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PHP - Dropdown Box Populating The Bgcolour Of A Table
Hi,
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> Similar Tutorialshello. 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> 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> 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. 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 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 /////
<?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 $movie; ?> I m using above code but it only display the last record please debug the code How would I force a certain table format and then populate it with the SQL results in their respective columns and rows? i.e. Code: [Select] +------------------+----------+----------+----------+ | | Col 1 | Col 2 | Col 3 | +-------------------+----------+----------+----------+ | row 1 lab data | 16777216 | | 573 | +-------------------+----------+----------+----------+ | row 2 lab data | 23454235 | 87247247 | 65743 | +-------------------+----------+----------+----------+ | row 3 lab data | 16777216 | 47832364 | | +-------------------+----------+----------+----------+ Currently I'm just using a while loop to populate it horizontally, but that doesn't work for row 1. Code: [Select] <?PHP while($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 | | +-------------------+----------+----------+----------+ 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] 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 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] Hi, I've got a basic sign up form but I want a drop down list which will list different catergories that relate to different tables which when selected will input the sign up information into that table which was selected from the catergory drop down. This is the signup form <html><head><title>Birthdays Insert Form</title> <style type="text/css"> td {font-family: tahoma, arial, verdana; font-size: 10pt } </style> </head> <body> <table width="300" cellpadding="5" cellspacing="0" border="2"> <tr align="center" valign="top"> <td align="left" colspan="1" rowspan="1" bgcolor="64b1ff"> <h3>Insert Record</h3> <form method="POST" action="test.php"> <? print "Enter Company Name: <input type=text name=company_name size=30><br>"; print "Enter Contact Name: <input type=text name=contact_name size=30><br>"; print "Enter Telephone: <input type=text name=telephone size=20><br>"; print "Enter Fax: <input type=text name=fax size=30><br>"; print "Enter Email: <input type=text name=email size=30><br>"; print "Enter Address: <input type=text name=address1 size=20><br>"; print "Enter Address: <input type=text name=address2 size=30><br>"; print "Enter Postcode: <input type=text name=postcode size=30><br>"; print "Enter Town / City: <input type=text name=town_city size=20><br>"; print "Enter Website: <input type=text name=website size=30><br>"; print "Enter Company Type: <select name='table'> <option>stationary</option><option>reception</option></select><br>"; print "<br>"; print "<input type=submit value=Submit><input type=reset>"; ?> </form> </td></tr></table> </body> </html> This is the part which I can't figure out and is probably totally wrong! Im trying to use this script to sort the drop down list to then run the correct script to insert the form data. <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Hello!</title> </head> <body> <?php if($_POST['table']=='stationary' 'birthdays_insert_record.php') else if($_POST['table']=='reception' 'insert_reception.php') ?> </body> </html> This is the script which works! that inserts the form data into a specific table <html><head><title>Birthdays Insert Record</title></head> <body> <? /* Change db and connect values if using online */ $company_name=$_POST['company_name']; $contact_name=$_POST['contact_name']; $telephone=$_POST['telephone']; $fax=$_POST['fax']; $email=$_POST['email']; $address1=$_POST['address1']; $address2=$_POST['address2']; $postcode=$_POST['postcode']; $town_city=$_POST['town_city']; $website=$_POST['website']; $db="myflawlesswedding"; $link = mysql_connect('localhost', 'root' , ''); if (! $link) die(mysql_error()); mysql_select_db($db , $link) or die("Select Error: ".mysql_error()); $result=mysql_query("INSERT INTO reception (company_name, contact_name, telephone, fax, email, address1, address2, postcode, town_city, website) VALUES ( '$company_name', '$contact_name', '$telephone', '$fax', '$email', '$address1', '$address2', '$postcode', '$town_city', '$website')") or die("Insert Error: ".mysql_error()); mysql_close($link); print "Record added"; ?> <form method="POST" action="birthdays_insert_form.php"> <input type="submit" value="Insert Another Record"> </form> <br> <form method="POST" action="birthdays_dbase_interface.php"> <input type="submit" value="Dbase Interface"> </form> </body> </html> I hope somebody can help me out here! or can point me in a better way to sort this problem! Thanks for any advice! 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>
