如何从mysql / php中的json对象中选择数据？

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":"a@a.a"}" 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.

Answer 1

尝试使用$obj->fname代替$obj->fname $obj->{'fname'} 。

Answer 2

Here is the correct syntax for your SQL Statement

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