I wonder if it is possible to run a PHP file in the background on my website?
What I want to do is to show a form with inputs and dropdowns. The content of the dropdown is taken from a table in a MySQL database. Next to the dropdown is an Edit button which opens a Bootstrap Modal and shows all the content in the table for the dropdown. Here I can make changes to the table. Then I want to go back to the form and apply the changes I jsut made (i.e. select a new post from the dropdown).
I know how to do all of this - BUT - when I save the changes to the dropdown table, a php file is called to perform the SQL syntax and then redirects the user back to the form. Every input you have made in the form is lost since the page is reloaded.
Is there a way to make it work like I want?
I understand that you need to implement Ajax to do the background operation. Ajax request runs in background, its entry and exit is handled through Javascript functions . See here : http://www.w3schools.com/ajax/
As some already noted, Ajax is the way to go in this case.
An idea would be to use ajax to send the modal data to the server (saving the new options for the dropdown) when the user clicks in the "save" button (or anything equivalent). When this ajax call finish successfully, you could reload the dropdown with another ajax call (or even better, save the modal data in a temporary structure in JavaScript and use it to update the dropdown once the first ajax call is successful, hence avoiding some work on the server)
Note that the page won't reload in the whole process.
The simplest and the easiest way to do this is to open the modal in a new window.
example
Modal
And you can attached this to your select option value with a on-click event.
Related
That's the task I am currently facing:
Starting point: I have a form, which accepts data and saves it into the database.
What needs to be done: I need to include a button called "show data" (a different one than the "submit" button in the form, it can even be a clickable div), which will then display the data from the database on the same page, under the form. The button should also enable refreshing the list after new data was inserted into the form.
Now, if the data from the database had to be displayed on a different page, that could be handled easily with routing. However, for this particular task, I need to display the data on the same page.
My question: Is there a pure Laravel/PHP way to solve that problem? I am considering using AJAX (to which I am also new), but if there is some Laravel specific solution then I would prefer that.
The solution needs to work on localhost.
For PHP, without having any help from JavaScript, showing any changes to the database will require a browser refresh. On the front-end, you can use VueJS and Axios to help you with these requests, which do not require the browser to be refreshed.
I am trying to create a delete-button for each printed row in a php while loop that opens a modal window. The delete query will be executed from the modal.
How can I pass the corresponding row's id, title etc to the modal?
The way it is now, I only get the data from the first table row no matter what delete-button I press.
There are many solutions out there, but I can't make them work.
Hoping someone can help me understand this...
PHP is server side, so it really needs done in Javascript.
I created a Javascript lib if you are interested called bs-delete-confirm. you add a class to your delete button, and when you click it, it stops an href from leaving the page and brings up a Bootstrap modal asking if you are sure you wish to perform the action. Clicking ok follows the link, clicking cancel doesn't. Simple!.
Check it out here https://github.com/delboy1978uk/bs-delete-confirm
X
$(document).ready(function(){
$('.delete_class').deleteConfirm();
});
and that's it! you now have a delete confirm modal :-)
I tend to create a hidden modal in the bottom of my html and clone it into a new variable, then change the data for that new modal, append it and fire it up.
That way when you delete it or need to create a second one, you always have access to your template/blueprint.
It also allows you to keep a separate reference to every modal for later use.
eg $modal1.find('.inputEmail') and $modal2.find('.inputEmail')
You can even make that dynamic afterwards if you have many inputs
I am designing a web page. I want to have the options(of another drop-down list on same page) based on the selection made by the user.
I am providing a drop-down list to the user called Application. Depending on the Application, I want to query sql and want to show only the options that are valid for selected Application in another drop-down list.
I want to get the value of Application on the same page (one that user has selected) and by getting that value will query the sql accordingly.
Once the page is loaded, the PHP cannot do anything more.
You cannot use PHP based on user interaction.
The only way to do so is through javascript.
Using ajax, you can retrieve data from a certain URL on your javascript.
A procedure would be like this:
User selects something
The javascript(ajax) loads a URL
In that URL, use your PHP to fetch a certain query
The result would be sent to javascript, so display the result.
Here's a good page explaining ajax:
http://wabism.com/ajax-tutorial-with-jquery/
you have to call an ajax function on the onchange event of dropdown that gets data from database.
ajax is your friend. Either you can use raw javascript or jquery( easier ). Your steps would be..
