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PHP - How To Use Onchange In A Dropdown Box With Php?
Hi,
I need to develop a code that allows me to choose how I want to see a list of data from the database. For example, I could choose to see "the most recent first" or "most recent last". I have this example on how work the "onChange" event: <html> <head> <title>Select Example</title> <script> function onSelectChange(){ var dropdown = document.getElementById("developer"); var index = dropdown.selectedIndex; var ddVal = dropdown.options[index].value; var ddText = dropdown.options[index].text; if(ddVal != 0) { output = "You Selected " + ddText; } document.getElementById("output").innerHTML = output; } </script> </head> <body> <h3>Developers</h3> <select id="developer" onChange="onSelectChange();"> <option value="0">the most recent first</option> <option value="1">the most recent flast</option> </select> <br / <div id="output"></div> </body> </html> Now I need to apply this to a SELECT from the database that will use this method to list data. I have tried to google it, but I can't find any example. Can someone give me a clue on how to find the way of doing this? Best Regards, Similar TutorialsThis topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=333787.0 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) This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=320136.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=305968.0 Good day, I new to PHP I am having problems with a two dropdowns on a form, can someone please tell me where I'm going wrong. [attachment deleted by admin] Hey guys I wasnt sure if this is a php topic or java script im trying to make my drop down menu's activate onchange. what confuses me is how do I make php know that the java script is being run. Or have the java script trip my php code. To know a button has been pressed i use "if (isset['_POST'])" echo '<select name="siege_list" id="siege_list">'; foreach($name as $key => $value) { echo '<option value="' . $value['name'] . '" ' . ($value['name'] == $current1 ['name'] ? 'selected="selected"' : '') . '> ' . $value['name'] . '</options>'; } echo '</select><input type="submit" id="siege_planet" onchange="this.form.submit()">'; This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=330989.0 Hey phpFreaks, im having some troubles getting my script to work correctly and im also not sure if this issue is in the right section of the forum. but heres what i have going on. I have a query result that displays a list of images with a checkbox and a couple buttons for edit and delete. everything works fine other than the checkbox stuff. I had it working when i was using a submit button, but i wanted to get rid of that button cuz it was only dealing with the checkboxes. so whats going on when using the checkboxes is that when checked or uncheck it would submit the form. It works to submit but its not submitting any data to the database. heres what i got for code for this section of checkbox. Code: [Select] <?php echo '<form method="post">'; if(isset($_POST['submit'])){ foreach($_POST['id'] as $id){ $value = (isset($_POST['location'][$id]) && $_POST['location'][$id]=="0" ? '0' : '1'); $insert = mysql_query("UPDATE items SET location='$value' WHERE id='$id'") or die('Insert Error: '.mysql_error()); } } $result = mysql_query("SELECT * FROM items") or die("Query Failed: ".mysql_error()); $counter = 0; echo '<div class="specialcontainer">'; while($row = mysql_fetch_array($result)){ list($id, $item_info, $item_img, $price, $sale, $location) = $row; if($location == '0'){ $set_checked = 'checked="checked"'; }else{ $set_checked = ''; } if($counter % 5==0) { echo '</div>'; echo '<div class="specialcontainer">'; } echo '<div class="special"><img src="../images/items/'.$item_img.'" width="130" /><br />'.$item_info.'<br />$'.$price.'<br />$'.$sale.'<br />Slide Show: <input type="checkbox" id='.$id.' value="0" name="location['.$id.']" '.$set_checked.' onchange="this.form.submit()"/><br /><input type="button" value="Edit" name="edit" id="'.$id.'" onclick="window.location.href=\'specials.php?action=edit&id='.$id.'\'"><input type="button" value="Delete" name="Delete" id="'.$id.'" onclick="window.location.href=\'specials.php?action=delete&id='.$id.'\'"><input type="hidden" name="id[]" value='.$id.' /></div>'; $counter++; } echo '</div>'; echo '</form>'; ?> Hello: I have a SELECT dropdown menu that works like this: Code: [Select] <?php $states_arr = array('AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",'AR'=>"Arkansas",'CA'=>"California",'CO'=>"Colorado",'CT'=>"Connecticut",'DE'=>"Delaware",'DC'=>"District Of Columbia",'FL'=>"Florida",'GA'=>"Georgia",'HI'=>"Hawaii",'ID'=>"Idaho",'IL'=>"Illinois", 'IN'=>"Indiana", 'IA'=>"Iowa", 'KS'=>"Kansas",'KY'=>"Kentucky",'LA'=>"Louisiana",'ME'=>"Maine",'MD'=>"Maryland", 'MA'=>"Massachusetts",'MI'=>"Michigan",'MN'=>"Minnesota",'MS'=>"Mississippi",'MO'=>"Missouri",'MT'=>"Montana",'NE'=>"Nebraska",'NV'=>"Nevada",'NH'=>"New Hampshire",'NJ'=>"New Jersey",'NM'=>"New Mexico",'NY'=>"New York",'NC'=>"North Carolina",'ND'=>"North Dakota",'OH'=>"Ohio",'OK'=>"Oklahoma", 'OR'=>"Oregon",'PA'=>"Pennsylvania",'RI'=>"Rhode Island",'SC'=>"South Carolina",'SD'=>"South Dakota",'TN'=>"Tennessee",'TX'=>"Texas",'UT'=>"Utah",'VT'=>"Vermont",'VA'=>"Virginia",'WA'=>"Washington",'WV'=>"West Virginia",'WI'=>"Wisconsin",'WY'=>"Wyoming"); function showOptionsDrop($array){ $string = ''; foreach($array as $k => $v){ $string .= '<option value="'.$k.'"'.$s.'>'.$v.'</option>'."\n"; } return $string; } ?> <form> <select name="states"> <option value="0">Choose a state</option> <?php echo showOptionsDrop($states_arr); ?> </select> </form> I would like to show all the towns/zipcodes for each individual state with an "onchange" from the SELECT menu. The table is called "zip_codes" and the City/Zip Codes are called "city" and "zip" What I would like to do is display them as links, so a user can click into each town and either create a new entry (a restaurant) or update an existing entry. What would be the best approach for this? I appreciate the help. How can I make my on change in the form pass through the year thats been selected, then use that in the second query? Code: [Select] <select name="mySelect" onchange=""> <?php $result= mysql_query('SELECT DISTINCT Year FROM MonthlySales'); ?