How to properly use JSON to get number of rows from MYSQL database from php file?

Question

I am having difficulty with properly setting up JSON post to get number of rows from MYSQL database from a php file. I am getting "undefined" alert, when I am looking to alert the number of rows as an integer. I have been using a different stackoverflow post to try to build this Get variable from PHP file using JQuery/AJAX.

Here is the ajax call:

// check number of records in Mine
$.ajax({
    url: 'pyrAddToMine.php',
    type: 'POST',
    success : function (result) {
    alert(result['ajax']); // "Hello world!" alerted
    console.log(result['numRec']) // The value of your php $row['numRec'] will be displayed
    },
    error : function () {
        alert("error");
    }
});

Here is the php code in pyrAddToMine.php: mysql_select_db($database_cms_test, $cms_test); $query = "SELECT * FROM $favUserTableName"; $result = mysql_query($query) or die(); $row = mysql_fetch_array($result); $num_records = mysql_numrows($result);

IF ($num_records >= 15){
    $numRec = array(
        'ajax' => 'Hello world!',
        'numRec' => $num_records,
            );
    echo json_encode($numRec);
    exit;
}

Here are more details on the php file:

<?php
require_once('../Connections/cms_test2.php');

...

mysql_select_db($database_cms_test, $cms_test);
$query = "SELECT * FROM `$favUserTableName`";
$result = mysql_query($query) or die();
$row = mysql_fetch_array($result);
$num_records = mysql_numrows($result);

IF ($num_records >= 15){

    $numRec = array(
        'ajax' => 'Hello world!',
        'numRec' => $num_records,
    );
    echo json_encode($numRec);

    exit;
}
...

?>

Mihai Vilcu · Accepted Answer · 2013-06-08 04:29:03Z

1

Hello you forgot to specify the datatype

   $.ajax({
        url : 'myAjaxFile.php',
        type : 'POST',
        data : data,
        dataType : 'json',
        success : function (result) {
           alert(result['ajax']); // "Hello world!" alerted
           console.log(result['advert']) // The value of your php $row['adverts'] will be displayed
        },
        error : function () {
           alert("error");
        }
    })

answered Jun 8, 2013 at 4:29

Mihai Vilcu

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8 Comments

Kylie Over a year ago

Just a question...I thought jquery tried to do this automatically if you leave it out....doesnt it?

Mihai Vilcu Over a year ago

in the docs it does say dataType (default: Intelligent Guess (xml, json, script, or html)) but just to be sure i would specify it

om123 Over a year ago

if I include the data and datatype per above, I get "data is not defined" error in firebug

Mihai Vilcu Over a year ago

@om123 are you sure you pointed to the correct json data source ?

om123 Over a year ago

Hello @ionutvmi, I think you may be onto something since when I extracted the above code into standalone test files, Hello World worked. But when it is in the main code not working. The URL pyrAddToMine.php is definitely correct. If you have suggestions please let me know.

|

iman · Accepted Answer · 2013-06-08 04:37:03Z

1

if you do not set required options for accepting json, you recieve an string that contain ajax format and you did not recieve an object in javascript

$.ajax({
    url: 'pyrAddToMine.php',
    type: 'POST',
    /* required for accept json for ajax */
    accepts:'application/json',
    dataType:'json',
    /* */
    success : function (result) {
    alert(result['ajax']); // "Hello world!" alerted
    console.log(result['numRec']) // The value of your php $row['numRec'] will be displayed
    },
    error : function () {
        alert("error");
    }
});

answered Jun 8, 2013 at 4:37

iman

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2 Comments

om123 Over a year ago

Hi, this returns error alert. Firebug says JSON.parse: unexpexted character

iman Over a year ago

@om123 use this one and say what is your php returned value to javascript $numRec = (object)array( 'ajax' => 'Hello world!', 'numRec' => $num_records );

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