Question

所以，我在我的php页面中有这个jquery代码，它从mysql表中获取数据：

$(function() {
    $("#tracking_num").autocomplete({
        source: "func/populate.php",
        minLength: 2,
        select: function(event, ui) {
            $('#name').val(ui.item.name);
            $('#particulars').val(ui.item.particulars);
            $('#remarks').val(ui.item.remarks);
            $('#location').val(ui.item.location);
        }
    });
});

我很难将数据立即放入字段而不从下拉框中选择。你们有任何想法怎么做吗？这是我的func / populate.php文件：

$return_arr = array();

$fetch = mysql_query("SELECT * FROM tbl_document WHERE tracking_number = '" . mysql_real_escape_string($_GET['term']) . "' ORDER BY id DESC LIMIT 1");

while ($row = mysql_fetch_array($fetch) ) {
    $row_array['name'] = $row['name'];
    $row_array['particulars'] = $row['particulars'];
    $row_array['remarks'] = $row['remarks'];
    $row_array['location'] = $row['location'];
    $row_array['value'] = $row['tracking_number'];

    array_push($return_arr,$row_array);
}

mysql_close();

echo json_encode($return_arr);

Answer 1

你试过了吗？

$(function() {
    $("#tracking_num").autocomplete({
        source: "func/populate.php",
        minLength: 2,
        create: function(event, ui) {
            $('#name').val(ui.item.name);
            $('#particulars').val(ui.item.particulars);
            $('#remarks').val(ui.item.remarks);
            $('#location').val(ui.item.location);
        }
    });
});

自动完成jquery多个字段onchange或tab事件

1 个答案: