Question

我正在做一个项目，我遇到了window.open的问题。问题是window.open在我的架构中不起作用。我需要一个帮助，

swal({
    title: "Submit Data ?",
    text: "Process only if you are sure",
    type: "warning",
    showCancelButton: true,
    confirmButtonColor: "#DD6B55",
    confirmButtonText: "Yes, Submit!",
    cancelButtonText: "No, Cancel!",
    closeOnConfirm: false,
    closeOnCancel: false
},
function(isConfirm){
    if (isConfirm) {
        insertData();
        swal("SUCCESS", "Data Has Been Submitted", "success");
        window.open("http://188.109.156.21/execution.php?str=james");
    } else {
        swal("CANCELLED", "", "error");
    }
});

function insertData(){
    $.ajax({
        type: 'POST',
        url: "../../../html/main/divpages/submit_data.php",
        data: sentReq,
        dataType: 'JSON',
        success: function (response, textStatus, jqXHR) {
            if (response.indexOf("GAGAL") == -1) {
                window.location.href = "main.php";
            } else {
                alert("GAGAL INSERT");
            }
        }
    });
}

所以我可以毫无问题地执行insertData（）。但问题在于在swal（）内执行windows.open时。我没有看到任何窗口弹出窗口正在打开。

Answer 1

这是浏览器使用的弹出式阻止逻辑。您必须直接在onClick事件上附加window.open。

使用类似的东西：

var checkSuccess = false;

$('#button').on("click", function(){
    $.ajax({
      type: 'POST',
      url: "your url",
      async:false,
      success: function(){ 
         checkSuccess = true;
         //YOUR LOGIC
      }
    });
    if(checkSuccess){
      window.open("http://188.109.156.21/execution.php?str=james");
    }
})

打开ajax成功的页面弹出窗口

1 个答案: