hello i have two tables joined , and i'm trying to fetch some values into a select box, making it so that when a user selects one of the options , when submitting the form it will post the other values associated
<select name="matricula[]">
<?php
// use a while loop to fetch data
// from the $all_categories variable
// and individually display as an option
while ($matricula = mysqli_fetch_array(
$result,MYSQLI_ASSOC)):;
?>
<option value="<?php echo $matricula["matricula"];
// The value we usually set is the primary key
?>">
<?php echo $matricula["matricula"];
// To show the category name to the user
?>
</option>
<?php
endwhile;
// While loop must be terminated
?>
</select>
in this code i need to add the value automatically and insert on form submission : "tipocomb" it has already got a table named "combustivel" created with the values and when i use json encode it displays the two arrays connected in the view source
it's just a question of how to echo the "tipocomb"?
If you want to do something like that you need js to post form that has hidden inputs with related data. Simpler solution would be when you read the post in your controller fetch related data and do further logic there.
Related
I want to create a form with two fields. The first field lists all pages and the second field lists all child pages (if they exist) of the chosen page from the first field. When the form is submitted the site redirects the user to the chosen page. How can I implement this form?
You're going to need to do an AJAX request to get the second drop down list.
There is a function built into WordPress which gets the pages and you can specify parent ID's and also to show only top level pages.
Heres an example of the code but you'll need to then implement it in a way that best suits your requirements
<?php
$top_level_pages = get_pages(array('parent'=> 0));
?>
<select>
<?php
foreach($top_level_pages as $top_level_page) {
echo '<option>'.$top_level_page['post_title'].'</option>';
}
?>
</select>
You will then need to determine what option the user has clicked and run an AJAX request to fill the second select.
The second select should look something like this:
<?php
$id = $_GET['page_id']; // get the id of the page from the first select
$child_pages = get_pages(array('child_of'=>$id));
?>
<select>
<?php
foreach($child_pages as $child_page) {
echo '<option>'.$child_page['post_title'].'</option>';
}
?>
</select>
Read more about the get_pages() function here: http://codex.wordpress.org/Function_Reference/get_pages
I'm new to Yii framework and I'm designing a form (I'm using create form) to create a row in database.
Let me explain about the scenario succintly, so that I can make it clear for what I want-
I have 10 fields in this form. Out of this 10 fields, five fields change dynamically.
I created two div's basically div A and div B and repeated the fields as required for the two cases.
Say some fields which are textfields in div A will be dropdownlists in div B.
<div id="A">
<?php echo $form->labelEx($model,'selectionList'); ?>
<?php echo $form->textArea($model,'selectionList',array('rows'=>6, 'cols'=>50)); ?>
<?php echo $form->error($model,'selectionList'); ?>
</div>
<div id="B">
<?php echo $form->labelEx($model,'selectionList'); ?>
<?php echo $form->dropdownList($model,'selectionList',array('rows'=>6, 'cols'=>50)); ?>
<?php echo $form->error($model,'selectionList'); ?>
</div>
I have two radiobuttons say Single and Multi. When I select Single radiobutton, div A should be considered and div B discarded and when I select Multi, vice versa should happen.
So, is this the right way to change the form fields dynamically. Or else how can I do this using Ajax validation.
Not so related to question:
1) You are using same name for all fields, so only last one will be submitted (check generated html, name is MyModel[selectionList] and not MyModel[][selectionList]
2) You will have lots of problems with Ajax validation and dynamic fields (generated via js), since CActiveForm will generate js code via php.
Related to question:
3) To use one or another field, i suggest you to have two separate field names and just hide with js/css one, that is not needed right now. After submitting, just check value of radio button and decide what fields will be saved.
To validate these inputs, you have to use model scenarios (will define in rules what to validate and what not to):
$model = MyModel();
if ($_POST['C'] == 'multi') {
$model->scenario = 'validateOnlyFirstField';
} else {
$model->scenario = 'validateOnlySecondField';
}
Also you can use second CActiveForm.validate() parameter to define what attributes to validate.
I am trying to write a wordpress plugin and I have hit a bump. I am new to PHP (coded in Java before) and javascript so I am not sure whats the best way to solve my problem.
The Background
I have some data in a mySQL DB that I am using (each row has a unique ID and some other information I have added). I am able to search the DB using
$headss = $wpdb->get_results("SELECT * FROM {$wpdb->prefix}costumesdb WHERE location = 'head'", ARRAY_A);
And display some of the information to the user using (this is one of 5 different drop-downs but they are all created in the same way)
Head: <select name="head">
<?php foreach ($heads as $head) { ?>
<option value="<?php echo $head['pieceName'] ?>"><?php echo $head['shopName'] . " - " . $head['pieceName'] ?></option>
<?php } ?>
</select>
For the moment I want the user to be restricted to choosing information that is already in the system.
The problem
The DB contains 2 pieces of information that the user does not need to know to fill in the form (a website URL and a picture URL). I need these 2 pieces of information once the form is submitted (I need to write some more code for that) to the server which spits out another page with the 2 URL's in it.
Whats the best way to send the data back to a PHP script? Am I able to access the row of data that the user has selected in the drop down and send the unique ID for that row back or do I need to do something else?
