PHP while in edit mode show selected value in to drop down - php

This question was asked already, but my question is very simple.
In the my account page, I have the employee country in a dropdown.
How to select a value in the combo, when in edit mode?

Let's assume you have the user's country in $user_country and the list of all countries in $all_countries array:
<select id="country">
<?php
foreach ( $all_countries as $country ):
$selected = "";
if ( $country == $user_country )
$selected = "selected";
?>
<option value="<?php echo $country; ?>"
selected="<?php echo $selected; ?>">
<?php echo $country; ?>
</option>
<?php
endforeach; ?>
</select>
should work.

An option tag will be the default for a select list when the selected attribute is set. In the following code option 2 will show up as the current selected option when the page loads:
<select>
<option value="1">1</option>
<option value="2" selected="selected">2</option>
<option value="3">3</option>
</select>
To achieve this in your PHP code conditionally display the selected attribute on your options against what the current value is:
<option value="1"<?php if($user['country'] == '1') { ?> selected="selected"<?php } ?>>1</option>
<option value="2"<?php if($user['country'] == '2') { ?> selected="selected"<?php } ?>>2</option>
<option value="3"<?php if($user['country'] == '3') { ?> selected="selected"<?php } ?>>3</option>

function p_edit_combo($cCurstatus,$h_code_default,$h_name=NULL){
<select name="<?php echo $cCurstatus;?>" id="<?php echo $cCurstatus;?>" class="main_form_select">
<option value="">Select</option>
<?php
$sql_h = "SELECT h_code,h_name FROM med_hl WHERE status = 1";
$sql_h_result = mysql_query($sql_h);
while($row=mysql_fetch_array($sql_h_result)){
$h_code = $row['h_code'];
$h_name = $row['h_name'];
?>
<option <?php if($h_code_default==$h_code){ ?> selected="selected" <?php }?> value='<?php echo $h_code; ?>' >
<?php echo $h_code."|".$h_name; ?>
</option>
<?php } ?>
</select>
<?php
}

**i have two table
" users" colmns(fname,lname,...as on ohther_infomation,hobbies datatype(int))
"info" columns (id (primary_key),hobbies(varchar 200)); in which i stored for hobbies name
In my case i am storing values in from (1,2,3,4) in hobbies (int) filled of users table which i matached them through join after time of fetch them,
in my info table i stored hobbies by their name (reading, writing,playing,gyming)
$row has our users selected hobbies (int)
$rows has list of our hobbies(varchar)
edit.php i need Dropdown value selected :==== And i am Doing Like this :--- (100% Working)**
<div class="form-control">
<label for="hobbies">Hobbies</label>
<select name="hobbies">
<?php
$query = "SELECT * FROM info";
$results = mysqli_query($connect, $query);
while ($rows = mysqli_fetch_array($results)) {
?>
<option <?php if ($rows['id'] == $row['hobbies']) { ?> selected="selected" <?php } ?> value='<?php echo $rows['id']; ?>'>
<?php echo $rows['hobbies']; ?>
</option>
<?php
}
?>
</select>
<span class="text-danger"><?php if (isset($err_hobbies)) echo $err_hobbies; ?></span>
</div>

Related

Update dependent dropdown not showing specific data

I am trying to create a UPDATE page which contain dependent dropdown.
When the 1st dropdown is selected, the dependent dropdown does not show data for the specific data selected in the 1st dropdown.
It is showing "Select Option".
Below is what i have try so far.
1st dropdown
> <label>DEPARTMENT</label>
<div class="col-sm-10">
<select name="dept_ID" class="form-control">
<option value="">Select Department</option>
<option value="1" <?php if($data['dept_ID'] == '1'){ echo 'selected'; } ?>>Department One</option>
<option value="6" <?php if($data['dept_ID'] == '6'){ echo 'selected'; } ?>>Department Six</option>
<option value="2" <?php if($data['dept_ID'] == '2'){ echo 'selected'; } ?>>Department Two</option>
<option value="3" <?php if($data['dept_ID'] == '3'){ echo 'selected'; } ?>>Department Three</option>
</select>
</div>
>
Dependent dropdown
<label>SUPERVISOR : </label>
<div class="col-sm-10">
<select class="form-control" name="op_supervisor" id="selectID">
<option>Select Option</option>
<?php $sql = "SELECT * FROM supervisor JOIN DEPARTMENT ON supervisor.dept_ID = DEPARTMENT.dept_ID ";
$result = mysqli_query($conn,$sql);
while($row = mysqli_fetch_assoc($result)) {?>
<option value="<?php echo $row['op_supervisor'] ?>"> <?php echo $row['dept_name'] ?> - <?php echo $row['op_supervisor'] ?></option>
<?php }?>
</select>
</div>
My expected result is the dependent dropdown will show specific data chosen in the department dropdown and what have been selected before.

