用PHP的ajax帖子

Question

why isn't this working?

jQuery AJAX Code:

$("header input").bind("keyup", function()
{
    var searchString= $("header input").val();
    var dataString = 'search=' + searchString;
    alert(dataString);
    $.ajax({
        type: "POST",
        url: "index.php",
        data: dataString,
        cache: false
    });
});

PHP Code(Just a test Code):

if($_POST["search"]) {
    echo "TEST MESSAGE!";
}

It doesn't show The echo :/

thanks for ur help ;)

Answer 1

You need to display the data you receive from the ajax call. Example, to put the result into a <div> called YourresultDiv:

Try with this

$("header input").on("keyup", function () {
    var searchString = $("header input").val();
    var dataString = 'search=' + searchString;
    alert(dataString);
    $.ajax({
        type: "POST",
        url: "index.php",
        data: dataString,
        cache: false,
        success: function (data) {
            $('#YourresultDiv').html(data);
            alert("Successful");
        }
    });
});