获取页面上显示的AJAX数据库结果

Question

Hello and thank you for readying my post. Im struggling with the question how to get my results on the screen via AJAX. I have a textbox and I want to make a selection in my table whenever something gets typed in (basicly selected from the autocomplete list). I cant seem to get it to work so that it shows the results on the screen via my table.

Im stuck on how to do this, any help would be great :)

This is my code at the moment:

HTML

<input type="text" name="naam_klant" size="20" id="naam_klant" onkeyup="lookup(this.value);" onblur="fill();" >

PHP

<?php 
$pdo = new PDO('mysql:host=localhost;dbname=records', 'root', '***');
$select = 'SELECT *';
$from = ' FROM overboekingen';

$where2 = ' WHERE naam_klant LIKE %$val% ';

$opts = (isset($_POST['filterOpts']) ? $_POST['filterOpts'] : FALSE);  

$val = (isset($_POST['txt']) ? $_POST['txt'] : FALSE);

  if (is_array($opts) || $val)  {

    $where = ' WHERE FALSE';
}

else {

$where = false;

}

$sql = $select . $from . $where;
$statement = $pdo->prepare($sql);
$statement->execute();
$results = $statement->fetchAll(PDO::FETCH_ASSOC);
$json = json_encode($results);
echo($json);
?>

SCRIPT

<script>

function makeTable(data) {
    var tbl_body = "";
    $.each(data, function() {
        var tbl_row = "";
        $.each(this, function(k , v) {
            tbl_row += "<td>"+v+"</td>";
        })
        tbl_body += "<tr>"+tbl_row+"</tr>";
    })
    return tbl_body;
}

function getEmployeeFilterOptions(){
    var opts = [];
    $checkboxes.each(function(){
        if(this.checked){
            opts.push(this.name);
        }
    });
    return opts;
}

$('naam_klant').change(function(){
updateEmployeesText($(this).val()); 
});

function updateEmployeesText(val){
    $.ajax({
    type: "POST",
    url: "submit.php",
    dataType : 'json',
    cache: false,
    data: {text: val},
    success: function(records){
        $('#employees tbody').html(makeTable(records));
    }
}); 
}

function updateEmployees(opts){
    $.ajax({
    type: "POST",
    url: "submit.php",
    dataType : 'json',
    cache: false,
    data: {filterOpts: opts},
    success: function(records){
        $('#employees tbody').html(makeTable(records));
    }
});
}

var $checkboxes = $("input:checkbox");
$checkboxes.on("change", function(){
    var opts = getEmployeeFilterOptions();
    updateEmployees(opts);
});

updateEmployees();

</script>

Answer 1

I don't really understand what are you doing, but maybe you need something like this:

<?php 
$pdo = new PDO('mysql:host=localhost;dbname=records', 'root', '***');
$select = 'SELECT *';
$from = ' FROM overboekingen';

$opts = (isset($_POST['filterOpts']) ? $_POST['filterOpts'] : FALSE);  

$val = (isset($_POST['txt']) ? $_POST['txt'] : FALSE);

  if (is_array($opts) || $val)  {

    $where = ' WHERE naam_klant LIKE %$val% ';
}

else {

$where = false;

}

$sql = $select . $from . $where;
$statement = $pdo->prepare($sql);
$statement->execute();
$results = $statement->fetchAll(PDO::FETCH_ASSOC);
$json = json_encode($results);
echo($json);
?>