CODEWITHSUNDEEP
phpjson
JPK
JPK

Reputation: 491

JSON PHP data returning null

I am trying to fetch a PHP script using AJAX and return the values as JSON. For some reason, my script fails and I am trying to figure out the problem. When I enter a value from the database into the address bar, like: 

www.someaddress/post.php?kinaseEntry=aValue

I get a JSON output like so:

{"kinaseSKU":null,"url":null,"molecularWeight":null,"tracerSKU":null,"antiSKU1":"antiSKU1","antiSKU2":"antiSKU2","bufferSKU":"bufferSKU","tracerConc":null,"assayConc":null}

My PHP file looks like so:

<?php

//Include connection to database
require_once 'connect.php';

$kinase = mysql_real_escape_string ($_POST["kinaseEntry"]);

mysql_query('SET CHARACTER SET utf8'); 

$findKinase = "SELECT * FROM kbaData where cleanSKU = '" .$kinase. "' ";

if ($result = mysql_query($findKinase)) {

    $row = mysql_fetch_array($result, MYSQL_ASSOC);

    $kinaseSKU = $row['cleanSKU'];
    $url = $row['url'];
    $molecularWeight = $row['molecularWeight'];
    $tracerSKU = $row['tracerSKU'];
    $antiSKU1 = $row['antiSKU1'];
    $antiSKU2 = $row['antiSKU2'];
    $bufferSKU = $row['bufferSKU'];
    $tracerConc = $row['tracerConc'];
    $assayConc = $row['assayConc'];


    /* JSON ROW */
    $json = array ("kinaseSKU" => $kinaseSKU, "url" => $url, "molecularWeight" => $molecularWeight, "tracerSKU" => $tracerSKU, "antiSKU1" => $antiSKU1, "antiSKU2" => $antiSKU2, "bufferSKU" => $bufferSKU, "tracerConc" => $tracerConc, "assayConc" => $assayConc );

} else  {

    /* CATCH ANY ERRORS */
    $json = array('error' => 'Mysql Query Error');
}

/* SEND AS JSON */
header("Content-Type: application/json", true);

/* RETURN JSON */
echo json_encode($json);

/* STOP SCRIPT */
exit;
?>

Am I going about this the wrong way? Or have I done this wrong?

EDIT: Here is my jQuery/Ajax that calls the PHP script:

$(document).ready(function()    {

$('#kinaseEntry').change(function ()    {

    var kinaseEntry = $('#kinaseEntry').val();
    var dataString = 'kinaseEntry' + kinaseEntry;

    $('#waiting').show(500);
    $('#message').hide(0);
    alert(kinaseEntry);

    //Fetch list from database
    $.ajax({
        type : "POST",
        url : "post.php",
        datatype: "json",
        data:   dataString,
        success : function(datas)   {
            alert("datas" + datas);
        },
        error : function(error) {
            alert("Oops, there was an error!");
        }
    });
    return false;
});
});

Upvotes: 1

Views: 1894

Answers (4)

pyronaur
pyronaur

Reputation: 3545

Yeah, you're getting back an array of rows, so that doesn't really work as you'd expect it to. This should fix it, but I don't think it is the best approach (but I think it should work anyway). 

$kinase = mysql_real_escape_string ($_POST["kinaseEntry"]);

mysql_query('SET CHARACTER SET utf8'); 

$findKinase = "SELECT * FROM kbaData where cleanSKU = '" .$kinase. "' , LIMIT 0,1";

if ($result = mysql_query($findKinase)) {

    $row = mysql_fetch_array($result, MYSQL_ASSOC);

    $kinaseSKU = $row[0]['cleanSKU'];
    $url = $row[0]['url'];
    $molecularWeight = $row[0]['molecularWeight'];
    $tracerSKU = $row[0]['tracerSKU'];
    $antiSKU1 = $row[0]['antiSKU1'];
    $antiSKU2 = $row[0]['antiSKU2'];
    $bufferSKU = $row[0]['bufferSKU'];
    $tracerConc = $row[0]['tracerConc'];
    $assayConc = $row[0]['assayConc'];


    /* JSON ROW */
    $json = array ("kinaseSKU" => $kinaseSKU, "url" => $url, "molecularWeight" => $molecularWeight, "tracerSKU" => $tracerSKU, "antiSKU1" => $antiSKU1, "antiSKU2" => $antiSKU2, "bufferSKU" => $bufferSKU, "tracerConc" => $tracerConc, "assayConc" => $assayConc );

} else  {

    /* CATCH ANY ERRORS */
    $json = array('error' => 'Mysql Query Error');
}

/* SEND AS JSON */
header("Content-Type: application/json", true);

/* RETURN JSON */
echo json_encode($json);

/* STOP SCRIPT */
exit;

Upvotes: 0

Mariusz Jamro
Mariusz Jamro

Reputation: 31692

The mysql_query function returns a true value even if 0 rows are returned. It returns false only when there was a database error. So i believe it executes a query and there are no results, so there are null values in the JSON. Try to use mysql_num_rows:

$findKinase = "SELECT * FROM kbaData where cleanSKU = '" .$kinase. "' ";

if ($result = mysql_query($findKinase)) {
      $json = array('error' => 'Mysql Query Error');
}else{
    $num_rows = mysql_num_rows($result);
    if($num_rows > 0){
         // Do sth with the results
    }else{
         $json = array('error' => 'No results');
    }
}

Upvotes: 1

Joshua
Joshua

Reputation: 543

First I am not sure if if (isset($_POST['kinaseEntry'])) will work for what you have shown. The URL you have shown is a get request, so if you want access to that variable you will have to use $_GET['kinaseEntry']. If you need to do POST change the method attribute in your form to <form method="POST"> that will give you the post variable.

Upvotes: 0

WWW
WWW

Reputation: 9870

Your problem appears to be that there wasn't actually a row returned from your query. Try:

$result = mysql_query($findKinase);
if ($result !== false && mysql_num_rows($result) > 0) {
    // your code
}

$result is false in this case if there was an error in the query. If there was no error, then check to see if there were actually rows returned. If there were more than 0 rows returned, then attempt to extract data.

Upvotes: 0

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