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I am using AJAX and trying to return variables that I got after querying the database in the modcomp.php file, I am then trying to input those values back into my main site to put into input boxes. Below is what I have tried but it is not working. Is this because I am getting variables in PHP and then bringing them back to use in JS / jQuery? Doing a few searches it looks like JSON might be the answer but looking at JSON I havent seen an example of how I could run code to query the database and then put that info in JSON format to pull back? My goal is to eventually pull back all 10 variables.

                $.ajax({
                method: "POST",
                url: "modcomp.php",
                data: {item: $(this).val()}
                success: function(data) {
                    $('#itemnumber').val($itemnumber);
                    $('#cost').val($cost);

                }
            });

The modcomp.php file

<?php

if(array_key_exists("item", $_POST)) {

    include("connection.php");

    $query="SELECT * FROM components WHERE itemname = '".mysqli_real_escape_string($link,$_POST['item'])."' LIMIT 1";

    if ($result=mysqli_query($link,$query)) {

        $row=mysqli_fetch_array($result);

        $id=$row[0];
        $itemnumber=$row[1];
        $itemname=$row[2];
        $cost=$row[3];
        $company=$row[4];
        $contact=$row[5];
        $address1=$row[6];
        $address2=$row[7];
        $phone=$row[8];
        $part=$row[9];

        //print_r ($id." ".$itemnumber." ".$itemname." ".$cost." ".$company." ".$contact." ".$address1." ".$address2." ".$phone." ".$part);

    } else {

        print_r("Issue with query");
    }

}

?>

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8
  • You need to json_encode your PHP array and echo it out. Then you can decode it on the JS side of things. Commented Nov 1, 2016 at 17:31
  • print(json_encode($row)) Commented Nov 1, 2016 at 17:31
  • 1
    Use mysql_fetch_assoc() so you get an associative array, then it will be turned into a Javascript object when you use JSON. Commented Nov 1, 2016 at 17:33
  • @Barmar ok so I change $row=mysqli_fetch_array($result); to $row=mysqli_fetch_assoc($result); then I change my output to echo json_encode($row); and then in my ajax coding under success I data=json.parse(data); but then how do I get my information out of data? Commented Nov 1, 2016 at 19:21
  • data.itemnumber, data.itemname, etc. Commented Nov 1, 2016 at 19:23

3 Answers 3

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2

The easiest way to do that is just set your jquery.ajax to expect a json as return:

 $.ajax({
     method: "POST",
     dataType: "JSON",
     /** others attributes **/

After, convert your return to a json and print it (just it, nothing more) at php script:

 //a better approach is return the column name (fetch_assoc)
 echo json_encode($row);

Now, your return can be used as json:

 success: function(data) {
     data.column_name
 }
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13 Comments

With this, I would need to decode in success before using the data correct? I tried this method with the fetch_assoc and for some reason its not working for me. I even tried console.log(data); to view whats going on and the variable isnt even showing up in colsole log, that means im not hitting success function at all then?
@Brandon no, you didn't. As ajax expect to receive a JSON data type, It'll be already a JSON (just have to print only the json at php script). After that, use as a JSON. If you didn't see the console at success event, use the error to see what is happening
I am getting a little closer to the issue. Apparently its in the success line. Im not sure what it is though? jsbin.com/rodaxuduke/edit?html
I figured out that it was a coma missing after the data line that was causing the error on the success line, but now im getting an undefined error for my console.log(data); line, why would that be?
AFAIK the T from dataType must be uppercase.
|
2

json_encode is the answer 

<?php

if(array_key_exists("item", $_POST)) {

    include("connection.php");

    $query="SELECT * FROM components WHERE itemname = '".mysqli_real_escape_string($link,$_POST['item'])."' LIMIT 1";

    if ($result=mysqli_query($link,$query)) {

        $row=mysqli_fetch_array($result);
        $array = ();
        $array['id']=$row[0];
        $array['itemnumber']=$row[1];
        $array['itemname']=$row[2];
        .
        .
        $array['part']=$row[9];
        $array['status'] = true;
       echo json_encode($array);

    } else {

        echo json_encode(array(status => false , msg => "Issue with query");
    }

}

Then in your js code use json as 

 $.ajax({
                method: "POST",
                url: "modcomp.php",
                data: {item: $(this).val()}
                success: function(data) {
                    data = JSON.parse(data); // to parse json string to object

                    if(data.status){
                     $('#itemnumber').val(data.itemnumber);
                     $('#cost').val(data.cost);
                    }

                }
            });
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8 Comments

for some reason $('#itemnumber').val(data.itemnumber); isnt working for me. I checked with console log to make sure json is being returned properly and double checked the spelling and capitalization of both the id on the input form and on the database column name. I am out of ideas on how to get this working, do you have any suggestions?
{"id":"13","itemnumber":"A26587-A1","itemname":"HMO Seasoning","cost":"$3.29","company":"Elite Spice","contact":"Joe Friert","address1":"4593 Tapica Rd","address2":"Harrisburg, PA. 17984","phone":"717-423-7698","part":""}
jsfiddle.net/axcl/k3nLgbeL/2 check this fiddle if you are trying to set value of input box .val() should work or may be there are more than one input boxes with same id that you are using plz check it once. if issue still persist i would like to check your HTML markup
I dont see the same id being used any more than 1 time. Here is a copy/paste of the view source of the page. jsbin.com/lozibidoti/edit?html
jsbin.com/vekayapeke/1/edit?html,css,js,output your code seems fine value of itemnumber is being set by the same code .. are you able set values of rest of the input boxes?
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1

you should return your value from your php file as

<?php
    if(array_key_exists("item", $_POST)) {
    include("connection.php");
    $query="SELECT * FROM components WHERE itemname = '".mysqli_real_escape_string($link,$_POST['item'])."' LIMIT 1";
    if ($result=mysqli_query($link,$query)) {
        $row=mysqli_fetch_array($result);
        return json_encode($row,true);
   } else {
        return json_encode("Some error occured");
   }
}

don't use print_r() function to return data from your php file

And set the datatype in ajax as JSON to parse the json data to be returned as

$.ajax({
   method: "POST",
   url: "modcomp.php",
   datatype:'json',
   data: {item: $(this).val()}
   success: function(data) {
      $('#itemnumber').val(data.itemnumber);
      $('#cost').val(data.cost);       
   }
});

Now you are able to use your json data using the dot(.) operator like

$('#cost').val(data.cost);
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1 Comment

I have done all of the above, checked through console to make sure the json came through and is usable and it is all there and correct but for some reason $('#itemnumber').val(data.itemnumber); is not working for me. I also double checked the spelling and capitalization to the column header of the database and the same for the id of the form input. No errors are being thrown at all when I inspect via the console either. Ignore the output window but here is my code to give you a better idea. jsbin.com/zabuwuwigi/edit?html

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