Leveraging AJAX for sending variables to PHP and fetching them through AJAX once more

Question

Leveraging AJAX for sending variables to PHP and fetching them through AJAX once more

I need to pass values to a PHP script using AJAX for passing and retrieving those values. However, I am facing an issue where the second AJAX call is not able to retrieve any value from the PHP file. Are there any reasons why this might be happening? How can I ensure that the variable passed to the PHP script is stored correctly so that it can be retrieved by the second AJAX call?

Below is the code snippet I am working with:

AJAX CODE:

$(document).ready(function() {    
    $("#raaagh").click(function(){    
        $.ajax({
            url: 'ajax.php', 
            type: "POST",
            data: ({name: 145}),
            success: function(data){
                console.log(data);
            }
        });  
        $.ajax({
            url:'ajax.php',
            data:"",
            dataType:'json',
            success:function(data1){
                var y1=data1;
                console.log(data1);
            }
        });
    });
});

PHP CODE:

<?php
$userAnswer = $_POST['name'];    
echo json_encode($userAnswer);    
?>

php jquery ajax

Answer 1

Answer №1

To retrieve JSON data, use dataType:"json"

$.ajax({
     url: 'ajax.php', // Refers to the current document
     type: "POST",
     dataType:'json', // Specify json datatype for receiving json
     data: ({name: 145}),
     success: function(data){
         console.log(data);
     }
});

For more information, refer to the documentation at http://api.jquery.com/jQuery.ajax/

In addition, here is how it can be implemented in PHP

<?php
  $userAnswer = $_POST['name']; 
  $sql="SELECT * FROM <tablename> where color='".$userAnswer."'" ;
  $result=mysql_query($sql);
  $row=mysql_fetch_array($result);
  // Assuming there is data in the table and only fetching the first row
  echo json_encode($row);  // Converting array into JSON format  
?>

Answer 2

To retrieve JSON data, use dataType:"json"

$.ajax({
     url: 'ajax.php', // Refers to the current document
     type: "POST",
     dataType:'json', // Specify json datatype for receiving json
     data: ({name: 145}),
     success: function(data){
         console.log(data);
     }
});

For more information, refer to the documentation at http://api.jquery.com/jQuery.ajax/

In addition, here is how it can be implemented in PHP

<?php
  $userAnswer = $_POST['name']; 
  $sql="SELECT * FROM <tablename> where color='".$userAnswer."'" ;
  $result=mysql_query($sql);
  $row=mysql_fetch_array($result);
  // Assuming there is data in the table and only fetching the first row
  echo json_encode($row);  // Converting array into JSON format  
?>

Answer 3

Answer №2

Avoid the need for a second ajax function by retrieving data back on success within one function. It can be challenging to determine when the first ajax call is complete, and using SESSION may not guarantee getting the data in the second AJAX call.

Therefore, it is advisable to make only one AJAX call and retrieve the value upon successful completion.

For example, during the first ajax call:

    $.ajax({
        url: 'ajax.php',
        type: "POST",
        data: ({name: 145}),
        success: function(data){
            console.log(data);
            alert(data);
            // If the data is JSON
            var jdata = jQuery.parseJSON(data);
        }
    });

Answer 4

Avoid the need for a second ajax function by retrieving data back on success within one function. It can be challenging to determine when the first ajax call is complete, and using SESSION may not guarantee getting the data in the second AJAX call.

Therefore, it is advisable to make only one AJAX call and retrieve the value upon successful completion.

For example, during the first ajax call:

    $.ajax({
        url: 'ajax.php',
        type: "POST",
        data: ({name: 145}),
        success: function(data){
            console.log(data);
            alert(data);
            // If the data is JSON
            var jdata = jQuery.parseJSON(data);
        }
    });

Answer 5

Answer №3

$(document).ready(function() {
    $("#clickme").on("click", function() {
        $.ajax({
            url: 'fetch_data.php',
            type: "POST",
            data: ({id: 145}),
            success: function(response) {
                console.log(response);
                $.ajax({
                    url:'process_data.php',
                    data: response,
                    dataType:'json',
                    success:function(newData) {
                        var result=newData;
                        console.log(newData);
                    }
                });
            }
        });
    });
});

To use this code, first make an ajax call to fetch some data from the server. Your PHP function will return the data which you will receive in the response variable. Then pass that data to a new ajax call for further processing.

