I am having trouble passing & displaying default value for select tag. So, I have tried different variants of select and option tag but I am not able to get this right.
I want to fetch different options in form of a drop down menu. I want to display a default value in it. ($categorytemp[$i] in code). On displaying the form
A user can update this choice; or
A user can select the same choice again; or
A user can not change it at all.
Expected result for the above activity should be update, update, default value for $category
Below is the snapshot of my code
<select name="category" value="<?php echo $categorytemp[$i] ?>">
<?php
$sql = "SELECT * FROM education_details";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result))
{
$education_category = $row["id"];
$education_name= $row["name"];
?>
<option value='<?php echo $education_category?>'><?php echo $education_name;?></option>
<?php
}
?>
</select>
I tried using the option tag with selected attribute but that is making things more complicated. For example, if I use
<select name="category" value="<?php echo $categorytemp[$i] ?>">
<option selected="selected"><?php echo $categorytemp[$i]; ?> </option>
<?php
$sql = "SELECT * FROM education_details";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result))
{
$education_category = $row["id"];
$education_name= $row["name"];
?>
<option value='<?php echo $education_category?>'><?php echo $education_name;?></option>
<?php
}
?>
</select>
It displays the form as desired but on making no selection, it passes null value in the category. I have gone through questions of similar type asked earlier, but they do not answer my query to my satisfaction
You do it with a "selected" attribute of a given <option> and not in the <select> tag with the value attribute.
Check these two:
http://www.w3schools.com/tags/tag_select.asp
http://www.w3schools.com/tags/tag_option.asp
Have you consider using Javascript on the form submit to achieve it?
Another option would be a hidden field with default value before the select and an hidden option of the select field:
<html>
<head>
<body>
<form method="POST" action="">
<input type="hidden" name="dropdown" value="default" />
<select name="dropdown">
<option selected disabled hidden style="display: none" value="default"></option>
<option>1</option>
<option>2</option>
<input type="submit" value="send" />
</form>
</body>
</html>
Is $categoryTemp[$i] an ID or a name? If it is the ID and you are wanting to check the result against it to determine the default option to display then something along these lines would select the option that is equal to the value of $categoryTemp[$i].
<option <?php echo ($categorytemp[$i] == $education_category ? 'selected' : ''); ?>
So if $categorytemp[$i] = 1 and $education_category = 1 then that option would be marked selected and be displayed when the page loaded
This did the trick. When no value is selected, pass the default value as value in option
<select name="category" >
<option value='<?php echo $categorytemp[$i]?>'><?php echo $categorytemp[$i]; ?> </option>
<?php
$sql = "SELECT * FROM education_details";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result))
{
$education_category = $row["id"];
$education_name= $row["name"];
?>
<option value='<?php echo $education_category?>'><?php echo $education_name;?></option>
<?php
}
?>
</select>
Related
I have edit form where I get info from database
<select name="table">
<?php
//fetch all tables from database
$user = $con->query("SELECT * FROM table") or die(mysql_error());
while($row = $table->fetch_object()) { ?>
<option value="<?php echo $row->tablename;?>">
<?php echo $row->tablename; ?>
</option> <?php }?> </select>
<label for="time">Time :</label>
<select name="time">
<option value="twotothree">2PM-3PM</option>
<option value="threetofour">3PM-4PM</option>
<option value="fourtofive">4PM-5PM</option>
<option value="fivetosix">5PM-6PM</option>
</select>
The names of the tables from drop-down menu are different. Everytime I select a table and a time slot and save the data, the selection goes back to the first row of the menu. Eg I select table 3 and 4PM-5PM after saving it goes back to table 1 and 2PM-3PM. I need to be fixed on the last selection as I might use 4PM-5PM for table 4 also. Any idea? Thanks
You can add selected attribute when rendering your select list, depending on $_POST variable, when it's available. For example for you table select element:
<select name="table">
<?php
$user = $con->query("SELECT * FROM table") or die(mysql_error());
while($row = $table->fetch_object()) { ?>
<option value="<?php echo $row->tablename;?>" <?php if (isset($_POST['table']) && $_POST['table'] == $row->tablename) echo 'selected'; ?> >
<?php echo $row->tablename; ?>
</option>
<?php }?>
</select>
In similar way you can do for time select element.
on calculatePC.php, I have this code to display the finish_product
Select product:
<select class="itemTypes">
<?php while ($row1 = mysqli_fetch_array($result)) { ?>
<option value="<?php echo $row1['finish_product']; ?>">
<?php echo $row1['finish_product']; ?>
</option>
<?php } ?></select>
<br/><br/>
Let's say I have chosen Table as the finish_product.
On docalculate.php, I would like to display what I've chosen based on the dropdown list I've selected.
I tried this but there is error.
<?php echo $_POST['finish_product'] ?>
May I know how to display the result?
This doesn't exist:
$_POST['finish_product']
because you don't have a form element named "finish_product" in your markup. Add that name to the form element:
<select name="finish_product" class="itemTypes">
You need to do two things:-
Create a form before select and give it an action
give name attribute to your select box, then only you can get data in $_POST
So do like below:-
<form method="POST" action = "docalculate.php"> // give the php file paht in action
<select class="itemTypes" name="finish_product"> // give name attribute to your select box
<?php while ($row1 = mysqli_fetch_array($result)) { ?>
<option value="<?php echo $row1['finish_product']; ?>">
<?php echo $row1['finish_product']; ?>
</option>
<?php } ?></select>
<br/><br/>
<input type ="submit" name="submit" value ="Submit">
</form>
Now on docalculate.php:-
<?php
echo "<pre/>";print_r($_POST['finish_product']); // to see value comes or not
Note:- through Jquery its also possible. Thanks
I've got a table that populates data from a MYSQL database and populates a drop-down menu from the same database. I have the drop down menu and table just fine, I would like to be able to choose which data I show in the table however.
<select name = 'peer-id' method='post' style = 'position: relative'>
<?php
while ($content = mysql_fetch_array($peer)) {
echo "<option value='" . $content['Peer'] . "'>" . $content['Peer'] . "</option>";
}
$results = mysql_query("SELECT Destination FROM rate ");
?>
</select>
That's what I have for the select box. How can I get the choice from that and save that as a variable and refresh the table data?
I need to clarify that this will change that current data
#Data#Data#Data
#Data#Data#Data
#Data#Data#Data
Then choose drop down choice and I want it to show new data
#Data2#Data2#Data2
#Data2#Data2#Data2
#Data2#Data2#Data2
So it's going to need to load a new page or refresh some how because it's changing via PHP and not javascript.
I think form may be better, for example
<form id="myform" method="post">
<select name = 'peer-id' style = 'position: relative' onchange="change()">
<option value="1">12</option>
<option value="2">15</option>
<option value="3">16</option>
<option value="4">18</option>
</select>
</form>
<script>
function change(){
document.getElementById("myform").submit();
}
</script>
In the above code, whenever you change the value of select, it will post to the backend, then according to the posted value, you can do want you want, to get the peer-id in php, you can use the following code
$peer-id = $_POST['peer-id'];
Hope helps!
apply this code in select tag hope this works
<select onchange="location = this.options[this.selectedIndex].value;" style="text-decoration:none;">
<option value="customers.php"></font></option>
</select>
insted of the static options, you can do it like this :) here you get all the options from the database. Just replace it with the static options
$peer = mysql_query("SELECT Peer FROM rate Group By Peer Where peer = 'variable'");
$result_peer = mysql_query($peer);
if($result_peer){
while($row_peer = mysql_fetch_array($result_peer)){
echo'<option value='.$row_peer['Peer'].'>'.$row_peer['Peer'].'</option>';
}
I agree in using form, and with this you can echo back onto the page with a submit button (code tested):
<form id="myForm" method="POST">
<select name="select" onchange="<?php echo $_SERVER['PHP_SELF'];?>">
<option value="N">No</option>
<option value="Y">Yes</option>
</select>
<input type="submit" name="formSubmit" value="Submit" >
</form>
<?php
if(isset($_POST['formSubmit']) ){
$var = $_POST['select'];
$query = "SELECT * FROM table_name WHERE DesiredField='$var'";
$result = mysql_query($query)
or die(mysql_error());
while($row = mysql_fetch_array($result)){
$var2 = $row['FieldName'];
echo "First field: " . $var2 . "<br>";
// and so on for what you want to echo out
}
}
?>
I've been trying to find the best way to do this for a while now but cannot figure it out.
I have a simple dropdown which auto populates based on an SQL query. When you press "search" I need it to go to a page with the url extension ?id=x where x is the id of the option they selected.
$locations = $wpdb->get_results( $wpdb->prepare(
"SELECT * FROM wp_places_index WHERE visible='Y' ORDER BY place_name ASC") );
?>
<form id="find-buses-form" method="post" action="places_index.php">
<select class="default-value" type="text" name="from">
<option>Please select...</option>
<?php
foreach($locations as $location)
{
echo "<option>".$location->place_name."</option>";
// maybe this? echo "<input type=\"hidden\" name=\"id\">".$location->id."</input>";
}
?>
</select>
<input type="submit" name="submit" value="Show me" />
</form>
I think I may need to make it go to an external page which uses $_POST to pull that hidden field but I'd rather do it on one page.