1. Write a php code block, most probably a function, which takes the selected application option from post, queries the database and outputs the result.
2. In your current page, write a js code that fires the ajax request each time user selects an option. This ajax will send the selected option to the php script by post method and u can grab the output upon the success of the request.
$('.selectBox').on('change', function(){
// do your ajax here.
});
something like that..
I need to take form data from user input- i.e. radio buttons and use that to apply search filters to a database search.
However there are specific parameters which are giving me difficulty.
Specifically: The search filter options pane is a static fixture on the main page of the site. The query to be modified by the search filters is a separate php page which is called by an ajax function to display search results in the middle of the page without page refresh.
Is it even possible to submit variable values to another php page without going to that page and processing the php immediately? Or will the variables not be stored like that?
The code is too long but I'll give basic pseudocode:
Form action="Query Page to receive user input.php"
Some radio buttons:
20
15
10
Submit button--> Submits the radio button value to QueryPage.php but does not redirect
User clicks a category link (i.e. fitness) that calls the ajax function which displays the output of QueryPage.php. At this point QueryPage.php should perform the search with the specific user input filters that were selected earlier.
Is this possible?
Let me see if I understand correctly:
You're basically saying that your radio buttons will modify the search results, based on the what user selected?
If that's the case, I can think of 2 options:
1- When you make the ajax request for the search, first, grab the user input and send it to the QueryPage.php file with the search query. Do you have access to that function?
2- Post the user input using ajax (are you using jQuery or some other library for this?) to a UserInput.php file, where you'll store that data on the session, and then from QueryPage.php you just access the session and grab the values sent previously.
Does that answer your question? Sorry if it doens't, it's a bit hard to understand the problem.
You can use JQUERY and its events method. For example change,click,hover. In your case you want to use radio buttons, so you might wanna use the click event for that.
$("#radio").click(function () {
// SEND HTTP REQUEST
});
http://www.mkyong.com/jquery/how-to-select-a-radio-button-with-jquery/
Yes. You could just use .ajax(), or .post() or cURL. This will post your data to the specified page without redirecting.
Example using .post()
$.post("test.php", { name: "John", time: "2pm", fieldname: "your value" } );
I have a form on one page, some of the form data is pulled from various tables in my mysql database -- the tables hold the options for different selects, so the following is the way my data will look after the php scripts are run:
<select name="vehicleStyle">
<option value="empty" selected="selected">- select -</option>
<option value="coupe">Coupe</option>
etc....
</select>
I have a second form that loads over this page in an iframe, this 2nd form allows me to update the data included in this mysql table so that I can add options to the form on the fly. Everything works absolutely fine and wonderful however, if I update (add a value) to this table on the fly, I have to refresh the page in order for the new data to show up in the first form on the main page. I've played with auto-refreshing the page, but that's kind of obnoxious.
Is there a way to add this form data directly into my original page via jquery?
I think twitter does something like this when you add a tweet (I don't use twitter) but it adds your recent tweet onto the top of the page without actually reloading the entire page.
Thanks so much!
I believe the best way is to use the load jQuery function (http://api.jquery.com/load/).
It lets you load a page from a url directly from javascript.
Here is what I would do:
move the creation of the select to a new url (fillselect.php for exemple)
load the page without creating the select: create a div tag instead (name it divforselect for exemple)
in the onload javascript event of the page, write this:
$("#divforselect").load("fillselect.php") This will result in the div tag innerText set to the content of the fillselect.php page. So the url fillselect.php will not be a complete html page but will only contain the select tag.
After that you can call $("#divforselect").load("fillselect.php") whenever you want to refresh the select!
I think you want to do DOM manipulation using JS. All you have to do is append a new OPTION tag with the user-input value to the SELECT like:
var option = $('<OPTION/>');
$(option).html('INSERT YOUR VALUE HERE');
$('SELECT').append(option);
Remember: this is adding an OPTION just on the client side. So if your actual form submission (in your iframe) fails to push the data to your database, this would represent an inconsistent view for your user( where he would think new OPTION is created but its actually failed).
So, you would have to tie your UI update with your backend response. Typically, you would refresh this UI in the AJAX response of the call updating the backend.
Maybe you should be using AJAX to populate the boxes? Then you could have it refresh the data anytime you want.
jQuery Ajax