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PHP - How To Retrieve Value From Drop-down List?
I'm doing this activity where a user chooses a base timezone and when the user clicks convert the current time in the selected GMT will be converted to GMT-11 to GMT+13.
As of now, I have these codes: act09_view.php: Code: [Select] <?php session_start(); $s="GMT "; echo "Select the base time zone:</br>"; for($n=-11;$n<=13;$n++) { if ($n>=0) $s="GMT +"; $gmt[]=$s . $n . "</br>"; } echo "<select name='gmt'>"; foreach ($gmt as $value) { echo '<option value="' . $value . '">' . $value . '</option>\n'; } echo '</select>'; //$_SESSION['value']=$value; ?> <form action="act09_process.php"> </br><input type='submit' value='Convert'/> </form> act09_process.php: Code: [Select] <?php session_start(); //$value=$_SESSION['value']; //echo $value; date_default_timezone_set('Asia/Manila'); $gmttime=date('M j, Y g:i:s A'); echo "The current date and time at" . " is " . $gmttime; ?> I've tried using session variables. I think I executed them incorrectly. The output is supposed to look like this: Similar TutorialsI have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information. How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> Hi I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form. Sales.php Page <form action="saleprocess.php" method="GET"> <?php echo 'Product Model:'; $query="SELECT * FROM products"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=product_model value=Select>Product Model</option>"; // printing the list box select command while($rows=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>"; /* Option values are added by looping through the array */ } echo "</select><br>"; ?> <input type='submit' name='submit' value='Create'></input> <br> </form> ******************************************************************************** Salesprocess.php page <?php include("connect.php"); if(isset($_GET['product_id'])){ $product_id = $_GET['product_id']; $query = mysql_query("SELECT * FROM products WHERE product_id= $product_id"); while($rows = mysql_fetch_assoc($query)) { echo 'Product Model<br>'; echo $rows['product_id']; echo $rows['product_model']; } } ?> Muchly appreciated if someone can help me Hi everyone, I am having trouble passing/displaying the values inside of a selected list. I created a add/remove list using Jquery and I tried to display the values passed using foreach and for loops but it is still not working. The values I am trying to get are $existing_mID[$j], which is inside of the option value attribute. Please kindly let me know what should I do in order to get the values and I really appreciate your help.
<?php
$selected = $_POST['selectto'];
if(isset($selected))
{
echo "something in selected<br />";
for ($i=0;$i<count($selected);$i++)
echo "selected #1 : $selected[$i]";
foreach ($selected as $item)
echo "selected: item: $item";
}
?>
This is the formAs a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> I have a "select"-drop down bar and I want to have a numbered list in it, i've tried but it doesn't seem to be possible. Is there any way that i'm able to do this? Hi ,
I am pretty new to php. I would like to create a php website based on folders.
I have a code so far that shows the names of folder in a selection box. What I would like to achieve now is how i get the images inside the folder after I click on submit. I know how to get the images with a glob function. What I don't know is how to get the value of each specific folder name. How would I do that?
I have this code so far to get the names of the folders in the selection box:
<form action="test.php" method="post">
<select name="myDirs">
<option value="" selected="selected">Select a genre</option>
<?php
$dirs = glob("*", GLOB_ONLYDIR);
foreach($dirs as $val){
echo '<option value="'.$val.'">'.$val."</option>\n";
}
?>
</select>
<input type="submit" name="submit2">
<?php
if (isset($_POST['submit2']))Helllo Every1 well im kinda new to php and i needed some help i was working on a page all my text boxes and check boxes are at the bottom of my file and like this //====================================================================== </script> </head> <body> <div style="margin-left: 170px"> <input type="checkbox" id="getitems" checked value="1">Run Plugin? <input type="checkbox" id="sellitm" checked value="1">Sell Items?<br><br> Item Name:<br><input type="text" id="itemd" value=""><br><br>How Many Cycles?:<br><input type="text" id="runtm" value="10"></div> <div style="margin-left: 170px"><br> <button id="btn_save" style="color:white;background-color:#00660F;border-width:1px;border-style:solid; "> Save settings </button> </div> '; echo $this->ObjectTable(); echo ' </body> </html> '; } } ?> and well i decided to do away wit one of the text boxes and use a drop down list instead that is populate from a text file and the only code that i could find that i was able to get working was this 1 <?php $text = file_get_contents("itemlist.txt"); $array = explode("\n",$text); echo "<select>"; foreach ($array as $value) { echo "<option value='$value'>$value</option>"; } echo "</select>"; and the problem im having is with that one it just stays at the top of the page when loaded like i have no way to position it... and im really stuck i tried saving it in another php and i tried using <?php include(); ?