I've achieved this before out of wordpress using something like this:
while ($row = mysql_fetch_array($result))
{
$picture = $row['Image'];
if($picture == ""){
$picture= "thumbnails/no-image-small.png";
}
$product_ID = $row["ProductID"];
But wordpress does not like the mysql_fetch_array :( Any advise?
You need to put whatever in a value attribute of your option tag. For example:
<?php
$locations = $wpdb->get_results( $wpdb->prepare(
"SELECT * FROM wp_places_index WHERE visible='Y' ORDER BY place_name ASC") );
?>
<form action="places_index.php" method="get">
<select name="x">
<?php foreach ($locations as $location): ?>
<option value="<?php echo $location->id; ?>"><?php echo $location->place_name; ?></option>
<?php endforeach; ?>
</select>
</form>
When submitted, the form will go to http://example.com/places_index.php?x=1 presuming the name of your select is x and the option selected has 1 in its value attribute.
Hope this helps.
Im trying to implement the search feature in my website.
when the search keyword is entered in the textbox, and the category combo is selected, the form will be Posted and the result will be shown on the same page.
what i want is to keep the selected category of the combo by default in the form after posted
For eg., If i select the category 'Automobiles' in the combo and click search, after form submit, the combo should show the automobiles as default selected option. Please help me. Any help will be appreciated
I assume you get categories from database.
you should try:
<?php
$categories = $rows; //array from database
foreach($rows as $row){
if($row['name'] == $_POST['category']){
$isSelected = ' selected="selected"'; // if the option submited in form is as same as this row we add the selected tag
} else {
$isSelected = ''; // else we remove any tag
}
echo "<option value='".$row['id']."'".$isSelected.">".$row['name']."</option>";
}
?>
Assuming that by "combo" you mean "A regular select element rendering as a drop down menu or list box" and not "A combobox that is a combination of a drop down menu and free text input":
When outputting the <option> elements, check the value against the submitted data in $_POST / $_GET and output selected (in HTML) or selected="selected" (in XHTML) as an attribute of the option element.
Here is the JQuery way I am using.
<select name="name" id="name">
<option value="a">a</option>
<option value="b">b</option>
</select>
<script type="text/javascript">
$("#name").val("<?php echo $_POST['name'];?>");
</script>
But this is only if you have jquery included in your webpage.
Regards
<?php
$example = $_POST["friend"];
?>
<form method="POST">
<select name="friend">
<option value="tom" <?php if (isset($example) && $example=="tom") echo ' selected';?>>Thomas Finnegan</option>
<option value="anna" <?php if (isset($example) && $example=="anna") echo ' selected';?>>Anna Karenina</option>
</select>
<br><br>
<input type="submit">
</form>
This solved my problem.
This Solved my Problem. Thanks for all those answered
<select name="name" id="name">
<option value="a">a</option>
<option value="b">b</option>
</select>
<script type="text/javascript">
document.getElementById('name').value = "<?php echo $_GET['name'];?>";
</script>
$countries_uid = $_POST['countries_uid'];
while($row = mysql_fetch_array($result)){
$uid = $row['uid'];
$country = $row['country_name'];
$isSelected = null;
if(!empty($countries_uid)){
foreach($countries_uid as $country_uid){//cycle through country_uid
if($row['uid'] == $country_uid){
$isSelected = 'selected="selected"'; // if the option submited in form is as same as this row we add the selected
}
}
}else {
$isSelected = ''; // else we remove any tag
}
echo "<option value='".$uid."'".$isSelected.">".$country."</option>";
}
this is my solutions of multiple select dropdown box after modifying Mihai Iorga codes
After trying al this "solves" nothing work. Did some research on w3school before and remember there was explanation of keeping values about radio. But it also works for Select option. See here an example. Just try it out and play with it.
<?php
$example = $_POST["example"];
?>
<form method="post">
<select name="example">
<option <?php if (isset($example) && $example=="a") echo "selected";?>>a</option>
<option <?php if (isset($example) && $example=="b") echo "selected";?>>b</option>
<option <?php if (isset($example) && $example=="c") echo "selected";?>>c</option>
</select>
<input type="submit" name="submit" value="submit" />
</form>
Easy solution:
If select box values fetched from DB then to keep selected value after form submit OR form POST
<select name="country" id="country">
<?php $countries = $wpdb->get_results( 'SELECT * FROM countries' ); ?>
<option value="">
<?php if(isset($_POST['country'])){echo htmlentities($_POST['country']); } else { echo "Select Country *"; }?>
</option>
<?php foreach($countries as $country){ ?>
<option <?php echo ($_POST['country'] == $country->country_name ? 'selected="selected"':''); ?> value="<?php echo $country->country_name; ?>"><?php echo $country->country_name; ?>
</option>
<?php } ?>
</select>