How to return data after submitting using PHP and mysql

Question

i need one help that is returning data just after submit inside the database.I am explaining my code below.

addCourse.php:

<?php
$course_name=stripslashes($_POST['course_name']);
$course_short_name=stripslashes($_POST['course_short_name']);
$semester=stripslashes($_POST['semester']);
$con = mysql_connect('localhost', 'root', '******');
mysql_select_db('go_fasto', $con);
$qry ='INSERT INTO db_course (course_name,short_name,semester) values ("' . $course_name . '","' . $course_short_name . '","' . $semester . '")';
$qry_res = mysql_query($qry);
if ($qry_res) {
        echo "Course has added successfully";
    } else {
        echo "course could not added ";
    }
?>

This file is called using ajax and all returning value will appear in ajax's success function.Here i need to return both the submitted data and success message to ajax's success function.Please help me to resolve this issue.

Return a json array with a response and data array contained within — Rasclatt
– Rasclatt, Commented Sep 29, 2015 at 5:00

Happy Coding · Accepted Answer · 2015-09-29 05:04:58Z

1

You can use json_encode to achieve the expected output.

    $result['course_name'] = $course_name;
    $result['course_short_name'] = $course_short_name;
    $result['semester'] = $semester;
    if ($qry_res) {
    $result['message'] = 'Course has added successfully';
    } else {
        $result['message'] = 'course could not added';
    }

echo json_encode($result);

You can retrieve all the submitted data along with the message.

answered Sep 29, 2015 at 5:04

Happy Coding

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6 Comments

satya Over a year ago

@ raveenanigam : Yes, your answer gave the solution but here one small problem,i need to return data from database directly because i will append those data on view page for display purpose so the database column name and here the variable are diffrent.

Happy Coding Over a year ago

@satya you just want to return the single record which is currently inserted

Happy Coding Over a year ago

@satya...$result['course_name'], $result['short_name'],$result['semester'] no issues...just change the indexes

satya Over a year ago

because i also need the id of each column that is the reason why i want return value from database.

Happy Coding Over a year ago

append those id's in $result array

|

Suman Singh · Accepted Answer · 2015-09-29 05:03:03Z

0

use below code in you ajax script:

$.ajax({
    type: 'POST',
    url: ajaxURL,
    data: datastring,
    dataType: 'json',
    success: function(data){
        console.log(data);
        alert(data.msg)
    }
});

and PHP file like:

<?php
$course_name=stripslashes($_POST['course_name']);
$course_short_name=stripslashes($_POST['course_short_name']);
$semester=stripslashes($_POST['semester']);
$con = mysql_connect('localhost', 'root', 'Oditek123@');
mysql_select_db('go_fasto', $con);
$qry ='INSERT INTO db_course (course_name,short_name,semester) values ("' . $course_name . '","' . $course_short_name . '","' . $semester . '")';
$qry_res = mysql_query($qry);
$result = $_POST;
if ($qry_res) {
        $result['msg'] = "Course has added successfully";
    } else {
         $result['msg'] =  "course could not added ";
    }
return $result;
?>

answered Sep 29, 2015 at 5:03

Suman Singh

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2 Comments

satya Over a year ago

I tried your answer but in console i did not get any value.It is not returning any value.

Suman Singh Over a year ago

use the echo json_encode($result); instead of return $result;

Suchit kumar · Accepted Answer · 2015-09-29 05:26:24Z

0

try using this:

 $result['postValue'] = $_POST;
    if ($qry_res) {
            $result['msg'] = "Course has added successfully";
        } else {
             $result['msg'] =  "course could not added ";
        }
    echo  json_encode($result);

////////////////////in script//////////////

$.ajax({
    type: 'POST',
    url: ajaxURL,
    data: datastring,
    success: function(data){
        var data =JSON.parse(data);
        console.log(data.postValue);
        alert(data.msg)
    }
});

answered Sep 29, 2015 at 5:26

Suchit kumar

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