Hello everyone !!! I am sorry if my vocabulary is not exact because english is not my first language. Also i am a newbie at PHP. I am doing this project for myself and if it work might be able to use it at work. But i am doing this to learn. I have been stuck on this problem for 2 weeks and i can t figure it out on my own. Ihave spend many hours searching forums but no success. Oh yeah i almost forgot some part of code are from me, some are scripts from internet i adapted. I have a form with a dropdown menu and when i submit the form the value selected in the dropdown would be inserted in a table. The problem i have is that whatever the value i select it always inserts the last value of the dropdown in the table??? The form is made with the dropdown as an include. It is populated with values from an another table. Here is the code for the dropdown list: Code: [Select] <form> <select name="nom_pcu_form" method="post"> <?php $SQL = "SELECT * FROM pcu ORDER BY nom_pcu"; $res = mysql_query($SQL); while($val=mysql_fetch_array($res)) { $nom=$val["nom_pcu"]; $prenom=$val["prenom_pcu"]; $nom_complet = $prenom . $nom; echo "<option>".$val["nom_pcu"].", ".$val["prenom_pcu"]."</option>\n"; $nom_pcu_form="".$val["nom_pcu"].", ".$val["prenom_pcu"].""; } ?> </select> </form> Here is the part of the form wich calls the dropdown: Code: [Select] <form name="form2" method="post" action="Grille ecoute Permanent.php"> <p>Nom : <?php include 'liste_deroulante_pcu.php' ;?> <p>no carte appel <input name="no_carte_appel" type="text" id="no_carte_appel"> </p> <?php echo date("Y/m/d"); ?> </form> And this is the part where it is inserted in the table: Code: [Select] <?php if($_POST['doSubmit'] == 'Create') mysql_query("INSERT INTO grille_ecoute_pcu_permanent (`user_name`,`nom_pcu`,`no_carte_appel`,`question_1`,`question_2`,`question_3`,`question_4`, `question_5`,`question_6`,`question_7`,`question_8`,`question_9`,`question_10`,`question_11`,`question_12`,`question_13`,`question_14`,`question_15`, `question_16`,`question_17`,`question_18`,`question_19`,`question_20`,`question_21`,`question_22`,`question_23`,`question_24`,`question_25`,`question_26`, `question_27`,`question_28`,`question_29`) VALUES ('$user_name','$nom_pcu_form','$no_carte_appel','$question_1','$question_2','$question_3','$question_4','$question_5','$question_6', '$question_7','$question_8','$question_9','$question_10','$question_11','$question_12','$question_13','$question_14','$question_15','$question_16','$question_17', '$question_18','$question_19','$question_20','$question_21','$question_22','$question_23','$question_24','$question_25','$question_26','$question_27','$question_28', '$question_29') ") or die(mysql_error()); Note:$user_name and all $question are inserted correctly in the table. $nom_pcu_form is the dropdown and it only records the last value of the dropdown even if it s not the value selected. $no_carte_appel are not recorded at all thanks for your time Hi? im just a beginner in php i just want to ask how to insert a data into a table from a dropdown list. I have concatenate the itemid and description to form the dropdown list. But when i viewed my item_table the itemid and description columns are null. can you help me with this.. this is my php code for the dropdown list... <?php $query = "SELECT CONCAT(itemid,' ', '-',' ', description) AS Item FROM item_table"; $result = mysql_query($query) or die(mysql_error()); $dropdown = "<SELECT CONCAT(itemid,' ' '-',' ', description) AS Item FROM item_table>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['Item']}'>{$row['Item']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> this my code for inserting data into the item_table... <?php if(isset($_POST ['submit'])) { $itemid = $_POST['itemid']; $description = $_POST['description']; $datein = $_POST['datein']; $qtyin = $_POST['qtyin']; $unitprice = $_POST['unitprice']; $unit = $_POST['unit']; $category = $_POST['category']; $empid = $_POST['empid']; $message =''; if(($itemid && $description == "")||($itemid && $description == null)) { header("location:IncomingEntry.php?msg=Incorrect"); exit(); } else { $link = mysql_connect('localhost', 'root', '') or die(mysql_error()); $db_selected = mysql_select_db('inventory', $link); $message=''; $query = "INSERT INTO incoming_table (itemid , description, datein, qtyin, unitprice, unit, category, empid) VALUES ('".