Edit:
This is the script that I am using to submit the code:
$('#createacostume').form({
success:function(data){
$.messager.alert('Info', data, 'info');
}
});
'
And then the page to display the information returned is:
$cname = $_POST['cname'];
$head = $_POST['head'];
echo "Data Returned Name $cname head $head
I think this is what you are asking:
User has to choose an item from a drop down and submit a form. You have to display the website URL and the image for that item in a second page. You want to know how this is typically accomplished.
If that's the case, you should pass the row id of the item to the second page like so:
<option value="<?php echo $head['ROW_ID'] ?>"><?php echo $head['shopName'] . " - " . $head['pieceName'] ?></option>
Then use the ROW_ID in the second page to access the data from the database and print out the website URL and the image.
Submit the first form (without the two field), INSERT the data into the database, get the ID of insert.
Pass the ID to the next page which would set the ID into a hidden form field (or GET or POST parameter, plenty of choices) of the new form (with the two fields and just UPDATE the database upon submitting the second form.
If you like to show the original data in the second form, just pull the data from the database and use it to render the form instead of passing just the ID into a hidden field.
I need to create a drop-down menu in Codigniter and the values will pulled from a database table. The data contains a group of movies, I need to be able to populate the data into the menu.
How should I do this?
just iterate through your data and create a select list
<select name="movie_id">
<?php foreach ($rows as $row): ?>
<option value="<?= $row['id']; ?>"><?= $row['name']; ?></option>
<?php endforeach; ?>
</select>
This will obviously be different based on your table schema
According to Codeigniter documentation
see this url
Code Igniter - form_dropdown selecting correct value from the database
Codeigniter form_helper getting database rows to be values in select menu
The first parameter will contain the name of the field, the second parameter will contain an associative array of options, and the third parameter will contain the value you wish to be selected. You can also pass an array of multiple items through the third parameter, and CodeIgniter will create a multiple select for you.
Your admin controller should have something like
$data['selected'] = $this->salary_expectation->get_salary_selected();
According to this, the admin view should be like this
<?php echo form_dropdown('salaries', $salaries, $selected_value); ?>
I have two <select> one is category and the second is subcategory.
here is the first <select> for category.
<select name="category" size="10">
<?php foreach($categories->fetch(array('table' => 'categories')) as $category) { ?>
<option value="<?php echo $category['id']; ?>"><?php echo $category['name']; ?></option>
<?php } ?>
</select>
now the second <select> i.e subcategory should be hidden initially and when a user click on category <select> based on the value it should populate the value in subcategory.
One way of doing this is via AJAX by passing categoryId as POST Request and getting HTML as response.
however i would like to know if there is any other alternative so that it automatically pass the categoryId value to PHP and unhide the second <select> here is the code of second <select>
<select name="subcategory" size="10">
<?php foreach($categories->fetch(array('table' => 'subCategories', 'categoryId' => $categoryId)) as $subCategory) { ?>
<option value="1"><?php echo $subCategory['name']; ?></option>
<?php } ?>
</select>
the only thing i need here is $categoryId to be populated dynamically. is there any way of doing this?
thank you..
No, there is no way to do what you are suggesting. PHP is only run on the server, so by the time the page is rendered on the client the PHP has already been run.
Your best bet would be what you already suggested, running some AJAX after the first select is changed, sending back the category ID to the server and retrieving what you need to build the second select.
Is there a reason why you don't want to do it this way?
Sukumar has probably suggested the best and most intuitive solution to make it appear as if the data is being loaded dynamically to the user.
The other alternative would be to submit the form when the select box is changed. Once the form has been submitted PHP would pick up the ID from the POST array and then re-populate the sub-category select box. This is often used as a fallback in case the user doesn't have JavaScript enabled.
Structurally, there are three choices to solve this problem:
Use an ajax call to fetch the required data when a user selection is made as jbruno has described.
Submit the whole page to the server, let your PHP see the newly selected option and fill in the newly desired data in a returned page. This will cause the page to refresh so is less ideal than option 1.
Pre-populate the page with all possible data in a javascript data structure so you can use Javascript to just look up the desired category ID in a local data structure, modify the page and never have to talk to the server in order to update the page.
In my opinion, option 3) is the most desirable if the data set required for local lookup is not too large (say under 100k) and it's not too expensive on the server to collect all that data for inclusion in the original page and if the data doesn't change real-time or having data as of the page load time is OK.
If option 3) isn't feasible for any reason, then option 1) is next best. Option 2) is not as good a user experience so it should only be the last resort if you really can't implement options 1) or 3).
You asked more specifically about option 3. I don't really yet understand what the whole data you need looks like. If you really only have four total data types residential_plot, residential_apartment, office_space and showroom, then you can just make those be four keys on an object and store their data that way:
var data = {
"residential_plot": 1,
"residential_apartment": 2,
"office_space": 3,
"showroom": 4
};
The 1, 2, 3 and 4 are just whatever data you want to store for that type. It can be numbers, strings, arrays of data, other objects of data, anything.
To access this, you would do like this:
var id = data.residential_plot;
or
var index = "residential_plot";
var id = data[index];
If you wanted to store the notion of categories and sub-categories, you would need an extra level of objects:
var data = {
"residential": {"residential_plot": 1, "residential_apartment": 2},
"commercial": {"office_space": 3, "showroom": 4}
};
Then, you would access it like this:
var id = data.residential.residential_plot;
or like this:
var category = "residential";
var catType = "residential_plot";
var id = data[category][catType];