onchange this.form.submit（）不适用于Web表单 [英] onchange this.form.submit() not working for web form

问题描述

这种方式工作的时间太长了......但似乎无法确定问题所在。已经阅读了有关stackoverflow和其他地方的数十篇文章。

been working on this way too long...but can't seem to identify the problem. Already read dozens of articles on stackoverflow and elsewhere.

当我点击并更改该值时，它不会自动提交：

when I click and change the value, it doesn't auto-submit:

   <form id="orderbyfrm" name="orderbyfrm" action="http://staging.whiterabbitexpress.com/" method="post" class="orderbyfrm">
            <input name="s" value="<?php echo $wre_search_txt?>" type="hidden">
            <label for="orderby" class="sortByLabel">Sort by&nbsp;</label>
            <select class="sortByDropdown" name="orderby" id="orderby" onchange="this.form.submit();">
                <option value="Relevance">Relevance</option>
                <option value="likes" selected="selected">Likes</option>
            <option value="comments" selected="comments">Comments</option>
            </select>
</form>

我看到错误
未捕获的TypeError：无法调用方法'提交' nullonchange

in Chrome inspector I see an error "Uncaught TypeError: Cannot call method 'submit' of null" onchange

我也尝试过onchange =javascript：document.orderbyfrm.submit但这也无效。

I also tried onchange="javascript:document.orderbyfrm.submit" but that didn't work either.

推荐答案

可能你有名为表格的元素或JS对象或提交某处，与真实形式冲突。

Probably you have element or JS object called form or submit somewhere, conflicting with the real form.

最安全的方法是使用document.getElementById：

Most safe way is using document.getElementById:

<select onchange="SubmitForm('orderbyfrm');">

以及JavaScript：

And the JavaScript:

function SubmitForm(formId) {
    var oForm = document.getElementById(formId);
    if (oForm) {
        oForm.submit(); 
    }
    else {
        alert("DEBUG - could not find element " + formId);
    }
}

使用良好的旧警报进行进一步调试..而不是 alert（DEBUG ... 有这个：

Further debugging with good old alert.. instead of the alert("DEBUG ... have this:

var sDebugInfo = "found " + document.forms.length + " forms: \n";
for (var i = 0; i < document.forms.length; i++) {
    var curForm = document.forms[i];
    sDebugInfo += "name: " + curForm.name + ", id: " + curForm.id;
    sDebugInfo += "\n";
}
alert(sDebugInfo);

根据你得到的结果，调试应该继续。

Depending on what you get, debug should continue.

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