$itemid."', '".$description."', '".$datein."', '".$qtyin."', '".$unitprice."', '".$unit."', '".$category."', '".$empid."')"; if (!mysql_query($query,$link)) { die('Error: ' . mysql_error()); } header("location: IncomingEntry.php?msg=1 record added"); } } ?> Hi all, firstly apologies as this is a cross post from another forum and we have hit a block.. I am hoping that opening this up to another set of gurus we can get a resolution. What I am trying to achieve is this... I have 2 tables Main and FinancialYear. Main holds all data which I use a form to post the data to it..(all works fine). I use this code to create a drop down in the insert.php form. again this works. Code: [Select] <tr><td>Financial Year: xxxx/xxxx</td><td> <!-- pulls the data from the table variable to populate the dropdown menu --> <?php $database = 'Projects_Main'; $fintable = 'FinancialYear'; if (!mysql_connect($db_host, $db_user, $db_pwd)) die("Can't connect to database 'cos somethin' is wrong"); if (!mysql_select_db($database)) die("Can't select database"); $result = mysql_query("SELECT FinancialYear_id, FinancialYear FROM {$fintable} order by FinancialYear"); $options=""; while ($row=mysql_fetch_array($result)) { //$id=$row["FinancialYear_id"]; $thing=$row["FinancialYear"]; $options.="<OPTION VALUE=\"$thing\">".$thing.'</option>'; } ?> <SELECT NAME="FinancialYear"> <OPTION VALUE=0>Choose</OPTION> <?=$options?> </SELECT> </td></tr> What I have done is built another form which list all records in the database and creates an update url for every record that passes the field Project_id where i use $_get to retrieve the Project_id to retrieve the relevant data into the update.php form. I am able to populate the form with all the correct information BUT I am looking to introduce some dropdowns to aid updating the data and provide consistency to the data. . Code: [Select] // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db('Projects_Main')or die("cannot select DB"); // get value of id that sent from address bar $Project_id=$_GET['Project_id']; //define vars $FinancialYear=$_POST['FinancialYear']; // other vars defined here also.. about 30 // Retrieve data from database $sql="SELECT * FROM Main WHERE Project_id='$Project_id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result); ?> //update_record_ac.php posts the data to the dbase. <form name="form1" method="post" action="update_record_ac.php"> <center> <table> <tr><td><b>Store Details<b></td></tr> <tr><td>Financial Year:</td><td> // takes the data from $rows and present to form <input name="FinancialYear" type="text" id="FinancialYear" value="<?php echo $rows['FinancialYear']; ?>"> // this is where I need to create the drop down.. see my other comments in the post..... </td></tr> the financialYear table consists if the following; financialyear_id - pri, auto inc. ---- data format is 2010/2011, 2011/2012.... financialyear the main table contains 30 fields .. won't list em all... Project_id - pri, auto inc financialyear I need the drop down to pull the data from the financialyear table and then to present or focus on the currently stored data... so if the store value in the table Main is 2010/2011 if Ii was to select the update url in the list_record.php it will pull all the relevant data into update_record.php form. the financialyear field in the form should be a dropdown with all the financial years listed but the 2010/2011 is selected or focused. I still need to be able to change the entry and post this back to the table Main..... So the dropdown contains the list of years from the financialyear table but when the record is pulled from table main the year that is stored in table Main should be highlighted in the dropdown and I should be able to select a new record and post back to the table Main.. any thoughts... please don't slate for the cross post, I haven't sanatised the data at any stage. I know i'm open to injection attacks. and yes my code is a little dirty... all these will be rectified as i finalise the process and ensure the consept works. Thanks for taking the time to read and hopefully you are able to understand the requirement and are able to assist. thanks Balgrath Hello all,
For the UPDATE portion of my CRUD WebApp what I would like to do is to bring in (and display) the values (of a selected row) from my transaction table.