> <?php while($row= mysql_fetch_assoc($result)) { ?> <option value="<?php echo htmlspecialchars($row['Year']);?>"> <?php echo htmlspecialchars($row['Year']); ?> </option> <?php } ?> </select> <?php $query="SELECT * FROM MonthlySales WHERE Year = PASS YEAR SELECTED VARIABLE TO THIS BIT HERE"; $results = mysql_query ( $query, $conn); Hi. Need some help on one simple task yet so hard. I have puzzled on this for days and no luck. I have a great script i got from somewhere. Code: [Select] function get_time_difference( $start, $end ) { $uts['start'] = strtotime( $start ); $uts['end'] = strtotime( $end ); if( $uts['start']!==-1 && $uts['end']!==-1 ) { if( $uts['end'] >= $uts['start'] ) { $diff = $uts['end'] - $uts['start']; if( $days=intval((floor($diff/86400))) ) $diff = $diff % 86400; if( $hours=intval((floor($diff/3600))) ) $diff = $diff % 3600; if( $minutes=intval((floor($diff/60))) ) $diff = $diff % 60; $diff = intval( $diff ); return( array('days'=>$days, 'hours'=>$hours, 'minutes'=>$minutes, 'seconds'=>$diff) ); } else { trigger_error( "Ending date/time is earlier than the start date/time", E_USER_WARNING ); } } else { trigger_error( "Invalid date/time data detected", E_USER_WARNING ); } return( false ); } Now the other part is that i have 3 textboxes and i woul like to add the first one like 15:00:00 the second one 18:30:00 and the third would display the difference in lets say seconds. At least the function displays the results in seconds. But the problem is how to get the textbox to work whitout posting the info but to use onchange Any ideas? Code: [Select] <input type="text" value="" name="start_time"/> <input type="text" value="" name="end_time"/> <input type="text" value="______" name="difference"/> Hi there, I am having a PHP form with simple combo box. Here is the code: <form name="form1" method="get" action=""> <select name="select" onChange="form1.submit()"> <option value='value1'>My Value1</option> <option value='value2'>My Value2</option> <option value='value3'>My Value3</option> </select> </form> Now the form is successfully being submitted upon 'onChange' event. The only thing I am trying to do is to retain the value of the new option. As every time I select the value from the combo, it goes back and displays the 'value 1' even when I select value 3 or 2. How do I make the combo box retain the value as 3 (when I select option 3) after the form is submitted. Please reply. Thank you! Hi everyone. I have a combo box which lists usernames and onchange, the username value is passed to a textbox. however, I have 3 textboxes i need to populate based on the selection of the combobox: username. department and email. i have the username going into a textbox, but i'm not sure how to pass department and email into two other textboxes. I'd appreciate any help you could provide. Thanks. Code: [Select] <script> function CBtoTB() {document.getElementById("username").value=document.getElementById("usernameselect").value} </script> <?php $result=mysql_query("select Username, EMail, Department from users"); $options=""; while ($row=mysql_fetch_array($result)) { $username=$row["Username"]; $options.="<OPTION VALUE=\"$username\">".$username.'</option>'; } ?> <select name="usernameselect" id="usernameselect" onchange="CBtoTB()"> <option value="">< select user ><?php echo $options ?></option> </select> <input name="username" type="text" id="username" value="<?php echo $username ?>" size="25" readonly="readonly" /> 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 /////
This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=323252.0 Hello, first time poster.. I've looked the web over for a long time and can't figure this one out. - Below is basic code that successfully checks MySQL for a match and displays result. I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem. That's good.. - I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction) - In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems. It would simply change one select-option value at a time til they get down the form and click submit... but in WordPress I'm having trouble making even ONE single onchange work! - I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor. That works with basic tests like my first bullet point above. - I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work... tried forcing it into an <option value=""> tag.. didn't work. I'm just not sure. I've obviously put xx's in place of private info.. Why does this form give me Page Cannot Be Displayed in WordPress every time? It won't do anything no matter how simple.. using onchange.. Code.. $con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("xxxxxx", $con); $myprodwidth=24; $myprodheight=36; $result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'"); while($row = mysql_fetch_array($result)) { echo $row['prodprice']; } mysql_close($con); <form method="post" action=""> <select name="myheight" onchange="this.form.submit();"> <option selected="selected" value="">select height</option> <option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option> <option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option> </select> Hi. I have a page that allows users to upload some files, using <input type="file">... What I would like is to be able to upload the file "onchange" of this input field using ajax so that (1) users can upload their second file while their first one is uploading, and (2) so that they can see their file on the page as soon as it's done uploading. But I can't seem to find a way to populate the $_FIlLES array without actually submitting the form. I've seen some stuff online where people use iframes to do this kind of thing, but I was hoping there was another way to do it. Any help would be great. Hi, I have a dropdown box, and I want it so that if dropdownbox.value='selected:' then show error message. The checkbox is dynamically populated. Pleaee Advise, - stuart |
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