How can I select data from a json object in mysql/php?

Question

I'm sending JSON to my server via jQuery's ajax method. Here's my jquery, and im pretty sure this is fine:

function stuffs() {
    this.fname = document.getElementById("fname").value;
    this.lname = document.getElementById("lname").value;
    this.email = document.getElementById("email").value;    
}
$(function() {
function ajaxhelper(data){
    //console.log(JSON.stringify(data));
    //this returns "{"fname":"mike","lname":"smith","email":"[email protected]"}" which is what i expect
        $.ajax({
        url: 'postdb.php',
        type: 'POST',
        data: {data : JSON.stringify(data)},
        success: function(data) {
            console.log(JSON.stringify(data));
            console.log("Success");
        },
        error: function(e) {
            console.log(e);
        }

    });
}
$("form").on('submit', function(e){
    e.preventDefault();
    var data = new stuffs();
    ajaxhelper(data);
    //window.location = "postdb.php";
    });
    });
</script>

I get back an 500 server error. Here's my php code. (yes i'm only sending it an fname that i've preloading into my database, $con is valid i just didnt share the code to connect to my database)

$obj = json_decode($_POST['data']);

$sql = "SELECT * FROM testtable WHERE fname = \"$obj->{'fname'}\"";
$query = $con->query($sql);

I think my sql is incorrect due to the quotes? This is where im stuck.

Please don't directly query user input and try $sql = "SELECT * FROM testtable WHERE fname = '{$obj->fname}'"; Otherwise please check error logs. — Matt
– Matt, Commented Jan 15, 2016 at 23:13
That didn't work. How can i view the php page with data sent over? with ajax I just stay on one page, otherwise I would do like print_r($_POST) on the php page to see what I'm getting sent. — Taylor's Designs
– Taylor's Designs, Commented Jan 15, 2016 at 23:20
Andreas, what do you mean? just have $obj = $_POST['data'], same sql tho? — Taylor's Designs
– Taylor's Designs, Commented Jan 15, 2016 at 23:22

Ajin · Accepted Answer · 2016-01-15 23:13:23Z

2

Try using $obj->fname instead of $obj->{'fname'}.

answered Jan 15, 2016 at 23:13

Ajin

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3 Comments

Ajin Over a year ago

Is your DB working properly? Try calling a query for SELECT that you try to run, but with data written into the query instead of data from the JSON.

Ajin Over a year ago

Also, you stuffs() function doesn't seem to return any value, try changing it to something like that:

function stuffs() {     var data = {         fname: document.getElementById("fname").value,         lname: document.getElementById("lname").value,         email: document.getElementById("email").value     };          return data; }

Taylor's Designs Over a year ago

YES!!! THANK YOU SO MUCH! So the database was broken. permission issues, blah blah blah. Thank you for bringing up that i should just hard code something to test it!!

darrylp · Accepted Answer · 2016-01-15 23:16:01Z

0

Here is the correct syntax for your SQL Statement

$sql = "SELECT * FROM testtable WHERE fname = '".$obj->fname."'";

answered Jan 15, 2016 at 23:16

darrylp

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1 Comment

Taylor's Designs Over a year ago

This looks good! but it still gives a 500 error, which does not happen when the php doesn't have last 2 lines. it must be something else now

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