This is working just fine for all fields which are of the "input type". The problem I'm having is with two fields which are of the "select type" i.e. dropdown listboxes.
For those two fields, I would like to bring in all the valid choices from the respective lookup/master tables, but then have the default/selected value be shown based on what's in the transaction table. The way I have it right now, those two fields are showing (and updating the record with) the very first entry's in the two lookup tables/select query.
The attached picture might make things a little bit clearer. You'll notice in the top screenshot that the first row (which is the one I'm selecting to update) has a "Store Name" = "Super Store" and an "Item Description" = "Old Mill Bagels". Now, when I click the "update" botton and I'm taken to the update screen, the values for those two fields default to the very first entries in the SELECT resultset i.e. "Food Basics" and "BD Cheese Strings". Cricled in green (to the top-left of that screenshot) is the result of an echo that I performed, based on the values that are in the transaction record.
I cannot (for the life of me) figure out how to get those values to be used as default/selected values for the two dropdown's...so that if a user does not touch those two dropdown fields, the values in the transaction table will not be changed.
Your help will be greatly appreciated.
Here's a portion of the FORM code:
<form class="form-horizontal" action="update.php?idnumber=<?php echo $idnumber?>" method="post">
<?php
// Connect to Store_Name (sn) table to get values for dropdown
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT DISTINCT store_name FROM store_master ORDER BY store_name ASC";
$q_sn = $pdo->prepare($sql);
$q_sn->execute();
$count_sn = $q_sn->rowCount();
$result_sn = $q_sn->fetchAll();
Database::disconnect();
// Connect to Item_Description (id) table to get values for dropdown
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT DISTINCT product_name FROM product_master ORDER BY product_name ASC";
$q_id = $pdo->prepare($sql);
$q_id->execute();
$count_id = $q_id->rowCount();
$result_id = $q_id->fetchAll();
Database::disconnect();
foreach($fields AS $field => $attr){
$current_store_name = $values['store_name'];
$current_item_description = $values['item_description'];
//Print the form element for the field.
if ($field == 'store_name')
{
echo $current_store_name;
echo '<br />';
echo $current_item_description;
echo '<div class="control-group">';
echo "<label class='control-label'>{$attr['label']}: </label>";
echo '<div class="controls">';
//Echo the actual input. If the form is being displayed with errors, we'll have a value to fill in from the user's previous submission.
echo '<select class="store-name" name="store_name">';
// echo '<option value="">Please select...</option>';
foreach($result_sn as $row)
{
echo "<option value='" . $row['store_name'] . "'>{$row['store_name']}</option>";
}
// $row['store_name'] = $current_store_name;
echo "</select>";
echo '</div>';
echo '</div>';
}
elseif ($field == 'item_description')
{
echo '<div class="control-group">';
echo "<label class='control-label'>{$attr['label']}: </label>";
echo '<div class="controls">';
echo '<select class="item-desc" name="item_description">';
// echo '<option value="">Please select...</option>';
foreach($result_id as $row)
{
echo "<option alue='" . $row['product_name'] . "'>{$row['product_name']}</option>";
}
echo "</select>";
echo '</div>';
echo '</div>';
}
else
{
echo '<div class="control-group">';
echo "<label class='control-label'>{$attr['label']}: </label>";
echo '<div class="controls">';
//Echo the actual input. If the form is being displayed with errors, we'll have a value to fill in from the user's previous submission.
echo '<input type="text" name="'.$field.'"' . (isset($values[$field]) ? ' value="'.$values[$field].'"' : '') . ' /></label>';
echo '</div>';
echo '</div>';
}
And here's some other declarations/code which I think might be required.
