Can anyone help me what is wrong in this code ajax using with php?

Question

This is the code which inserts data into the database. and the result is sent back to the ajax code.

<?php

if (isset($_FILES['files']['name'])) {
    $files = $_FILES['files']['name'];
    $desc = $_POST['description'];
    $subject = $_POST['subject'];
    $path = 'Lectures/' . $files;
    move_uploaded_file($_FILES["files"]["tmp_name"], $path);
    $date = date('d-M-y');

    $query = "INSERT INTO content(file_name,course_code,description,file_path,upload_date) VALUES ('$files','$subject','$desc','$path','$date')";

    $cm = sqlsrv_query($conn, $query);
    if ($cm) {
        $result['status'] = "Succsess";
    } else {
        $result['status'] = "failed";
    }
    echo json_encode($result);
}

And this is the Ajax Success: function.it executes only the else portion even if the condition is true.

success: function(output) {
    alert(output);
    if(output.status == "Succsess")
    { 
        alertify.success('Success message');
    }else{
        alertify.set('notifier','delay', 2);
        alertify.set('notifier','position', 'top-right');
        alertify.error('Error message');
    }
    readRecords();
    $('#form1').trigger("reset");
}

`output.status == "Succsess"` Just an idea, proper spelling can help reduce bugs and make code easier to understand. Have you checked what `output` contains in the `success` handler? — CertainPerformance, Oct 22 '18 at 07:46
Check the request in the console. From the logic it's not hard to work out that `output.status` is not equal to `Succsess`. You need to determine exactly what the value is. — Rory McCrossan, Oct 22 '18 at 07:47
you're also open to SQL injection - should look to resolve asap — treyBake, Oct 22 '18 at 07:47
@Ingus — Is case sensitive, but the case matches throughout the code. — Quentin, Oct 22 '18 at 07:48
in your php, try creating an array before appending status. meaning `$result = array();` — Rotimi, Oct 22 '18 at 07:50
did you have `dataType :json` in your `ajax` code? if not then you have to parse the coming response first and then only you are able to compare — Alive to die - Anant, Oct 22 '18 at 07:51
one important question. Does the insert statement actually succeed? — Rotimi, Oct 22 '18 at 07:52
@Akintunde-Rotimi — No need. PHP will implicitly create arrays. https://i.imgur.com/fRbxsFu.png — Quentin, Oct 22 '18 at 07:52
@AlivetoDie jquery will automagically determine the returning type - so you don't *have* to have dataType:json, *but* if you do have dataType:somethingelse then it will fail. — freedomn-m, Oct 22 '18 at 07:52
@freedomn-m — It's not magic, it is based on the content-type response header, which isn't being set in the code we've been shown. (It might be set elsewhere). — Quentin, Oct 22 '18 at 07:53
@Quentin I didn't say it was "magic"...you've just expanded on what "automagically" means :) — freedomn-m, Oct 22 '18 at 07:54
@Quentin alert(output); shows the success message if the data is inserted. but the condition didn't work — Waqas Insafian, Oct 22 '18 at 07:56
@WaqasInsafian You should try to answer the questions that are asked in the comments instead of commenting the same that you already have in the question. — jeroen, Oct 22 '18 at 07:58
But... what? There's no useful information we can use to help you here. — Rory McCrossan, Oct 22 '18 at 08:17

score 1 · Answer 1 · answered Oct 22 '18 at 08:02

alert(output); shows the success message if the data is inserted.

It should show [object Object].

You are looking at a string, and not an object parsed from JSON.

Since it is a string, it doesn't have a status property.

const output = '{ "status": "Succsess" }';
alert(output);
alert(output.status);

You need to tell jQuery that the response is JSON so it will parse it into an object.

header("Content-Type: application/json");
echo json_encode($result);

Note that you must set the headers before sending any other output.

Without setting the Content-Type header explicitly, PHP will default to claiming that the JSON is HTML … which it isn't.

Note that the below is for the sake of example. You should not add JSON.parse to your client-side code. jQuery will do that behind the scenes.

const output = JSON.parse('{ "status": "Succsess" }');
alert(output);
alert(output.status);

Can anyone help me what is wrong in this code ajax using with php?

1 Answers1