Load mysqli php data via ajax call

What I'm trying to do is calling some database data via ajax and php. But the ajax call doesn't work, and I can't find out a solution on the web.

So here is my code:

test.php

<?php

include_once 'db_class.php';

$cat = $_GET['cat'];  

$dbconn = new dbconn('localhost', 'root', 'somepsw', 'blog');

 $dbconn->set_query("select * from posts where category = '".$cat."'");

 echo '<br/>'.$dbconn->query.'<br/>';

 $result = $dbconn->result;

 $num = $dbconn->num_results;

 $array = mysqli_fetch_assoc($result);

 echo json_encode($array);
?>

If i type that url on browser: http://127.0.0.1:82/blog/ws/test.php?cat=css

The data returned via jsonEncode is correct, but when i'm loading it on a html page with jquery he can't read the data.

test.html

<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<script>
function ajaxCall() {

var css;

$.ajax({                                      
      url: 'test.php',
      type: "GET",     
      data: {cat: css},              
      dataType: 'json',    
      success: function(rows)         
      {

     alert(rows);

      },
      error: function() { alert("An error occurred."); }

    });

    }

    ajaxCall();

</script>
</head>
<body></body>
</html>

Thanks in advance.

Your variable css has no value. You wanted to use the string 'css'. Maybe you want to be able to load other categories, too. So change your ajaxCall function to

function ajaxCall(category)
{
    $.ajax({
        url: 'test.php',
        type: "GET",
        data: {cat: category},
        dataType: 'json',    
        success: function(rows) {
           alert(rows);
        },
        error: function() {
           alert("An error occurred.");
        }
    });
}

and call it using

ajaxCall('css');

I just rewrote the php code using PDO, should be more safe now.

db.php

<?php

$dbhost = "localhost";  

$dbuser = "root";

$dbpsw = "somepsw";    

$dbname= "blog"; 

try {

    @$dbh = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpsw);
    $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);  
}

catch(PDOException $e) {

    echo "Connection failed, an error occured! Please contact server administrator."; //user friendly message
    getErrorsLog($e->getMessage());
 }

 function closeDbConn () {

    $dbh = null;

 }

 function getErrorsLog($message) {

    $file = 'dberrors.log';
    $date = date("d/m : H:i :");

    // Open the file to get existing content
    $current = file_get_contents($file);
    // Append a new error message to the file
    $current .= $date.$message;
    $current .= "\r\n";
    // Write the contents back to the file
    file_put_contents($file, $current);
    exit();

 }

?>

blogdata.php

<?php

include_once "db.php";

$tableName = "posts";
$data = array();
@$view = $_GET["view"];

if (isset($_GET["view"])) { 

    $stmt = $dbh->prepare("SELECT * FROM $tableName WHERE category =? ORDER BY created DESC"); 
 }
 else {  

    try {

    $stmt = $dbh->prepare("SELECT * FROM $tableName ORDER BY created DESC");

    }

    catch (PDOException $e) {

        getErrorsLog($e->getMessage());

    }

 }

$stmt->bindValue(1, $view, PDO::PARAM_STR);

$stmt->execute();

$affected_rows = $stmt->rowCount(); //Rows count

 if ($affected_rows == 0) {

     echo "The data you looking for no longer exist, please contact the administrator.";
     exit();
 }

foreach ($stmt->fetchAll(PDO::FETCH_ASSOC) as $row) {

    $data[] = $row;

 }

echo json_encode($data);

closeDbConn();

?>

来源：https://stackoverflow.com/questions/16637076/load-mysqli-php-data-via-ajax-call

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mysqli