2015-11-04 20:23
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I'm using AJAX to send a request to another php page where it returns the result of a query. I was calling it through xmlhttprequest to javascript to get the contents of that php page, but since i was mixing presentation logic "e.g. echo blah blah" with actual code logic I wanted to find a way where I could leave the php logic with my query alone, store the result on an array, and through ajax pass it down to my js code to use in a function.

I'm trying to populate a drop down list with the contents of this array. I've tried json encode but I'm not having any luck.

Here's code for the php page requested:


$con = mysqli_connect('*******','********','******','****');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));

$j = mysqli_real_escape_string($con, $_GET['j']);

$sql="SELECT nombreClub FROM club4h WHERE oficinaLoc LIKE '%".$j."%'";
$result = mysqli_query($con,$sql);

while($row = $result->fetch_array()) {
    $response[] = $row;



Here's my js function:

function loadDoc(url, cfunc, sel){
    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = function(){
        if (xhttp.readyState == 4 && xhttp.status == 200){
    xhttp.open("GET", url +sel, true);

function selClub(xhttp) {
    document.getElementById("nombreClub").style.display = "inline";
    var optData = <?php file_get_contents('getnombre.php'); ?>;
    var newClub = document.getElementById("addClub");
    newClub.type = "text";
    document.getElementById("addOption").style.display = "inline";
            $("#nombreClub").append('<option value="' + $("#addClub").val() + '">' + $("#addClub").val() + '</option>');

I call it through an onchange event when an user clicks a previous drop down list so this drop down list gets populated with options specific for the previous list. I know how to add the options to the list, I'm just having trouble getting the data from the php page to js without having to do something like this:

$a = mysqli_real_escape_string($con, $_GET['a']);

$sql="SELECT nombre FROM personal4h WHERE unidadProg LIKE '%".$a."%'";
$result = mysqli_query($con,$sql);

echo '<select name="agenteExt" id="agenteExt">
    <option value="">Seleccione Agente</option>';
while($row = mysqli_fetch_array($result)) {
    echo "<option value =" . $row['nombre'] . ">" . htmlentities($row['nombre']) . "</option>";

echo "</select>";
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1条回答 默认 最新

  • dongse5408 2015-11-05 15:55

    I'll attempt to explain how to do this...

    PHP Query Page

    Query your data and return it as JSON. PHP has a json_encode() function to do this.

    // ...your code to query database which should make an array like this
    $query_results = array(
        array('agent'=>'Agent 1'),
        array('agent'=>'Agent 2'),
        array('agent'=>'Agent 3')
    return json_encode($query_results);

    HTML Page should have your select

    <select name="agenteExt" id="agenteExt"></select>


    A function to get data via AJAX

    function loadDoc(url, callback){
        var xhttp = new XMLHttpRequest();
        xhttp.onreadystatechange = function(){
            if (xhttp.readyState == 4 && xhttp.status == 200){
                if( callback ) callback(xhttp)
        xhttp.open("GET", url, true);

    The function to get the data to fill your html

    function fillOptions(){
        loadDoc('your-query-page.php', function(xhttp){
            // parse the JSON data into a JavaScript object
            var data = JSON.parse(xhttp.responseText);
            // get the `<select>` element
            var el = document.getElementById('agenteExt');
            var html = '<option>Make a selection</option>';
            // loop through data and create each `<option>` for the select
            for(i = 0; i < data.length; i++){
                html += '<option value="'+data[i].agent+'">'+data[i].agent+'</option>'
            // update the select with the new options
            el.innerHTML = html;
    解决 无用
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