How to add option fetched from the database to the end in select php

I have a categories table where I have different categories. I have also add Other in the category. When I'm getting the categories to populate the select I want the other option to appear at last. how can I do that?
<select name="category" id="category" class="form-control <?php if (isset($errors['category'])) echo 'form-error'; ?>">
<option value="">Select a category...</option>
<?php while ($category = mysqli_fetch_assoc($categories)) { ?>
<option value="<?php echo h($category['id']); ?>" <?php if (isset($_POST['category']) && h($category['id']) === $_POST['category']) echo 'selected'; ?>><?php echo h($category['name']); ?></option>
<?php } ?>
</select>
If the Id of Other category is fixed, then you have to ignore it in your database query.
Let's say that the Id of that option is 1, our query will be like this:
select * from categories where Id <> 1
This query brings all categories except "other".
After that do the following:
<select name="category" id="category" class="form-control <?php if (isset($errors['category'])) echo 'form-error'; ?>">
<option value="">Select a category...</option>
<?php while ($category =
mysqli_fetch_assoc($categories)) { ?>
<option value="<?php echo h($category['id']); ?>" <?php
if (isset($_POST['category']) && h($category['id']) ===
$_POST['category']) echo 'selected'; ?>><?php echo
h($category['name']); ?></option>
<?php } ?>
<option value="1">Other</option>
</select>
hope it helps

How will it show the selected value (SELECTED) in CodeIgniter 3.x?

I am trying to populate a drop-down list of the database. In my view file I have the following code
Here is my controller
$query = $this->interprete_model->interpreteID($this->session->userdata('user_id'));
print_r($query);
$data['interprete'] = $query;
Aqui esta mi vista, usa set_select.
<select class="form-control" name="regionI" id="regionI">
<option value="">- Select -</option>
<?php foreach($result as $row):?>
<option value="<?php echo $row->id;?>"
<?php echo set_select('regionI', $row->id, TRUE); ?>><?php echo $row->name;?></option>
<?php endforeach; ?>
</select>
Result:
enter image description here
Many selected, I need one selected to modify (update) the data.
You can try this :
<select class="form-control" name="regionI" id="regionI">
<option value="">- Select -</option>
<?php foreach($users as $row):
$selected = FALSE;
// 1 is the id u want to be selected u can change it according to you
if ($row->id == 1){
$selected = TRUE;
}
?>
<option value="<?php echo $row->id;?>"
<?php echo set_select('regionI', $row->id, $selected); ?>><?php echo $row->name;?></option>
<?php endforeach; ?>
</select>
You can also use form_dropdown as
// FOR ids
$ids = array(1,2,3,4); // array of user ids
echo form_dropdown('regionI',$ids,1,array('class'=>'form-control'));
// FOR name
$names= array('name1','name2','name4','name3'); // array of user names
echo form_dropdown('regionI',$names,'name1',array('class'=>'form-control'));
For More :
https://www.codeigniter.com/user_guide/helpers/form_helper.html
i write this way for edit time selection
<?php foreach ($select_single as $select_single_show):?>
<select class="form-control" name="regionI">
<?php foreach ($users as $row):?>
<option <?php if($row->id==$select_single_show->regionI)echo "selected";?> value="<?php echo $all_branch_show->id?>"><?php echo $row->name?>
</option>
<?php endforeach;?>
</select>
<?php endforeach;?>

