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PHP - Javascript Dropdown Image Change And Zooming
i am new to web designing,
i don't know much about javascript.
<select id="plan" align="center" valign="center">
Similar TutorialsI'm new to this form and php/mysql so sorry if this isn't the right place to post this. This is what is in the first dropdown box. Code: [Select] $selValues['john'] = "a, b, c, d, e"; These are the different lists I want it to put in the second drop down box depending on what they choose in the first. Code: [Select] $selValues['list1']= " a1, a2, a3, a4, a5"; $selValues['list2']= " b1, b2, b3, b4, b5"; This is how I made an attempt to make it work before posting here. Along with plently of other ways. Code: [Select] if ($selValues['john']=a) { $chan_input = selectBuilder($selValues['$list1'] }; else if ($selValues['john']=b) { $chan_input = selectBuilder($selValues['$list2'] }; Basicly how I need it to work is there are two drop down boxes. First they will chose between a,b,c,d,e. If they chose a then on the second drop down box I only want them to be able to select a1,a2,a3,a4,a5. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=305930.0 Hi I'm trying to create a form where people first select the number of children they have, and then a table should appear where they fill in the extra information (name, sex, dob) about every child. The code works except for that i don't know how to add the year of the date of birth of every child with a loop. Appartly i cannot use php in javascript? Any solutions? thanks! Code: [Select] function addKindForms(aantal) { if (aantal != "-") { var output = ""; output = output + "<table cellpadding='2' cellspacing='0'>"; output = output + "<tr><td><b>Naam:</b></td><td><b>Geslacht:</b></td><td><b>Geboortedatum:</b></td><td></td><td></td></tr>"; for(i=1;i<=aantal;i++) { output = output + "<tr><td><input type='text' name='kind"+i+"_name' size='15' value=''></td><td><select name='kind"+i+"_sex'><option value='m'>jongen</option><option value='f'>meisje</option></select></td><td><select name='kind"+i+"_gebdatumdag'><option value=''></option><option value='1'>1</option><option value='2'>2</option><option value='3'>3</option><option value='4'>4</option><option value='5'>5</option><option value='6'>6</option><option value='7'>7</option><option value='8'> 8</option><option value='9'>9</option><option value='10'>10</option> <option value='11'> 11</option><option value='12'>12</option><option value='13'>13</option><option value='14'>14</option><option value='15'>15</option><option value='16'>16</option><option value='17'>17</option><option value='18'>18</option><option value='19'>19</option><option value='20'>20</option><option value='21'>21</option><option value='22'>22</option></select></td><td><select name='kind'+i+'_gebdatummaand'><option value=''></option><option value='1'>jan</option><option value='2'>feb</option></select></td><td><select name='kind'+i+'_gebdatumjaar'><option value=''></option><?php for ($y=date('Y');$y>=(date('Y')-125);$y--){if($_POST['bdaykind_jaar']==$y){$selected='selected';}echo'<option value=''.$y.'' '.$selected.'>'.$y.'</option>';$selected = '';}?></select></td></tr>"; } output = output + "</table>"; document.getElementById("kindforms").innerHTML = output; } } I have this code where I add records: Code: [Select] <?php ini_set('display_errors',1); error_reporting(-1); require_once ('./includes/config.inc.php'); require_once (MYSQL); $add_cat_errors = array(); if ($_SERVER['REQUEST_METHOD'] == 'POST') { // Check for a name: if (empty($_POST['product'])) { $add_cat_errors['product'] = 'Please enter the name!'; } // Check for a description: if (empty($_POST['prod_descr'])) { $add_cat_errors['prod_descr'] = 'Please enter the description!'; } // Check for a category: if (!isset($_POST['cat']) || !filter_var($_POST['cat'], FILTER_VALIDATE_INT, array('min_range' => 1))) { $add_cat_errors['cat'] = 'Please select a category!'; } // Check for a price: if (empty($_POST['price']) || !filter_var($_POST['price'], FILTER_VALIDATE_FLOAT) || ($_POST['price'] <= 0)) { $add_cat_errors['price'] = 'Please enter a valid price!'; } // Check for a category: if (!isset($_POST['directory']) || !filter_var($_POST['directory'], FILTER_VALIDATE_INT, array('min_range' => 1))) { $add_cat_errors['directory'] = 'Please select a directory!'; } // Check for an image: if (is_uploaded_file ($_FILES['image']['tmp_name']) && ($_FILES['image']['error'] == UPLOAD_ERR_OK)) { $file = $_FILES['image']; $size = ROUND($file['size']/1024); // Validate the file size: if ($size > 512) { $add_cat_errors['image'] = 'The uploaded file was too large.'; } // Validate the file type: $allowed_mime = array ('image/jpeg', 'image/JPG', 'image/jpg'); $allowed_extensions = array ('.jpg', 'jpeg'); $image_info = getimagesize($file['tmp_name']); $ext = substr($file['name'], -4); if ( (!in_array($file['type'], $allowed_mime)) || (!in_array($image_info['mime'], $allowed_mime) ) || (!in_array($ext, $allowed_extensions) ) ) { $add_cat_errors['image'] = 'The uploaded file was not of the proper type.'; } // Move the file over, if no problems: if (!array_key_exists('image', $add_cat_errors)) { // Create a new name for the file: $new_name = (string) sha1($file['name'] . uniqid('',true)); // Add the extension: $new_name .= ((substr($ext, 0, 1) != '.') ? ".{$ext}" : $ext); //$new_name = $dir . '/' . $new_name; $dest = "../db/images/$new_name"; // Move the file to its proper folder but add _tmp, just in case: //$dest = "../db/images/$new_name"; $dirs = array('full_heads', 'human_hair', 'lip_tattoos', 'ponytails', 'synthetic_hair'); if (move_uploaded_file($file['tmp_name'], $dest)) { // Store the data in the session for later use: $_SESSION['image']['new_name'] = $new_name; $_SESSION['image']['file_name'] = $file['name']; // Print a message: echo '<h4>The file has been uploaded!</h4>'; } else { trigger_error('The file could not be moved.'); unlink ($file['tmp_name']); } } // End of array_key_exists() IF. } elseif (!isset($_SESSION['image'])) { // No current or previous uploaded file. switch ($_FILES['image']['error']) { case 1: case 2: $add_cat_errors['image'] = 'The uploaded file was too large.'; break; case 3: $add_cat_errors['image'] = 'The file was only partially uploaded.'; break; case 6: case 7: case 8: $add_cat_errors['image'] = 'The file could not be uploaded due to a system error.'; break; case 4: default: $add_cat_errors['image'] = 'No file was uploaded.'; break; } // End of SWITCH. } // End of $_FILES IF-ELSEIF-ELSE. // Check for a stock: if (empty($_POST['stock']) || !filter_var($_POST['stock'], FILTER_VALIDATE_INT, array('min_range' => 1))) { $add_cat_errors['stock'] = 'Please enter the quantity in stock!'; } if (empty($add_cat_errors)) { $query = "INSERT INTO product (product, prod_descr, catID, price, dirID, image, stock) VALUES (?, ?, ?, ?, ?, ?, ?)"; // Prepare the statement: $stmt = mysqli_prepare($dbc, $query); // For debugging purposes: // if (!$stmt) echo mysqli_stmt_error($stmt); // Bind the variables: mysqli_stmt_bind_param($stmt, 'ssssssi', $name, $desc, $_POST['cat'], $_POST['price'], $_POST['directory'], $_SESSION['image']['new_name'], $_POST['stock']); // Make the extra variable associations: $name = strip_tags($_POST['product']); $desc = strip_tags($_POST['prod_descr']); // Execute the query: mysqli_stmt_execute($stmt); if (mysqli_stmt_affected_rows($stmt) == 1) { // If it ran OK. // Print a message: echo '<h4>The product has been added!</h4>'; // Clear $_POST: $_POST = array(); // Clear $_FILES: $_FILES = array(); // Clear $file and $_SESSION['image']: unset($file, $_SESSION['image']); } else { // If it did not run OK. trigger_error('The product could not be added due to a system error. We apologize for any inconvenience.'); unlink ($dest); } } // End of $errors IF. } else { // Clear out the session on a GET request: unset($_SESSION['image']); } // End of the submission IF. require_once ('./includes/form_functions.inc.php'); ?> <form enctype="multipart/form-data" action="add_product.php" method="post" accept-charset="utf-8"> <input type="hidden" name="MAX_FILE_SIZE" value="524288" /> Product<br /><?php create_form_input('product', 'text', $add_cat_errors); ?> Description<br /><?php create_form_input('prod_descr', 'textarea', $add_cat_errors); ?> Category<br /><select name="cat"<?php if (array_key_exists('cat', $add_cat_errors)); ?>> <option>Select One</option> <?php // Retrieve all the categories and add to the pull-down menu: $q = 'SELECT catID, cat FROM category ORDER BY cat ASC'; $r = mysqli_query ($dbc, $q); while ($row = mysqli_fetch_array ($r, MYSQLI_NUM)) { echo "<option value=\"$row[0]\""; // Check for stickyness: if (isset($_POST['cat']) && ($_POST['cat'] == $row[0]) ) echo ' selected="selected"'; echo ">$row[1]</option>\n"; } ?> </select><?php if (array_key_exists('cat', $add_cat_errors)) echo $add_cat_errors['cat']; ?