Answer 6

$(document).ready(function() {
    $("#clickme").on("click", function() {
        $.ajax({
            url: 'fetch_data.php',
            type: "POST",
            data: ({id: 145}),
            success: function(response) {
                console.log(response);
                $.ajax({
                    url:'process_data.php',
                    data: response,
                    dataType:'json',
                    success:function(newData) {
                        var result=newData;
                        console.log(newData);
                    }
                });
            }
        });
    });
});

To use this code, first make an ajax call to fetch some data from the server. Your PHP function will return the data which you will receive in the response variable. Then pass that data to a new ajax call for further processing.

Answer 7

Answer №4

Within your PHP script, be on the lookout for a variable named $_REQUEST. This variable holds an array that includes all the information sent from JavaScript to PHP via AJAX.

To confirm receipt of the data, try running: var_dump($_REQUEST);

Answer 8

Within your PHP script, be on the lookout for a variable named $_REQUEST. This variable holds an array that includes all the information sent from JavaScript to PHP via AJAX.

To confirm receipt of the data, try running: var_dump($_REQUEST);

Answer 9

Answer №5

Make sure to use single quotes when passing values

$(document).ready(function() {    
    $("#clickme").click(function(){    
        $.ajax({
            url: 'ajax.php', //This is the current doc
            type: "POST",
            data: ({name: '145'}), //values should be passed like this
            success: function(data){
                console.log(data);
                           }
        });  
        $.ajax({
    url:'ajax.php',
    data:"",
    dataType:'json',
    success:function(data1){
            var y1=data1;
            console.log(data1);
            }
        });

    });
});

Give it a try, it might just work.......

Answer 10

Make sure to use single quotes when passing values

$(document).ready(function() {    
    $("#clickme").click(function(){    
        $.ajax({
            url: 'ajax.php', //This is the current doc
            type: "POST",
            data: ({name: '145'}), //values should be passed like this
            success: function(data){
                console.log(data);
                           }
        });  
        $.ajax({
    url:'ajax.php',
    data:"",
    dataType:'json',
    success:function(data1){
            var y1=data1;
            console.log(data1);
            }
        });

    });
});

Give it a try, it might just work.......

Leveraging AJAX for sending variables to PHP and fetching them through AJAX once more

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Similar questions

Obtain the key's value from a JSON object that is nested within another

What is the best way to fetch images from a JSON object in a React application?

Invalid JSON Primitive encountered when making an AJAX call using jQuery

Establish a connection to couchDB using JavaScript on the server side

Error in syntax for jQuery's .find() method

Handling Errors in Node.js with Express

Creating a notification for specific choices in a dropdown menu

I have a json file that I need to convert to a csv format

Enhancing PDF templates in SugarCRM with associated fields using PDF Manager

What is the best way to populate a JSP session with a jQuery value?

Pop-up windows, the modern day digital version of fortune cookies

Prevent Cross-Site Scripting attacks in Laravel by sanitizing user input, even when allowing HTML tags

Encountering a Laravel Axios error when using Vue: status code 419 issue

Each row should contain an editing button that executes specific code based on the device's name

Leveraging the orWhere clause in Eloquent

Is there a way to identify if you are at the bottom row of a column defined by CSS styling?

Utilizing JQuery in Wordpress to manipulate content within dynamically generated AJAX divs

Creating a single object from an array using Zend_Soap_Client

Eliminating .php from the URL and the GET request

Ways to limit access to the jQuery AJAX method from the browser console window