> function but it did not work if any1 could help me out that would be awsome. T.I.A Basically I have a form and What I want it to do is when a user selects the 'For Sale' radio button the minimum and maximum prices will be displayed in the drop down list as "70,000"+ but when the user selects the 'To Rent' radial button then the minimum and maximum drop down list prices will change to e.g. "200"+. How would I do this? Heres my Code for the form: Code: [Select] <div id="ptype"> <input type="radio" class="styled" name="ptype" value="forsale"/> For Sale <p class="increase"> <input type="radio" class="styled" name="ptype" value="forrent"/> To Rent </p> <p class="increase"> <input type="radio" class="styled" name="ptype" value="any"/> Any </p> </div> </td> </tr> </table> <div id="table2"> <table id="NBtable"> <tr> <td><p class="NBS">Number of bedrooms:</p></td> <td><div id="NB"> <select name="min bedrooms"> <?php $roomLimit = 5; for($minRooms = 0; $minRooms <= $roomLimit; $minRooms++){ if ($minRooms == 0){ echo"<option value=\"0\" selected=\"selected\">No Min</option>"; } else{ echo "<option value=\"$minRooms\">$minRooms</option>"; } } ?> </select> to <select name="max bedrooms"> <?php $roomLimit = 5; for($maxRooms = 0; $maxRooms <= $roomLimit; $maxRooms++){ if ($maxRooms == 0){ echo"<option value=\"0\" selected=\"selected\">No Max</option>"; } else{ echo "<option value=\"$maxRooms\">$maxRooms</option>"; } } ?> </select> </div> </td> <td><p class="PR">Price range:</p></td> <td><div id="PR"> <select name="min price" style='width:80px;'> <?php $priceLimit = 350; for($minPrice = 40; $minPrice <= $priceLimit; $minPrice =( $minPrice+ 10)){ if ($minPrice == 40){ echo"<option value=\"0\" selected=\"selected\">No Min</option>"; } else{ echo "<option value=\"$minPrice000\">£$minPrice,000</option>"; } } ?> </select> to <select name="max price" style='width:80px;'> <?php $priceLimit = 350; for($maxPrice = 40; $maxPrice <= $priceLimit; $maxPrice =( $maxPrice+ 10)){ if ($maxPrice == 40){ echo"<option value=\"0\" selected=\"selected\">No Max</option>"; } else{ echo "<option value=\"$maxPrice000\">£$maxPrice,000</option>"; } } ?> </select> Thanks in advance. Hi, I am trying to create a drop down list in php and I want the data to come from a table that I have created in phpmyadmin. The code that I have created allows me to select values from the drop down list and insert the rest of the data. However when I check the the table the SID and Cid are set to 0 and the grade field is empty and the comments field contains the grade. The SID and Cid are both composite keys. <?php $sql = "SELECT Cid FROM course"; $db1 = new DBStudent_Course(); $db1->openDB(); $result = $db1->getResult($sql); echo"<select name = Cid>"; while ($row = mysql_fetch_object($result)) { echo "<option value = '" . $row->Cid . "'>$row->Cid</option>"; } echo"</select>"; echo "</p>"; ?> <?php $sql = "SELECT SID FROM student"; $db1 = new DBStudent_Course(); $db1->openDB(); $result = $db1->getResult($sql); echo"<select name = SID>"; while ($row = mysql_fetch_object($result)) { echo "<option value = '" . $row->SID . "'>$row->SID</option>"; } echo"</select>"; echo "</p>"; ?> <?php if (!$_POST) { //page loads for the first time ?> <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post"> grade:<input type="text" name="grade"/><br/> comments:<input type="text" name="comments" /><br /> <input type="submit" value="Save" /> </form> <?php } else { $Cid = $_POST["Cid"]; $SID = $_POST["SID"]; $grade = $_POST["grade"]; $comments = $_POST["comments"]; $db1 = new DBStudent_Course(); $db1->openDB(); $numofrows = $db1->insert_student_course("", $SID, $Cid, $grade, $comments); echo "Success. Number of rows affected: <strong>{$numofrows}<strong>"; $db1->closeDB(); } ?> I have two basic dropdown lists in a form, my question is this: After selecting an item from the 1st dropdown list, the second dropdown list should automatically load only fields that are connected with the first selected item. (ex Site1 has Tank01 and Tank06, but Site2 has Tank03, Tank04 and also a Tank06). What is the script needed / or how do I change my script to accommodate this? Here is the basic script I use for the second dropdown list: <?php include("../xxx.php"); $cxn = mysqli_connect($host,$user,$password,$dbname); $query = "SELECT DISTINCT `diesel_tank_id` FROM `diesel_tank` ORDER BY `diesel_tank_id`"; $result = mysqli_query($cxn,$query); while($row = mysqli_fetch_assoc($result)) { extract($row); echo "<option value='$diesel_tank_id'>$diesel_tank_id</option>\n"; } ?> hi, I'm sruggling to get a drop down list to post the chosen value from an sql database.. could someone please point me in the right direction many thanks! I'm useing a simple HTML drop-down list box. But now I want to be able to and an editable drop-down box so that if the particular item in the drop-down is not changed it won't update the database. But if a new value is selected it will update the DB with the new value.
