I am confused as to why my select is not passing any post data to the results page. If I run var_dump($_POST); the select (#boxdest or #boxdest2) is not being displayed. Is there some special way to pass select to php results page. What I normally do is as an example: $var=$_POST['boxdest'];. Problem is the select is not being sent from original code. I have posted my code and would be grateful if someone could show me where I have gone wromg. Many thanks.
<?php
$conn = mysql_connect("localhost","root","");
mysql_select_db("sample",$conn);
$result = mysql_query("SELECT * FROM boxes where department = '{$_GET['dept']}'");
?>
<select id="boxdest" name="boxdest[]"size="7">
<?php
$i=0;
while($row = mysql_fetch_array($result)) {
?>
<option value="<?php echo $row["custref"];?>"><?php echo $row["custref"];?></option>
<?php
$i++;
}
?>
</select>
<input type="button" id="submit2" name="submit2" value=">" />
<input type="button" id="submit3" name="submit3" value="<" />
<select id="boxdest2" name="boxdest2[]" size="7"></select>
<script type="text/javascript">
$("#submit2").click( function()
{
//alert('button clicked');
$box1_value=$("#boxdest").val();
$box1_text=$("#boxdest option:selected").text();
$("#boxdest2").append('<option value="'+$box1_value+'">'+$box1_text+'</option>');
$("#boxdest option:selected").remove();
});
$("#submit3").click( function()
{
//alert('button3 clicked');
$box2_value=$("#boxdest2").val();
$box2_text=$("#boxdest2 option:selected").text();
$("#boxdest").append('<option value="'+$box2_value+'">'+$box2_text+'</option>');
$("#boxdest2 option:selected").remove();
}
);
</script>
Related
I got some problems with my code. I want to delete multiple data from MySQL database that populate from Select Option.
Example: I select data with id 5, 2, 4, then press the delete button, it only deletes the latest id which is 5.
Can I know what is the problem? Below is my code:
index.html
<?php
include("configPDO.php");
$smt = $conn->prepare("SELECT * FROM frame_list ORDER BY framework_name ASC");
$smt->execute();
$results = $smt->fetchAll();
?>
<form method="post" id="multiple_select_form">
<select name="framework[]" id="framework" class="form-control selectpicker" data-live-search="true" multiple>
<?php foreach ($results as $row2): ?>
<option value= <?php echo $row2["framework_id"]; ?>><?php echo $row2["framework_name"];?></option>
<?php endforeach ?>
</select>
<br><br>
<input type="hidden" name="framework_id" id="framework_id" />
<input type="submit" name="submit" class="btn btn-info" value="Submit" />
</form>
<script>
$(document).ready(function(){
$('.selectpicker').selectpicker();
$('#framework').change(function(){
$('#framework_id').val($('#framework').val());
});
$('#multiple_select_form').on('submit', function(event){
event.preventDefault();
if($('#framework').val() != '')
{
var form_data = $(this).serialize();
$.ajax({
url:"delete.php",
method:"POST",
data:form_data,
success:function(data)
{
//console.log(data);
$('#framework_id').val('');
$('.selectpicker').selectpicker('val', '');
alert(data);
}
})
}
else
{
alert("Please select framework");
return false;
}
});
});
</script>
delete.php
<?php
include("configPDO.php");
$smt = $conn->prepare("DELETE FROM frame_list WHERE framework_id = '".$_POST["framework_id"]."'");
$smt->execute();
if($smt){
echo "Data DELETED";
}else{
echo "Error";
}
?>
Appreciate it if anyone can solve my problem.
Thank you very much.
this is where framework id got overwritten:
$('#framework').change(function(){
$('#framework_id').val($('#framework').val());
});
delete.php will only run when your ajax request send(in this case, when you click submit button).
so maybe you'll want to pass several id at a single request if you want delete them at the same time. or you may want to delete them one by one by sending the ajax request each time you choose a id.
Now I create a system that can delete multiple data using select option. But here I got some issues. When i only select one data, and press button delete, it will delete. But if I choose more than one data, for example, 3 data, it will only delete the latest id of the data. Below is my the image
And below is my code:
index.php
<form method="post" id="multiple_select_form">
<select name="framework" id="framework" class="form-control selectpicker" data-live-search="true" multiple>
<?php foreach ($results as $row2): ?>
<option value= <?php echo $row2["framework_id"]; ?>><?php echo $row2["framework_name"];?></option>
<?php endforeach ?>
</select>
<br /><br />
<input type="hidden" name="framework_id" id="framework_id" />
<input type="submit" name="submit" class="btn btn-info" value="Submit" />
</form>
<script>
$(document).ready(function(){
$('.selectpicker').selectpicker();
$('#framework').change(function(){
$('#framework_id').val($('#framework').val());
});
$('#multiple_select_form').on('submit', function(event){
event.preventDefault();
if($('#framework').val() != '')
{
var form_data = $(this).serialize();
$.ajax({
url:"insert.php",
method:"POST",
data:form_data,
success:function(data)
{
//console.log(data);
$('#framework_id').val('');
$('.selectpicker').selectpicker('val', '');
alert(data);
}
})
}
else
{
alert("Please select framework");
return false;
}
});
});
</script>
insert.php
<?php
include("configPDO.php");
$smt = $conn->prepare("DELETE FROM frame_list WHERE framework_id = '".$_POST["framework_id"]."'");
$smt->execute();
if($smt){
echo "Data DELETED";
}else{
echo "Error";
}
?>
Can anyone knows how to solve this problem? Thanks
framework will hold one value every time.
Use input arrays -
Change name="framework" to name="framework[]".
and in query -
WHERE framework_id in ('". implode("','", $_POST["framework_id"]) ."')"
Try to use parameter binding for security.
