jQuery AJAX Call to PHP Script with String Return

Question

jQuery AJAX Call to PHP Script with String Return

8.7k views Asked by Razor At 30 December 2024 at 00:08

My aim here is to replicate the Hot Towel SPA framework from ASP.NET in PHP.

I have two files in my project: show.js and displaystring.php. What I basically want to do is return whatever is present in the displayString() function. But due to a series of unsuccessful attempts, I inserted the debugger statement in the shows.js file (as marked in the source code below) to find out what's wrong. When I run the project in Google Chrome and open the developer tools (Ctrl + Shift + i), I see that the program flow stops at the debugger statement (as expected) and when I hover my mouse over the data variable in the success property, I see that the displaystring.php file returns me the entire html source code instead of just returning the ".$_POST('string'); statement.

I have gone through many solutions on the internet but to no avail. Can anyone please tell me where exactly am I going wrong? Thank you.

show.js

define(function (require) {

    var val = "";
    //var moviesRepository = require("repositories/moviesRepository");

    $.ajax({
      url: "controllers/displaystring.php",
      type: "post",
      data: {input:"show"},
      datatype: 'json',
      success: function(data){
            val = data;      
        debugger; // Debugger inserted here.
      }});

    return {
        result: ko.observable(),

        activate: function() {
            this.result(val);
        }
    };
});

displaystring.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>String Display</title>
</head>

<body>
<?php
    function displayString() {
        return "The string passed is: ".$_POST('string');   
    }

    displayString();
?>
</body>
</html>

Original Q&A

There are 5 answers

Ankit Tyagi On 16 December 2013 at 11:59

Try this way

displaystring.php :

<?php
    function displayString() {
        return "The string passed is: ".$_POST['input'];   
    }
    echo displayString();
?>

Vijayakumar Selvaraj On 16 December 2013 at 12:04

Have only PHP script like @tyagi mentioned & change $_POST('string') to $_POST['string].

RDK On 16 December 2013 at 12:04

First of all you must specify headers:

header('Content-type: text/json');
header('Content-type: application/json');

Second you must to formmat your content for this kind of headers:

<?php
    function displayString() {
        return json_encode($_POST);   
    }

    echo displayString();
?>

And also you must to read official documentation:

You php file must be something like that:

<?php
     header('Content-type: text/json');
     function displayString() {
                return json_encode($_POST);   
            }

            echo displayString();

Arslan Bilal On 04 May 2014 at 15:37

Make Sure You didn't print the html in config file or constructor or something like that. I got the the same problem. Solution is to create separate file to handle the ajax request.

**Razor** · Accepted Answer · 2013-12-16T15:14:40+00:00

Okay after banging my head against a stone wall for the whole day, I figured out the solution myself.

I just modified the code a bit to serve the purpose of finding the answer to my question.

show.js

define(function (require) { 
    return {
        result: ko.observable(),

        activate: function() {
            //var moviesRepository = require("repositories/moviesRepository");
            var self = this;
            $.ajax({
                url: "controllers/displaystring.php",
                type: "get",
                data: {input:"show"},
                dataType: "html",
                success: function(data){
                    self.result(data);
                    debugger;
                }});
        }
    };
});

displaystring.php

<?php
    function display() {
        echo 'Hi I am some random '.rand().' output from the server.';
    }
    display();
    //echo "Hello World!";
?>

And that's it! You don't have to type the entire HTML code for the PHP file. Just include the PHP script and you will get the output.

Thank you all very much for your help. :)

TechQA.

jQuery AJAX Call to PHP Script with String Return

There are 5 answers

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