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Javascript 在jquery中提交时触发未捕获范围错误_Javascript_Jquery_Ajax - Fatal编程技术网

Javascript 在jquery中提交时触发未捕获范围错误

Javascript 在jquery中提交时触发未捕获范围错误,javascript,jquery,ajax,Javascript,Jquery,Ajax,我试图使用AJAX发布一个表单,为此,我使用jQuery。我为此编写的代码如下: $("#addStaffForm").on("submit", function(event) { event.preventDefault(); var formPost = $.post("/medewerkers-toevoegen", { "_token": $("input[name=_token]"), "voorletters": $("input[n

我试图使用AJAX发布一个表单,为此,我使用jQuery。我为此编写的代码如下:

$("#addStaffForm").on("submit", function(event) {
    event.preventDefault();

    var formPost = $.post("/medewerkers-toevoegen", {
        "_token": $("input[name=_token]"),
        "voorletters": $("input[name=voorletters]"),
        "tussenvoegsels": $("input[name=tussenvoegsels]"),
        "achternaam": $("input[name=achternaam]"),
        "gebruikersnaam": $("input[name=gebruikersnaam]"),
        "wachtwoord": $("input[name=wachtwoord]")
    });

    formPost.done(function(data) {
        console.log("data");
    });
});
但是,很简单,当我尝试实际提交表单时,我得到以下信息:

Uncaught RangeError: Maximum call stack size exceeded
    at Function.type (jquery.min.js:1)
我不知道为什么会有这个错误,我能想到的唯一解释就是我创建了一个无限循环,但我似乎找不到一个

这段小而简单的代码怎么会不能像我希望的那样工作呢

谢谢你应该试试这个

  var params = {
        "_token": $("input[name=_token]").val(),
        "voorletters": $("input[name=voorletters]").val(),
        "tussenvoegsels": $("input[name=tussenvoegsels]").val(),
        "achternaam": $("input[name=achternaam]").val(),
        "gebruikersnaam": $("input[name=gebruikersnaam]").val(),
        "wachtwoord": $("input[name=wachtwoord]").val()
  }
  var formPost = $.post("/medewerkers-toevoegen", params);

找到了答案!我忘记了我实际上发送的是整个jquery元素,而不是输入值

$("#addStaffForm").submit(function(e) {
        e.preventDefault();
        var token = $("input[name=_token]").val();
        var voorletters = $("input[name=voorletters]").val();
        var tussenvoegsels = $("input[name=tussenvoegsels]").val();
        var achternaam = $("input[name=achternaam]").val();
        var gebruikersnaam = $("input[name=gebruikersnaam]").val();
        var wachtwoord = $("input[name=token]").val();


        $.post("/medewerkers-toevoegen", {
            "_token": token,
            "voorletters": voorletters,
            "tussenvoegsels": tussenvoegsels,
            "achternaam": achternaam,
            "gebruikersnaam": gebruikersnaam,
            "wachtwoord": wachtwoord
        }, function(data) {
            console.log(data);
        }, function(data) {
            console.log(data);
        });
    });
为我工作