo $row[1]; ?> and so forth <?PHP } ?> It ends up producing this: Code: [Select] +------------------+----------+----------+----------+ | | Col 1 | Col 2 | Col 3 | +-------------------+----------+----------+----------+ | row 1 lab data | 16777216 | 573 | | +-------------------+----------+----------+----------+ | row 2 lab data | 23454235 | 87247247 | 65743 | +-------------------+----------+----------+----------+ | row 3 lab data | 16777216 | 47832364 | | +-------------------+----------+----------+----------+ Ok, so I've spent quite a bit of time piecing together this solution from a variety of sources. As such, I may have something in my code below that doesn't make sense or isn't neccessary. Please let me know if that is the case. I'm creating an administrative form that users will you to add/remove items from a MySQL table that lists open positions for a facility. The foreach loop generates all of the possible job specialties from a table called 'specialty_list'. This table is joined to a second table ('open_positions') that lists any positions that have been selected previously. Where I'm stuck is getting the checkbox to be checked if the facility_ID from the open_positions table matches the $id passed in the URL via ?facility_id=''. Here's where I am so far: $query = "SELECT specialty_list.specialty_displayname , specialty_shortname , open_positions.position , facility_ID FROM specialty_list LEFT OUTER JOIN open_positions ON open_positions.position = specialty_list.specialty_shortname ORDER BY specialty_list.specialty_shortname"; $results = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($results)) { $positions[$row['specialty_shortname']] = $row['specialty_displayname']; } echo "<form method='POST' action='checkbox.php'>"; foreach($positions as $specialty_shortname => $specialty_displayname) { $facility_ID = $row['facility_ID']; $checked = $facility_ID == $row['facility_ID'] ? ' checked' : ''; echo "<input type='checkbox' name='position[]' value=\"{$specialty_shortname}\"{$checked}> {$specialty_displayname}</input><br/>"; } echo "<input type='hidden' name='facility_ID' value='$id'>"; echo "<input type='submit' value='Submit Checkboxes!'>"; echo "</form>"; Any ideas how to get this working? I feel like I'm very close, but I just can't get it. I also tried starting from scratch with a WHILE statement instead of a FOREACH, but haven't tweaked it enough to prevent duplicate checkboxes. With that in mind, here it is, just in case that's a better direction: $query = "SELECT specialty_list.specialty_displayname , specialty_shortname , open_positions.position , facility_ID FROM specialty_list LEFT OUTER JOIN open_positions ON open_positions.position = specialty_list.specialty_shortname ORDER BY specialty_list.specialty_shortname"; $results = mysql_query($query) or die(mysql_error()); echo "<form method='POST' action='checkbox.php'>"; while($row=mysql_fetch_assoc($results)) { $facility_ID = $row['facility_ID']; $specialty_shortname = $row['specialty_shortname']; $specialty_displayname = $row['specialty_displayname']; if ($facililty_ID==$id) { $checked=' checked'; } echo "<input type='checkbox' name='position[]' value=\"$specialty_shortname\"$checked> $specialty_displayname</input><br/>"; } echo "<input type='hidden' name='facility_ID' value='$id'>"; echo "<input type='submit' value='Submit Checkboxes!'>"; echo "</form>"; Hi Everyone, I was looking for some direction on how to go about this. A client has asked to set up a small league table with 12 teams where simply they can add fixtures and the result of that fixture. So pending on what the result is if they win they get 2 points and if they loose they get nothing and this will determine what position they are in the league table. Any help on this matter would be greatly appreciated even a small step in the right direction Thanks Barry Hi I making some forms that write to mysql database, Im now in the process of making the update form so the user can update there details on the form, I want it to populate the form with existing data but its not doing it at all. Thanks in advance
Attached Files
delete.php 210bytes
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modify.php 4.03KB
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index.php 473bytes
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