> Price<br /><?php create_form_input('price', 'text', $add_cat_errors); ?> Directory<br /><select name="directory"<?php if (array_key_exists('directory', $add_cat_errors)); ?>> <option>Select One</option> <?php // Retrieve all the categories and add to the pull-down menu: $q = 'SELECT dirID, directory FROM directory ORDER BY directory ASC'; $r = mysqli_query ($dbc, $q); while ($row = mysqli_fetch_array ($r, MYSQLI_NUM)) { echo "<option value=\"$row[0]\""; // Check for stickyness: if (isset($_POST['directory']) && ($_POST['directory'] == $row[0]) ) echo ' selected="selected"'; echo ">$row[1]</option>\n"; } ?> </select><?php if (array_key_exists('directory', $add_cat_errors)) echo $add_cat_errors['directory']; ?> </select> Image<br /><?php // Check for an error: if (array_key_exists('image', $add_cat_errors)) { echo $add_cat_errors['image'] . '<br /><input type="file" name="image"/>'; } else { // No error. echo '<input type="file" name="image" />'; // If the file exists (from a previous form submission but there were other errors), // store the file info in a session and note its existence: if (isset($_SESSION['image'])) { echo "<br />Currently '{$_SESSION['image']['file_name']}'"; } } // end of errors IF-ELSE. ?> <br /> Stock<br /><?php create_form_input('stock', 'text', $add_cat_errors); ?> <input type="submit" value="Add This Product" class="button" /> </fieldset> </form> What I want to achieve is this; When a record is being inserted, a user will make a selection from the dropdown box with a list of directories. Then, a user will upload an image. When a image is uploaded and a directory is chosen, I want the image to be assigned to a specific directory from the list so when the record is inserted, that image will be placed in the specific directory. How do I achieve this please? Using php I have a dynamically filled select box with the names of images. How can I get an image box on my page and change upon selection? Hi all, I am dealing with an html page in which a dynamically created PNG image is shown with: <img src="createchart.php"> The problem arises when users try to download the image since the default name that appears is just "createchart.php". Is there any way to make the browser suggest something like "chart.png"? Than you guys for your help, Mam Main URL of website home page: www.mysite.com and this shows image "A" in the header. I want to have a marketing campaign which will send people to my site via this link: www.mysite.com/index.php?ref=att When people get to my site via the 2nd link, I want to show image "B" instead of image "A". I put this at top of home page... Code: [Select] require_once("includes/session.php"); if (isset($_GET['ref']) && ($_GET['ref'] == "att")) { $_SESSION['ref'] = "att"; } else { } and this code is in my header include file, a little further down on the page... Code: [Select] <?php if (isset($_SESSION['ref']) && $_SESSION['ref'] == "att") { echo "<span class='phone'><img src='/images/B.png' /></span>"; } else { echo "<span class='phone'><img src='/images/A.png' /></span>"; } ?> Does this look right? because I can't get it to work. It's showing image A regardless of which URL I use. I have to be missing something obvious because I really thought this would be super easy to do. If anyone sees any problems, let me know. But I'll keep testing and hopefully find my mistake. Thanks! This creates an image 1500x1500 pixels. But it creates it in black... how can i make it white. And i have to use imagecreatetruecolor() not imagecreate(). <?php $im = imagecreatetruecolor(1500,1500); // Output the image to the browser header('Content-type: image/jpg'); imagejpeg($im); imagedestroy($im); ?> Thanks! HI I'm thinking to do something like this , I've seen this in a website ,I'm a php designer but i don't know who something like this can be done . it's a website that sells sofas . we've the ability to change the color and fabric and see the result right away . for example ,this is the default sofa : then I can select a part of it and I can choose the fabric and color , then it make the result : thats it , How can I do that ? What should I do ? I'm really looking to hear from you King Regards Hi,
I want to show one image for the new visitors of my website and another for the returning visitors. What I want to do is that after the visitor reload the page he will see another image instead of the first one.