<form action="mydtr" method="post">
<select id="periodname">
<?php foreach($period as $period): ?>
<option><?= $period['period']; ?></option>
<?php endforeach; ?>
</select>
</form>
Im looking for ways to prevent selection options to reset after i select and submit. thank you in advance for the help
Not sure if this is what you asking for but here you go.
<?php $period = array("period1", "period2", "period3", "period4"); ?>
<form action="" method="post">
<select id="periodname" name="periodname">
<?php foreach($period as $period): ?>
<option><?= $period; ?></option>
<?php endforeach; ?>
</select>
<input value="submit" type="submit"></form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<?php if(isset($_POST["periodname"])): echo "Form was submitted and we set that value of the select box to selected ".$_POST["periodname"];?>
<script>
$( "#periodname option" ).each(function( index ) {
var item = $(this).val();
if(item == "<?php echo $_POST["periodname"]?>"){
$(this).prop('selected', true);
}
});
</script>
<?endif;?>
I have the following option menu in a form that will insert the fields into a table:
<option value="">select staff</option>
<?php
do {
?>
<option value="<?php echo $row_Staff['Staff_Name']."||".$row_Staff['Email']?>">
<?php echo $row_Staff['Staff_Name']?></option>
<?php
} while ($row_Staff = mysql_fetch_assoc($Staff));
$rows = mysql_num_rows($Staff);
if($rows > 0) {
mysql_data_seek($Categories, 0);
$row_Staff = mysql_fetch_assoc($Staff);
}
?>
</select>
I have 2 fields from source table in value of option from technique explained in How to post two values in an option field?: Staff_Name and Email.
I am trying to insert both fields from the form into a table using:
<input type="hidden" name="Staff_Name" class="form-control" id="Staff_Name" value=<?php
$staff =$_POST['Staff_Data'];
$staff_name = explode("||", $staff);
echo $staff_Name[0];
?> />
and
<input type="hidden" name="Email" class="form-control" id="Email" value=<?php
$staff =$_POST['Staff_Data'];
$email = explode("||", $staff);
echo $email[1];
?> />
Unfortunately, I can see the 2 fields separated by "||" in the table if I insert the option menu value but cannot seem to insert Staff_Name or Email into individual fields. On insert both fields are blank. Any help would be appreciated.
Instead of combine staffname and staffemail in the dropdown value. Please staffname in dropdown value and staffemail in the property of dropdown and onchange of the dropdown set those values in the hidden inputs so you will easily get those values on the form submission.
Please go through below code and let me know if you have any query.
//Dropdown
<select id="ddStaff">
<option value="">select staff</option>
<?php
do { ?>
<option value="<?php echo $row_Staff['Staff_Name']; ?>" staff-email = "<?php echo $row_Staff['Email'];?>">
<?php echo $row_Staff['Staff_Name']?>
</option> <?php
} while ($row_Staff = mysql_fetch_assoc($Staff));
$rows = mysql_num_rows($Staff);
if($rows > 0) {
mysql_data_seek($Categories, 0);
$row_Staff = mysql_fetch_assoc($Staff);
}
?>
</select>
//Input hidden fields to store staff name and staff email
<input type="hidden" id="txtStaffName" name="txtStaffName">
<input type="hidden" id="txtStaffEmail" name="txtStaffEmail">
//Jquery code.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#ddStaff").on('change',function(){
var staffName = $(this).val();
var staffEmail = $('option:selected', this).attr('staff-email');
$("#txtStaffName").val(staffName);
$("#txtStaffEmail").val(staffEmail);
});
});
</script>
Please check it on https://jsfiddle.net/z4a0fywp/
For testing purpose I have not made the inputs hidden in the fiddle.
This question already has answers here:
jQuery Ajax POST example with PHP
(17 answers)
Closed 8 years ago.
I developed a php form which initiates a database request to fetch data depending on a drop down choice.
PHP Form:
<form method="get" action="<?php echo $url = basename($_SERVER['PHP_SELF']); ?>">
<select name="town" onchange='this.form.submit()'>
<?php $result= mysql_query('Query'); ?>
<option value="x" selected>Select Choice</option>
<?php while($row= mysql_fetch_assoc($result)) { ?>
<option value="<?php echo htmlspecialchars($row['town']);?>" >
<?php echo htmlspecialchars($row['town']); ?>
</option>
<?php } ?>
<input type="hidden" name="action" value="submit" /><br>
</select>
</form>
Form action:
<?php
if(isset($_GET["action"])) {
$var1= $wpdb->get_results("Query");
$var2= $wpdb->get_results("Query");
Content to show once executed }
?>
How can I make the form fetch the Data using AJAX not to stay refreshing the whole page continuously but only the form part?
<!DOCTYPE html>
<html>
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
</head>
<body>
<form id="form_id" action="<?php echo $url = basename($_SERVER['PHP_SELF']); ?>" method="post">
<select id="town" name="town" onchange="send_to_server()">
<?php $result= mysql_query("Query"); ?>
<option value="x" selected>Select Choice</option>
<?php while($row= mysql_fetch_assoc($result)){ ?>
<option value="<?php echo htmlspecialchars($row['town']); ?>">
<?php echo htmlspecialchars($row['town']); ?>
</option>
<?php } ?>
<input type="hidden" name="action" value="submit" /><br>
</select>
</form>
<script type='text/javascript'>
/* attach a submit handler to the form */
function send_to_server(){
var value = $("#town").val();
/* get some values from elements on the page: */
var $form = $("#form_id"); var url = $form.attr('action');
/* Send the data using post */
var posting = $.post( url, { option_value: $("#town").val() } );
posting.done(function( data ) {
alert('success');
});
}
</script>
</body>
</html>
The above does exactly what you want. Check it in localhost