How do I Populate Multi Select drop dwon using php

Here I am listing all cars.customers want to compare car so they will select from this drop down. A person can select multiple cars. At the first time he is selecting 'Audi' and Saab' I will store it into data base next if he came I need to populate Saab and audi as select how I can do this using php
<select name="cars" multiple>
<option value="volvo">Volvo</option>
<option value="saab">Saab</option>
<option value="opel">Opel</option>
<option value="audi">Audi</option>
</select>
Here is my code
<select id="cars" class="multiselect" multiple="multiple" name="cars[]">
<?PHP
if($carslist->num_rows() >0)
{
foreach($carslist->result_array() as $entry):
?> <option value="<?php echo($entry['ID']); ?>" ><?php echo($entry['car_name']); ?></option>
<?php
endforeach;
}
?>
</select>
Following code I tried $resources contain select cars
<select id="cars" class="multiselect" multiple="multiple" name="cars[]">
<?PHP
if($carslist->num_rows() >0)
{
foreach($carslist->result_array() as $entry):
if($resources->num_rows() >0)
{
foreach($resources->result_array() as $car):
if($entry['ID'] == $employee['car_id'])
{
$select = 'selected="selected"';
}
else
{
$select = '';
}
endforeach;
}
?> <option value="<?php echo($entry['ID']); ?>" <?php echo $select;?> ><?php echo($entry['car_name']); ?></option>
<?php
endforeach;
}
?>
</select>
but it showing error
Here, try something like this, and see if it works:
Here is the controller:
<?php
function something(){
$data = array();
$data['cars'] = $this->some_model->some_function_to_return_cars_array();
$data['selected'] = $some_array_of_selected_cars();
$this->load->view('some_view', $data);
}
?>
And this is the view:
<select id="cars" class="multiselect" multiple="multiple" name="cars[]">
<option value="">Select:</option>
<?php
foreach( $cars as $key => $val ){
?>
<option value="<?php echo $val['some_id'] ?>"
<?php
if( in_array( $val['some_id'], $selected ) ) echo ' selected';
?>
><?php echo $val['some_name'] ?></option>
<?php
}
?>
</select>

Populating Select Field from Database

I'm trying to populate a select field with PHP. The problem is I can't figure out how to display them because I'm getting the one that's value matches in the database showing up twice because I'm echoing it as selected and then looping it all the results. How can I just display the selected one that matched the fields value and then all the ones that don't match the selected one?
TABLE CATEGORIES
cat_id cat_name
1 soccer
2 baseball
3 basketball
TABLE ARTICLES
art_id art_cat_id
1 1
PHP / HTML
<select name="category">
<?php
$sql = "SELECT cat_id cat_name, art_id, art_cat_id
FROM categories LEFT JOIN articles
ON categories.cat_id = articles.art_cat_id
WHERE art_id = 1";
$result = query($sql);
if($result===false) {
echo("Query Fail");
}
else {
?>
<option value="<?php echo $data['art_cat_id'] ?>" selected="selected"><?php echo $data['cat_name'] ?></option>
<?php
while( $data = mysqli_fetch_array($result)) {
?>
<option value="<?php echo $data['cat_id'] ?>"><?php echo $data['cat_name'] ?></option>
<?php
}
}
?>
</select>
What it's returning
<select name="category">
<option value="1" selected="selected">soccer</option>
<option value="1">soccer</option>
<option value="2">baseball</option>
<option value="3">basketball</option>
</select>
What I'm looking for
<select name="category">
<option value="1" selected="selected">soccer</option>
<option value="2">baseball</option>
<option value="3">basketball</option>
</select>
Skip the row if the value matches the first one.
A snippet:
?>
<option value="<?php echo $data['art_cat_id'] ?>" selected="selected"><?php echo $data['cat_name'] ?></option>
<?php
while( $data = mysqli_fetch_array($result)) {
if ($data['art_cat_id'] == $data['cat_id']) continue;
?>
<option value="<?php echo $data['cat_id'] ?>"><?php echo $data['cat_name'] ?></option>
<?php
}

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