Question

当用户单击jQuery中的按钮时，我希望运行另一个页面的php脚本。我已经尝试过了，但是没有用。

下面是我的脚本：

要从索引页面单击的按钮

 <button id="updatebutton" class="btn btn-sm btn-info" onclick="update_btn_click('params');"><span class="fa fa-refresh">&nbsp;Generate Cards</span></button>

PHP页面（card_generator.php）

function getRandomCodes(){
                        $ans = "0123456789";
                        $sus = strlen($ans) - 1;
                        return substr($ans, rand(0, $sus), 1) . substr($ans, rand(0, $sus), 1) . substr($ans, rand(0, $sus), 1) . substr($ans, rand(0, $sus), 1) . substr($ans, rand(0, $sus), 1) . substr($ans, rand(0, $sus), 1) . substr($ans, rand(0, $sus), 1). substr($ans, rand(0, $sus), 1). substr($ans, rand(0, $sus), 1). substr($ans, rand(0, $sus), 1). substr($ans, rand(0, $sus), 1). substr($ans, rand(0, $sus), 1);
                        }

                        function getRandomCode(){
                        $an = "01234abcdef5678ghijklm9nopqrxtuvwxyz";

                        $su = strlen($an) - 1;
                        return substr($an, rand(0, $su), 1) . substr($an, rand(0, $su), 1) . substr($an, rand(0, $su), 1) . substr($an, rand(0, $su), 1) . substr($an, rand(0, $su), 1) . substr($an, rand(0, $su), 1) . substr($an, rand(0, $su), 1). substr($an, rand(0, $su), 1). substr($an, rand(0, $su), 1). substr($an, rand(0, $su), 1). substr($an, rand(0, $su), 1). substr($an, rand(0, $su), 1);
                        }

                        for($i = 0; $i < 20; $i++){
                             $gen =  mysqli_query($mysqli, "insert into pins(serials, pins)values('" .getRandomCode(). "', '" .getRandomCodes(). "')");

                        }

                        if($gen){
                            echo "ok";
                        }else{
                            echo $mysqli->error;
                        }

jQuery

function update_btn_click(param) {
$('#updatebutton').prop("disabled",true);
$.ajax({
    url:"generate_cards/card_generate.php",
    type:"post",
    beforeSend: function()
    {
        $("#message").fadeOut();
        $("#updatebutton").html('Generating <img src="../img/processing.gif" width="30" />');
    },
    success: function(response){
        if(response=='Ok') {
            //location.reload();
            $("#message").fadeIn(1000, function(){
                $("#message").html('<div class="alert alert-success"> <span class="fa fa-check"></span> &nbsp; Successfully Generated </div>');
                       //$("#btn-submit").html('<i class="fa fa-plus"></i> Add again');

                   });
                   cardTable.ajax.reload(null, false);
        } else {
            $("#message").html('<div class="alert alert-danger"><span class="fa fa-info-circle"></span> &nbsp; '+response+' !</div>');
            $('#updatebutton').prop("disabled",false);
        }
    },
    error: function(response){
        console.log('could not fetch data');
    },
    complete: function(response){
        // hide loading
    }
});
}

有人可以指引我走正确的路吗？

Answer 1

我已经可以使用以下方法解决它：

function update_btn_click(param)
    {
        //$('#updatebutton').prop("disabled",true);
        var resp = $("#updatebutton");
        $.ajax({
            type: "POST", // Method type GET/POST           
            url: "generate_cards/cards.php", //Ajax Action url
            data: {},

            // Before call ajax you can do activity like please wait message
            beforeSend: function(xhr){
                resp.html('Generating <img src="../img/processing.gif" width="30" />');
            },

            //Will call if method not exists or any error inside php file
            error: function(qXHR, textStatus, errorThrow){
                resp.html("There are an error");
                $("#message").html('<div class="alert alert-danger"><span class="fa fa-info-circle"></span> &nbsp; '+data+' !</div>');
                //$('#updatebutton').prop("disabled",false);
            },

            success: function(data, textStatus, jqXHR){
                //resp.html(data);
                $("#message").fadeIn(1000, function(){
                    $("#message").html('<div class="text-success"> <span class="fa fa-check"></span> &nbsp; Successfully Generated </div>');
                           $("#updatebutton").html('<i class="fa fa-refresh"></i>&nbsp;Generate Cards');

                       });
                       cardTable.ajax.reload(null, false);
            }
        });
     }

用户单击按钮时如何在另一个页面上运行PHP代码？

1 个答案: