Ajax 和 PHP：SELECT 查询数据库

Question

I have an UI in which the user can add objects (stored as JSON in my database). So, when the user is clicking on a button, it calls the ajax wich calls my php to get the desired JSON code.

The problem is that my ajax returns me "success" but with no JSON code. I don't know where the problem is located.

Here's my JS/AJAX:

 var object, objectJson;
    var objectName = $(this).attr("attr-lib");
    var objectId = $(this).attr("attr-id");

    $.ajax({
        url: 'scriptObject.php',
        type: 'POST',
        data: {objectId: objectId},
        dataType: 'json',
        success: function(response) {
            console.log("success");
            console.log(response);
        }
    });

Here's my scriptObject.php :

$objectId = $_POST["objectId"];

    $query = "SELECT objectJson FROM object WHERE objectId = ' $objectId '";

    $result = mysql_query($query);

    if($result)
    {
        if(mysql_num_rows($result) > 0)
        {
            $objJSON = mysql_fetch_array($result);
            $res = array("status"=>"success", "objectJson" => $objJSON['objectJson']);
        }
        else
        {
            $res = array("status"=>"error", "message" => "No records found.");
        }
    }
    else
        $res = array("status"=>"error", "message" => "Problem in fetching data.");

    echo json_encode($res);

[EDIT]

It answers me:

{"status":"error","message":"Problem in fetching data."}

Answer 1

Try following

<script type="text/javascript">
    var object, objectJson;
    var objectName = $(this).attr("attr-lib");
    var objectId = $(this).attr("attr-id");

    $.ajax({
        url: 'scriptObject.php',
        type: 'POST',
        dataType : 'json',
        data: { objectId: objectId },
        success: function(response) {
            var json = $.parseJSON(response);
            if(json.status == 'error')
                console.log(json.message);
            else if(json.status == 'success')
                console.log(json.objectJson);
        }
    });
</script>

and in scriptObject.php

<?php 
    // your database connection goes here
    include 'config.php';

    $objectId = $_POST["objectId"];

    $query = "SELECT objectJson FROM object WHERE objectId = ' $objectId '";

    $result = mysql_query($query);

    if($result)
    {
        if(mysql_num_rows($result) > 0)
        {
            $objJSON = mysql_fetch_array($result);
            $res = array("status"=>"success", "objectJson" => $objJSON['objectJson']);
        }
        else
        {
            $res = array("status"=>"error", "message" => "No records found.");
        }
    }
    else
        $res = array("status"=>"error", "message" => "Problem in fetching data.".mysql_error());

    echo json_encode($res);
?> 

Answer 2

scriptObject.php

if(isset($_POST['objectId']))
{
    $objectId = $_POST["objectId"];
    $query = "SELECT objectJson FROM object WHERE objectId = ' $objectId '";
    $result=mysql_query($query);
        echo json_encode($result);
        die;
}

you are execution the php code inside a function. how does the compiler or the interpreter know to execute a function inside your php file.

Answer 3

Add dataType to your ajax options

$.ajax({
    url: 'scriptObject.php',
    type: 'POST',
    data: objectId,
    dataType : 'json',
    success: function(response) {
        console.log("success");
        console.log(response);
    }
});

then your php code

function getObjectJson() {
    $objectId = $_POST["objectId"];
    $query = "SELECT objectJson FROM object WHERE objectId = ' $objectId '";
    $result=mysql_query($query);

    echo json_encode($result);
}