I am not clear on a couple of things. should the valuse of the drop down be in a seperate DB table. And then the script to change the new value or keep the old.
I've googled this and can't seem to find a workable solution.
Thanks in advance,
how do i loop this so that the option in the drop down list loop. i tried this but get an error Code: [Select] <select name="age"><?php for ($i=10; $i<71; $i++;) { echo "<option value='$i'>$i</option>"; } ?></select> Hello all,
I am new in php and would like to help me in in code.
I would like to implement a html form that will upload to my linux server some files, some info as well and will run a bash command.
To be more clear, users will:
Send to the linux server two files (I have done it).
Select from a drop down list a specific directory, and once this specific directory has been selected to activate one other drop down menu with its childs.
Send all those info to a bash script.
For case b, assume that we have the following structure
my_dir/cars/brand1
my_dir/cars/brand2
my_dir/cars/brand3
my_dir/motors/brand5
my_dir/motors/brand6
the first drop down list should list only cars and motors while the second only brand1, brand2, and brand3 if cars is selected or brand5 and brand6 if motors is selected.
For case c, once the form is submitted, I would like the command:
./script.sh –a uploadedfile1 –b uploadedfile2 –c dropdown1 –d dropdpwn2
Any help is appreciated.
Best,
Hello, I'm trying to get a drop down list (ordinary html function, select option ...) but I want it to start from an image. When you click the image I want the drop down list to appear. Hove can I do this? /Pelle I have a drop down list that contains expiration dates in the format 01/12 and if the user selected a certain drop down list I would like that same date selected. Does anyone know the code to do this? Thanks. Hey guys, I made in php an sql droplist. But I need some help. I want it to work like this. When I select something from the list and click Add To Cart, that also the product gets send to the cart and not only the quantity. you can check it out here : http://fhcs.be/cart-demo2/index.php I also added my index.php as an attachment Greets I have a drop down list which retrieves site name acronym and the url from the database. I need help in 2 things. 1.I do not want the url to be displayed in the drop down list. 2. When I hit submit I need to echo the sitename acronym as well as the url in a page called display.php. Here is my code so far: Code: [Select] <form action="display.php" method ="post"> <tr> <td>Category</td> <td> <select name="new_id"> <option value="">=============</option> <?php foreach($acronym as $key=>$value){ ?> <option value="<?php echo $value['site_id']; ?>"><?php echo $value['acronym']; ?></option> <option value="<?php echo $value['site_id']; ?>"><?php echo $value['url']; ?></option> <?php }?> </td> </tr> <input type="submit" name="submit" value="submit" /> </form> for some reason my code only posts the first word of the username even tho the last names are in the database. www.checkmyweave.co.uk/new/dropdown.php dropdown.php Quote <?php mysql_connect("localhost","****","****"); mysql_select_db("checkmyw_database") or die("Unable to select database"); ?> <form action="/new/dropdown2.php" Method="POST"> User:<select name="username"> <?php $result = mysql_query("SELECT * FROM members") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { echo "<option value=".$row['username'].">"; echo $row['username']."</option>"; } ?> </select> <form action="/new/dropdown2.php" METHOD="POST"> <input type = "submit" value="New!"> </form> dropdown2.php Quote <?php $username=$_POST["username"]; echo $username ; ?> |
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