Make a call from a PHP page in javascript

Question

0.00/5 (No votes)

See more:

I must call a PHP page with a Javascript command.

I developed a function that does almost exactly what I needed:

View code: Example 1

But in the last part where it makes me "view" (document.getElementById('showResult')) only the result, I need to do it in a way that instead calls me another PHP page, where it will execute the instructions passed in this page.

What I have tried:

JavaScript

//Example 1

function getCheckedValues() 
			{
				return Array.from(document.querySelectorAll('input[type="checkbox"]'))
						.filter((checkbox) => checkbox.checked)
						.map((checkbox) => checkbox.value);
				}

				const resultEl = document.getElementById('result');

						document.getElementById('showResult').addEventListener('click', () =>{
						resultEl.innerHTML = getCheckedValues();

});

PHP

//Local where to call via the function:


'Location: ../backofficeNew/Tests/controller/sendToPhysicianSampleID.php?idList=BIY53423' 


//Button

<button id="showResult">SEND</button>
    <br><br>
	<div id="result"></div>

Posted 8-Dec-21 23:48pm

iMati

Updated 10-Dec-21 2:54am

v14

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

iMati · Accepted Answer · 2021-12-10T02:54:00

Dear All,

After some tests I came up with a solution I needed. I will leave her with you.

The part that returns the values that were selected in the checkboxes, I kept.

JavaScript

function getCheckedValues() 
			{
				return Array.from(document.querySelectorAll('input[type="checkbox"]'))
						.filter((checkbox) => checkbox.checked)
						.map((checkbox) => checkbox.value);
				}

What I would have to do: It was to call another PHP page through Javascript.
So I developed:

JavaScript

function sendToPhysician() {
	if (getCheckedValues() == "") {
      alert('Please, select at least one row');
	}
	else {
	  window.location.assign("./../Tests/controller/sendToPhysicianSampleID.php?idList=" + getCheckedValues());	
	}  
}

The alert I created at the last minute, considering that when pressing the button without selecting any of the options, some people could get lost, not knowing if the data was sent or not.