jQuery的形式检查返回true，但需要返回false [英] Jquery form check returns true, but need to return false

问题描述

我正在这种形式的检查，当表单提交：

I am running this form check when the form is submitted:

if((formData[4].value == 1 || formData[4].value == 2) && !formData[2].value) { 
    alert('Please fill out the key field');
    return false;
} else {
    $.ajax({
        url: "/ajax/key_check.php",
        cache: false,
        data: "key=" + formData[4].value,
        dataType: "html",
        success: function(data) {
            if(data == 1) {
                alert('Key already exists');
                return false;
            }
        }
    });

    return true;
}

该脚本，它提醒键已如果数据确实== 1，但形式仍然提交存在。我想通过返回false，如果数据== 1将停止处理的形式，但它继续并增加了关键反正和弹出了键已经存在的消息。如何从数据是否== 1停止提交表单？我试着即使这样做：

The script works, it alerts key already exists if data does == 1, however the form still submits. I thought by returning false if data == 1 would stop the form from processing, however it continues and adds the key anyway and popups up key already exists message. How can I stop the form from submitting if data == 1? I tried even doing this:

if(data == 1) {
    alert('Key already exists');
    return false;
} else {
    return true;
}

然后删除返回真正的脚本的底部，但同样的问题发生。弹出大作，但形式仍然得到处理。

Then removed the return true at the bottom of the script, but the same issue happens. Pop up comes up but the form still gets processed.

推荐答案

A 返回在回调函数返回的从回调的，不是父功能像你想... ...但是你的可以的得到你想要的效果，像这样的：

A return in the callback returns from that callback, not the parent function like you want...however you can get the effect you want, like this:

if((formData[4].value == 1 || formData[4].value == 2) && !formData[2].value) { 
    alert('Please fill out the key field');
    return false;
} else {
    $.ajax({
        url: "/ajax/key_check.php",
        cache: false,
        data: "key=" + formData[4].value,
        dataType: "html",
        success: function(data) {
            if(data == 1) {
                alert('Key already exists');
            } else {
              $("#someForm")[0].submit();
            }
        }
    });
    return false;
}

这里做的事情是从来没有直接提交，但是的如果的检查正常，将调用*本地的 .submit（）方法，提交表单，而不是再次运行此处理。

What this does is never submit directly, but if the check is ok, calls the *native .submit() method, submitting the form and not running this handler again.

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