$fields = array(
'store_name' => array('label' => 'Store Name',
'error' => 'Please enter a store name'),
'item_description' => array('label' => 'Item Description',
'error' => 'Please enter an item description'),
'qty_pkg' => array('label' => 'Qty / Pkg',
'error' => 'Please indicate whether it\'s Qty or Pkg'),
'pkg_of' => array('label' => 'Pkg. Of',
'error' => 'Please enter the quantity'),
'price' => array('label' => 'Price',
'error' => 'Please enter the price'),
'flyer_page' => array('label' => 'Flyer Page #',
'error' => 'Please enter the flyer page #'),
'limited_time_sale' => array('label' => 'Limited Time Sale',
'error' => 'Please enter the days for limited-time-sale'),
'nos_to_purchase' => array('label' => 'No(s) to Purchase',
'error' => 'Please enter the No. of items to purchase')
);
...
...
....
{
// If [submit] isn't clicked - and therfore POST array is empty - perform a SELECT query to bring in
// existing values from the table and display, to allow for changes to be made
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT * FROM shoplist where idnumber = ?";
$q = $pdo->prepare($sql);
$q->execute(array($idnumber));
$values = $q->fetch(PDO::FETCH_ASSOC);
if(!$values)
{
$error = 'Invalid ID provided';
}
Database::disconnect();
}
If there's anything I've missed please ask and I'll provide.
Thanks
hi, I have a form that has 2 lists, first one has the months from Jan to Dec which i am done with, for the second list i want to have years from current year for other say 5 years. Example, we are in 2020 so i should have the list populate 2020,2021,2022,2023,2024 and next year it should show 2021-2025, how to do that. Here is what i have as HTML: <html> <head> <link rel="stylesheet" type="text/css" href="reportstyle.css"> </head> <body> <div class="form-wrapper"> <form action="reportanalysis.php" method="post"> <div class="form-item"> <h4>Select a month:</h4> <select id="months" name="months" required="required"> <option selected="selected" value="chose">Choose one option...</option> <option value="jan">Jan - 01</option> <option value="feb">Feb - 02</option> <option value="mar">Mar - 03</option> <option value="apr">Apr - 04</option> <option value="may">May - 05</option> <option value="jun">Jun - 06</option> <option value="jul">Jul - 07</option> <option value="aug">Aug - 08</option> <option value="sep">Sep - 09</option> <option value="oct">Oct - 10</option> <option value="nov">Nov - 11</option> <option value="dec">Dec - 12</option> </select> </div> <div class="form-item"> <h4>Select a year:</h4> <select id="years" name="years" required="required"> <option selected="selected" value="chose">Choose one option...</option> <option value="2020">2020</option> <option value="2021">2021</option> <option value="2022">2022</option> <option value="2023">2023</option> <option value="2024">2024</option> </select> </div> <div class="button-panel"> <center><button type="submit" name="reporting" class="buttonstyle">Generate Graph Report</button></center> </div> <div class="reminder"> <p><a href="home.php">Return to Main Menu</a> </div> </form> </div> </body> </html> Edited April 3, 2020 by ramiwahdan typo error Hi, I'm trying to populate the previous and next links with an id from mysql. The code below works but also displays ids that do not exist in the database. I want the code to only show me the rows that exist not the ones that do not exist. For example, at this time I have 5 rows. The ids of the rows are 1, 2, 3, 4, 6 when I get to say, id 6 and click on next it displays id 7 instead of going back to 1 or greying out next - meaning there's no more to view. Can someone help?
<?php
include('connection.php');
if(isset($_GET['id'])){
$start = $_GET['id'];
}else{
$start = 0;
}
$sql = mysql_num_rows(mysql_query("SELECT * FROM thetable"));
$result = mysqli_query($con,$sql);
$rows = mysql_fetch_array($result);
echo $rows['thetable'];
if($start == 0){
echo "Previous «";
}else{
echo "<a href=\"./thepage.php?id=" . ($start - 1) . "\">« Previous </a>";
}
if($start == $sql-1){
echo "Next »";
}else{
echo "<a href=\"./thepage.php?id=" . ($start + 1) . "\">Next »</a>";
}
?>
<?php
// End while loop.
mysqli_close($con);
?>
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