I have created a flashcard program to teach Japanese vocabulary. I want to add sound files from native speakers so I have created a page with a recorder that
1. displays a word (on load)
2. allows the native speaker to record the word (on press of a button)
3. uploads the recording to the server (on press of another button)
I wrote the word display as a separate php script so that I could use Ajax to just change the word being recorded, so that I would not need to reload the entire page every time the speaker switched to a new word.
The js that gets the word to be recorded is:
function ajaxRequest(){
//declare the variable at the top, even though it will be null at first
var req = null;
//modern browsers
req = new XMLHttpRequest();
//setup the readystatechange listener
req.onreadystatechange = function(){
//right now we only care about a successful and complete response
if (req.readyState === 4 && req.status === 200){
//inject the returned HTML into the DOM
document.getElementById('cardData').innerHTML = req.responseText;
};
};
//open the XMLHttpRequest connection
req.open("GET","nextcard.php",true);
//send the XMLHttpRequest request (nothing has actually been sent until this very line)
req.send();
};
The section that uploads the file is:
$('#send').click(function(){
$.jRecorder.sendData();
}
and it gets the filename from another script:
$.jRecorder(
{
host : 'http://localhost:10088/NewFolder/japanese/jrecorder/acceptfile.php?filename=hello.wav'
The problem that I have is that when I get each word I want to change the name of the file being uploaded. It all works and if I reloaded the page each time I could do this:
$.jRecorder(
{
host : 'http://localhost:10088/NewFolder/japanese/jrecorder/acceptfile.php?filename=<?php echo $word.'_'.$reader_id;?>.wav'
but I want to stay on the Ajax path. So, I think that I need to convert the newcard program to return json with something like:
$response=array(filename=>$word.$reader_id, card_data=>$card_stuff); echo json_encode($response);but once I get the json back I don't see how I get the filename into the script that sets the upload url and make that script re-execute. Is there a reasonable way to do that? Hey guys! I have the following php code that grabs variables (and the browsed image) from Flash. //FLASH VARIABLES $Name = $_POST['Name']; $itemNumber = $_POST['itemNumber']; $filename = $_FILES['Filedata']['name']; $filetmpname = $_FILES['Filedata']['tmp_name']; $fileType = $_FILES["Filedata"]["type"]; $fileSizeMB = ($_FILES["Filedata"]["size"] / 1024 / 1000); list($filename, $extension) = explode('.', basename($_FILES['Filedata']['name'])); $filename = $Name; $target = $filename . $itemNumber . "." . $extension; // Place file on server, into the images folder move_uploaded_file($_FILES['Filedata']['tmp_name'], "images/".$target); This works perfect, but what I want to change is the width and height of the uploaded image. Any ideas/suggestions on how this could be done? Thanks in advance!! Cheers! Currently, I insert a small thumbnail image at the begging of each row in a set of records from a mysql query. But that image is the same image for each record in that array. Is there a way to conditionally change which image is used, based on a value either in that same table, or a related table. See attached screenshot... and here's my current code.... Code: [Select] <?php $query = "select address from nvc"; $result = mysql_query($query); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $details=''; $details.='<li class="menu"><a href=detail.php?address='.(urlencode($row['address'])).'><img src="thumbs/house.png" /><span class="name">'.$row['city'].'</a>'; $details.='</li>'; echo($details); } ?> Thank you for any input. I have a PHP script that uploads images to a folder on my server (attachments folder). Currently the folder sits within my webroot and is publicly accessible (I have to use chmod 777 due to permissions issue). So, I created the "attachments" folder outside of my webroot (so that it is not publicly accessible), but I do not know how to set the path in the PHP code to upload it to that "attachments" folder outside of the webroot. As you see in the snippet of PHP code below, the code currently uploads the the "attachments" folder within the www (webroot) directory. How do I make it upload to the "attachments" folder OUTSIDE of the www (webroot) directory? foreach($files[$form] as $file){ $str = $file[1]; if (eval("if($str){return true;}")) { $_values[$file[0]] = $_FILES[$file[0]]["name"]; $dirs = explode("/","attachments//"); $cur_dir ="."; foreach($dirs as $dir){ $cur_dir = $cur_dir."/".$dir; if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}} $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]); @unlink($_FILES[$file[0]]["tmp_name"]); }else{ $flag=true; if ($_isdisplay) { //$ExtFltr = $file[2]; //$FileSize = $file[4]; if (!eval("if($file[2]){return true;}")){echo $file[3];} if (!eval("if($file[4]){return true;}")){echo $file[5];} $_ErrorList[] = $file[0]; } } } This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=347058.0 I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$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) 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 /////
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