无法在Firefox上从Javascript提交表单

Question

我正在创建一个在javascript函数中提交的表单.我将函数绑定到select元素的onchange事件.事件在chrome和firefox中运行良好,我测试了它,它调用了函数.但问题是; 虽然chrome提交表单,但firefox没有.能否请你帮忙？谢谢.

Javascript功能:

function getStatementDetails()
{
    var stmtSelect = document.getElementById("statementSelect");
    var selectedId = stmtSelect.options[stmtSelect.selectedIndex].value;

    var stmtForm = document.createElement("form");
    stmtForm.setAttribute('method', "post");
    stmtForm.setAttribute('action', "mymiles-mystatement");

    var stmtId = document.createElement("input");
    stmtId.setAttribute('type', "hidden");
    stmtId.setAttribute('name', "statementID");
    stmtId.setAttribute('id', "statementID");
    stmtId.setAttribute('value', selectedId);

    stmtForm.appendChild(stmtId);
    stmtForm.submit();
};

选择输入:

<select id="statementSelect" name="statementSelect" class="select-miles select-miles-medium spacing-right-10" onchange="getStatementDetails()">

编辑:我已阅读建议的帖子并尝试过.还是行不通.功能最新状态:

function getStatementDetails()
{
    var stmtSelect = document.getElementById("statementSelect");
    var selectedId = stmtSelect.options[stmtSelect.selectedIndex].value;

    var stmtForm = document.createElement("form");
    stmtForm.setAttribute('method', "post");
    stmtForm.setAttribute('action', "mymiles-mystatement");

    var stmtId = document.createElement("input");
    stmtId.setAttribute('type', "hidden");
    stmtId.setAttribute('name', "statementID");
    stmtId.setAttribute('id', "statementID");
    stmtId.setAttribute('value', selectedId);

    var stmtSbmt = document.createElement("input");
    stmtSbmt.setAttribute('type', "submit");
    stmtSbmt.setAttribute('name', "tryMe");
    stmtSbmt.setAttribute('id', "tryMe");
    stmtSbmt.setAttribute('value', "try submit");

    stmtForm.appendChild(stmtId);
    stmtForm.appendChild(stmtSbmt);
    stmtForm.submit();
};

Answer 1

评论部分已满,可能是用户看不到正确的答案.@spuyet说这是解决方案:

Javascript form.submit()无法在Firefox中运行

似乎firefox没有发送没有任何按钮的表单.如果您的代码中包含提交按钮,则必须正常工作.

  var button = document.createElement("input");
  button.setAttribute('type', "submit");
  stmtForm.appendChild(button);

而且我认为这可以解决您的问题.

请告诉我们你是否解决了.